NCERT Solutions for Class 11 Physics Chapter 7

Q.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Ans.

The centre of mass (C.M.) of a body is a point where the mass of a body is supposed to be concentrated. A sphere, cylinder, ring and a cube are symmetrical objects and have their centre of mass at their geometric centres. C.O.M. of a body need not necessarily lie inside the body, for example; hollow sphere and a ring.

Q.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Ans.

$\begin{array}{l}\text{The situation\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}problem\hspace{0.17em}can be shown by\hspace{0.17em}the\hspace{0.17em}given\hspace{0.17em}figure.}\\ \text{Here,\hspace{0.17em}seperation between H and Cl atoms}=\text{1}.\text{27}\mathrm{\AA }\\ \text{Let\hspace{0.17em}mass of H atom}=\text{m}\\ \therefore \text{Mass of Cl atom}=\text{35}.\text{5\hspace{0.17em}m}\\ \text{Let the\hspace{0.17em}distance\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}centre of mass of the system from the Cl atom=x}\\ \text{Distance of the centre of mass from the H atom}\\ =(\text{1}.\text{27}–\text{x})\\ \text{Consider that the centre of mass of the given molecule lies at the origin.}\\ \therefore \frac{\mathrm{m}(1.27-\mathrm{x})+35.5\mathrm{mx}}{\mathrm{m}+35.5\mathrm{m}}=0\\ \mathrm{m}(1.27-\mathrm{x})+35.5\mathrm{mx}=0\\ 1.27-\mathrm{x}=-35.5\mathrm{x}\\ \therefore \mathrm{x}=\frac{-1.27}{(35.5-1)}\\ =-0.037\mathrm{\AA }\\ \text{Here, the negative sign shows that the centre of mass lies at the left of the molecule.}\\ \therefore \text{The centre of mass of the HCl molecule lies at\hspace{0.17em}a\hspace{0.17em}distance\hspace{0.17em}0.037\hspace{0.17em}}\mathrm{\AA }\text{from the Cl atom.}\end{array}$

Q.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Ans.

Internal forces do not affect the motion of the bodies on which they act. No external force is not involved in the system, so the child’s motion will not change the velocity of the centre of mass of the system.

Q.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

Ans.

$\begin{array}{l}\text{Consider two vectors}\overrightarrow{\text{a}}\text{\hspace{0.17em}and\hspace{0.17em}}\overrightarrow{\text{b}}\text{\hspace{0.17em}represented\hspace{0.17em}by\hspace{0.17em}}\overrightarrow{\mathrm{OK}}\text{\hspace{0.17em}and\hspace{0.17em}}\overrightarrow{\mathrm{OM}}\text{respectively, inclined at an angle θ, as shown in the\hspace{0.17em}given figure.}\\ \text{From}\mathrm{\Delta }\text{OMN,\hspace{0.17em}we\hspace{0.17em}have}:\\ \mathrm{sin\theta }=\frac{\mathrm{MN}}{\mathrm{OM}}=\frac{\mathrm{MN}}{\left|\overrightarrow{\mathrm{b}}\right|}\\ \mathrm{MN}=\left|\overrightarrow{\mathrm{b}}\right|\mathrm{sin\theta }\\ \text{By\hspace{0.17em}definition,}\\ |\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\mathrm{sin\theta }\\ =\mathrm{OK}.\mathrm{MN}\times \frac{2}{2}\\ =\text{2}\times \text{Area of}\mathrm{\Delta }\text{OMK}\\ \therefore \text{Area of}\mathrm{\Delta }\text{OMK=}\frac{1}{2}|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|\\ \text{Hence,\hspace{0.17em}proved.}\end{array}$

Q.5 Show that a. (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.

Ans.

$\begin{array}{l}\text{Consider\hspace{0.33em}a\hspace{0.33em}parallelepiped with\hspace{0.33em}origin\hspace{0.33em}O\hspace{0.33em}and\hspace{0.33em}sides\hspace{0.33em}a,\hspace{0.33em}b\hspace{0.33em}and\hspace{0.33em}c\hspace{0.33em}respectively.}\\ \text{Volume of the given parallelepiped = abc\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\\ \text{Here,\hspace{0.33em}}\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}\\ \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}\\ \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}\\ \text{Let unit vector perpendicular to the\hspace{0.33em}plane\hspace{0.33em}}\mathrm{containing}\widehat{\mathrm{b}}\text{and}\widehat{\mathrm{c}}=\widehat{\mathrm{n}}\\ \therefore \widehat{\mathrm{n}}\text{\hspace{0.33em}and}\overrightarrow{\mathrm{a}}\text{are\hspace{0.33em}along the same\hspace{0.33em}direction}.\\ \therefore \overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}=\mathrm{bcsin\theta }\widehat{\mathrm{n}}\\ =\mathrm{bcsin}{90}^{\mathrm{o}}\widehat{\mathrm{n}}=\mathrm{bc}\widehat{\mathrm{n}}\\ \overrightarrow{\mathrm{a}}.\left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)=\overrightarrow{\mathrm{a}}.\left(\mathrm{bc}\widehat{\mathrm{n}}\right)\text{= abc cos0° = abc}\\ \overrightarrow{\mathrm{a}}.\left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)\text{=Volume of the parallelepiped}\end{array}$

Q.6

\begin{array}{l}\text{Find the components along the x, y, z axes of the angular momentum l of a particle,}\\ \text{whose position}\text{vector is}\overrightarrow{\text{r}}\text{with components x, y, z and momentum is}\overrightarrow{\text{p}}\text{with}\\ {\text{components p}}_x,{\text{p}}_y{\text{and p}}_z.\text{Show that if the particle moves only in the x-y plane}\\ \text{the angular momentum has only a z-component}.\end{array}

Ans.

$\begin{array}{l}\text{The\hspace{0.17em}three\hspace{0.17em}rectangular\hspace{0.17em}components\hspace{0.17em}of\hspace{0.17em}angular\hspace{0.17em}momentum\hspace{0.17em}are\hspace{0.17em}given\hspace{0.17em}as:}\\ {\text{l}}_{\text{x}}={\text{yp}}_{\text{z}}–{\text{zp}}_{\text{y}}\\ {\text{l}}_{\text{y}}={\text{zp}}_{\text{x}}–{\text{xp}}_{\text{z}}\\ {\text{l}}_{\text{z}}={\text{xp}}_{\text{y}}–{\text{yp}}_{\text{x}}\\ \text{Linear momentum of the particle\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \overrightarrow{\mathrm{p}}={\text{p}}_{\text{x}}\stackrel{\wedge }{\mathrm{i}}+{\text{p}}_{\text{y}}\stackrel{\wedge }{\mathrm{j}}+{\text{p}}_{\text{z}}\stackrel{\wedge }{\mathrm{k}}\\ \text{Position vector of the particle\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \overrightarrow{\mathrm{r}}=\mathrm{x}\stackrel{\wedge }{\mathrm{i}}+\mathrm{y}\stackrel{\wedge }{\mathrm{j}}+\mathrm{z}\stackrel{\wedge }{\mathrm{k}}\\ \text{Angular momentum,\hspace{0.17em}}\overrightarrow{\mathrm{l}}=\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{p}}\\ =(\mathrm{x}\stackrel{\wedge }{\mathrm{i}}+\mathrm{y}\stackrel{\wedge }{\mathrm{j}}+\mathrm{z}\stackrel{\wedge }{\mathrm{k}})\times ({\text{p}}_{\text{x}}\stackrel{\wedge }{\mathrm{i}}+{\text{p}}_{\text{y}}\stackrel{\wedge }{\mathrm{j}}+{\text{p}}_{\text{z}}\stackrel{\wedge }{\mathrm{k}})\\ =\left|\begin{array}{l}\stackrel{\wedge }{\mathrm{i}}\stackrel{\wedge }{\mathrm{j}}\stackrel{\wedge }{\mathrm{k}}\\ \mathrm{x}\mathrm{y}\mathrm{z}\\ {\text{p}}_{\text{x}}{\text{\hspace{0.17em}p}}_{\text{y}}{\text{\hspace{0.17em}p}}_{\text{z}}\end{array}\right|\\ {\text{l}}_{\text{x}}\stackrel{\wedge }{\mathrm{i}}+{\text{l}}_{\text{y}}\stackrel{\wedge }{\mathrm{j}}+{\text{l}}_{\text{z}}\stackrel{\wedge }{\mathrm{k}}=\stackrel{\wedge }{\mathrm{i}}(\mathrm{y}{\text{p}}_{\text{z}}-\mathrm{z}{\text{p}}_{\text{y}})-\stackrel{\wedge }{\mathrm{j}}(\mathrm{x}{\text{p}}_{\text{z}}-\mathrm{z}{\text{p}}_{\text{x}})+\stackrel{\wedge }{\mathrm{k}}(\mathrm{x}{\text{p}}_{\text{y}}-\mathrm{y}{\text{p}}_{\text{x}})\\ \text{On\hspace{0.17em}comparison\hspace{0.17em}of coefficients of}\stackrel{\wedge }{\mathrm{i}},\stackrel{\wedge }{\mathrm{j}}\text{\hspace{0.17em}and\hspace{0.17em}}\stackrel{\wedge }{\mathrm{k}}\text{,\hspace{0.17em}we obtain}:\\ \begin{array}{l}{\text{l}}_{\text{x}}=\mathrm{y}{\text{p}}_{\text{z}}-\mathrm{z}{\text{p}}_{\text{y}}\\ {\text{l}}_{\text{y}}=\mathrm{x}{\text{p}}_{\text{z}}-\mathrm{z}{\text{p}}_{\text{x}}\\ {\text{l}}_{\text{z}}=\mathrm{x}{\text{p}}_{\text{y}}-\mathrm{y}{\text{p}}_{\text{x}}\end{array}\}\to \left(\mathrm{i}\right)\\ \text{As,\hspace{0.17em}the particle\hspace{0.17em}travels in the x}-\text{y plane,}\\ \therefore \text{the z-component of the position vector\hspace{0.17em}and linear momentum vector becomes zero, i.e.,}\\ \text{z}={\text{p}}_{\text{z}}=0\\ \text{Therefore, equation}\left(\text{i}\right)\text{becomes:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \begin{array}{l}{\text{l}}_{\text{x}}=0\\ {\text{l}}_{\text{y}}=0\\ {\text{l}}_{\text{z}}=\mathrm{x}{\text{p}}_{\text{y}}-\mathrm{y}{\text{p}}_{\text{x}}\end{array}\}\to \left(\text{ii}\right)\\ \therefore \text{When the particle is restricted to move in\hspace{0.17em}the x-y plane, the direction of\hspace{0.17em}angular momentum\hspace{0.17em}is along the z-direction.}\end{array}$

Q.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}each\hspace{0.17em}particle}=\mathrm{m}\\ \text{Velocity\hspace{0.17em}of\hspace{0.17em}each\hspace{0.17em}particle}=\mathrm{v}\\ \text{Let, at a definite instant two particles be at points S and V, as shown in the given figure.}\\ \text{Angular momentum of the system about point S\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ {\overrightarrow{\mathrm{L}}}_{\mathrm{S}}=\mathrm{mv}\times 0+\mathrm{mv}\times \mathrm{d}\\ {\overrightarrow{\mathrm{L}}}_{\mathrm{S}}=\mathrm{mvd}\to \left(\mathrm{i}\right)\\ \text{Angular momentum of the system about point\hspace{0.17em}V\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ {\overrightarrow{\mathrm{L}}}_{\mathrm{V}}=\mathrm{mv}\times \mathrm{d}+\mathrm{mv}\times 0\\ {\overrightarrow{\mathrm{L}}}_{\mathrm{V}}=\mathrm{mvd}\to \left(\mathrm{i}\right)\\ \text{Consider a point T at a distance y from \hspace{0.17em}point V, i.e.,}\\ \text{TV}=\text{y}\\ \therefore \mathrm{ST}=\text{d}–\text{y}\\ \text{Angular momentum of the system about point R\hspace{0.17em}is given\hspace{0.17em}as:}\\ {\overrightarrow{\mathrm{L}}}_{\mathrm{T}}=\mathrm{mv}\times (\mathrm{d}-\mathrm{y})+\mathrm{mv}\times \mathrm{y}\\ =\mathrm{mvd}-\mathrm{mvy}+\mathrm{mvy}\\ =\mathrm{mvd}\to \left(\mathrm{iii}\right)\\ \text{Comparing equations}\left(\text{i}\right)\text{,}\left(\text{ii}\right)\text{, and}\left(\text{iii}\right)\text{, we obtain:}\\ {\overrightarrow{\mathrm{L}}}_{\mathrm{S}}={\overrightarrow{\mathrm{L}}}_{\mathrm{V}}\\ ={\overrightarrow{\mathrm{L}}}_{\mathrm{T}}\to \left(\mathrm{iv}\right)\\ \text{From equation}\left(\text{iv}\right)\text{,\hspace{0.17em}we conclude\hspace{0.17em}that the angular momentum of a system does not\hspace{0.17em}depend on the point about which it is taken.}\end{array}$

Q.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long.
Calculate the distance d of the centre of gravity of the bar from its left end.

Ans.

$\begin{array}{l}\text{The free body diagram of the bar is shown in the given figure.}\\ \text{Here,\hspace{0.17em}length of the bar},\text{l}=\text{2 m}\\ {\text{Let\hspace{0.17em}T}}_{\text{1}}{\text{and T}}_{\text{2}}\text{be the tensions produced in the left and right strings respectively.}\\ \text{At translational equilibrium, we have:}\\ {\text{T}}_{\text{1}}\sin 36{.9}^{\mathrm{o}}={\text{T}}_{\text{2}}\sin 53{.1}^{\mathrm{o}}\\ \frac{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}=\frac{\sin 53{.1}^{\mathrm{o}}}{\sin 36{.9}^{\mathrm{o}}}\\ =\frac{0.800}{0.600}\\ =\frac{4}{3}\\ \Rightarrow {\text{T}}_{\text{1}}=\frac{4}{3}{\text{T}}_{\text{2}}\\ \text{For rotational equilibrium about\hspace{0.17em}the centre of gravity,we obtain:}\\ {\text{T}}_{\text{1}}\cos 36{.9}^{\mathrm{o}}\times \mathrm{d}={\text{T}}_{\text{2}}\cos 53{.1}^{\mathrm{o}}(2-\mathrm{d})\\ {\text{T}}_{\text{1}}\times 0.800\mathrm{d}={\text{T}}_{\text{2}}\times 0.600(2-\mathrm{d})\\ \frac{4}{3}\times {\text{T}}_{\text{2}}\times 0.800\mathrm{d}={\text{T}}_{\text{2}}[0.600\times 2-0.600\mathrm{d}]\\ 1.067\mathrm{d}+0.6\mathrm{d}=1.2\\ \therefore \mathrm{d}=\frac{1.2}{1.67}\\ =0.72\mathrm{m}\\ \therefore \text{The C.G.}\left(\text{centre of gravity}\right)\text{of the given bar lies at\hspace{0.17em}a\hspace{0.17em}distance\hspace{0.17em}0.72 m from its left end.}\end{array}$

Q.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of car, m}=\text{18}00\text{kg}\\ \text{Separation between front and back axles, d}=\text{1}.\text{8 m}\\ \text{Seperation between\hspace{0.17em}the C.G. and the back axle}=\text{1}.0\text{5 m}\\ \text{The several forces acting on the car are shown in the given figure.}\\ {\text{Here,\hspace{0.17em}R}}_{\text{f}}{\text{and R}}_{\text{b}}\text{\hspace{0.17em}are the forces applied by the level ground on the front and back wheels\hspace{0.17em}respectively.}\\ \text{For translational equilibrium,\hspace{0.17em}we\hspace{0.17em}have:}\\ {\text{R}}_{\text{f}}+{\text{R}}_{\text{b}}=\text{mg}\\ =\text{18}00\times \text{9}.\text{8}\\ =\text{1764}0\text{N}\dots \left(\text{i}\right)\\ \text{At rotational equilibrium, about\hspace{0.17em}C.G.,we have:}\\ {\text{R}}_{\text{f}}(1.05)={\text{R}}_{\text{b}}(1.8-1.05)\\ \frac{{\text{R}}_{\text{f}}}{{\text{R}}_{\text{b}}}=\frac{0.75}{1.05}\\ =\frac{5}{7}\\ \therefore {\text{R}}_{\text{b}}=1.4{\text{R}}_{\text{f}}...\text{\hspace{0.17em}(i)}\\ \text{By\hspace{0.17em}solving equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we obtain:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ 1.4{\text{R}}_{\text{f}}+{\text{R}}_{\text{f}}=\text{1764}0\\ 2.4{\text{R}}_{\text{f}}=\text{1764}0\\ \therefore {\text{R}}_{\text{f}}=\frac{\text{1764}0}{2.4}\\ =7350\mathrm{N}\\ \therefore {\text{R}}_{\text{b}}=1.4\mathrm{Rf}\\ =\text{1}0\text{29}0\text{N}\\ \therefore \text{Force applied on each front wheel\hspace{0.17em}=\hspace{0.17em}}\frac{\text{735}0}{2}\\ =3675\text{\hspace{0.17em}N}\\ \text{Force applied on each back wheel=}\frac{\text{1}0\text{29}0}{2}\\ =5145\text{\hspace{0.17em}N}\end{array}$

Q.10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR²/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR²/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Ans.

$$\begin{gathered} \begin{array}{l}\left(\text{a}\right)\text{Here, moment of inertia}\left(\text{M.I.}\right)\text{of a sphere about its diameter\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{2}{5}{\mathrm{MR}}^2\\ \text{As\hspace{0.17em}per the theorem of parallel axes, the M.I. of a body about any\hspace{0.17em}axis is equal to\hspace{0.17em}the sum} \\ \text{of the moment of inertia of the body about a parallel\hspace{0.17em}}\\ \text{axis\hspace{0.17em}passing through its centre of mass and the product of its} \\ \text{mass and the square of the\hspace{0.17em}distance between the two parallel axes.}\\ \text{M.I. about a tangent to the sphere\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{2}{5}{\mathrm{MR}}^2+{\mathrm{MR}}^2\\ =\frac{7}{5}{\mathrm{MR}}^2\end{array} \end{gathered}$$

$$\begin{gathered} \begin{array}{l}\left(\text{b}\right)\text{Here,\hspace{0.17em}M.I. of a disc about its diameter}=\frac{1}{4}{\mathrm{MR}}^2 \\ \text{As\hspace{0.17em}per the theorem of perpendicular axis, the M.I. of a planar\hspace{0.17em}body about an\hspace{0.17em}axis}\\ \text{normal to its plane is equal to the sum of its\hspace{0.17em}moments of inertia about two} \\ \text{perpendicular axes concurrent with perpendicular axis}\\ \text{and lying in the plane of the body.}\\ \text{M.I. of the disc about its centre=}\frac{1}{4}{\mathrm{MR}}^2+\frac{1}{4}{\mathrm{MR}}^2=\frac{1}{2}{\mathrm{MR}}^2\\ \text{This situation is shown in the figure(b). According\hspace{0.17em}to the theorem of parallel axes,the M.I.} \\ \text{about an axis perpendicular to the}\\ \text{disc and passing through a\hspace{0.17em}point\hspace{0.17em}on its edge}=\frac{1}{2}{\mathrm{MR}}^2+{\mathrm{MR}}^2=\frac{3}{2}{\mathrm{MR}}^2\end{array} \end{gathered}$$

Q.11 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is
free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Ans.

If m be the mass of the hollow cylinder and r is the radius of the solid sphere.

$$\begin{gathered} \begin{array}{l}\text{M.I. of hollow cylinder about its\hspace{0.17em}standard axis\hspace{0.17em}of\hspace{0.17em}symmetry},{\mathrm{I}}_1={\mathrm{MR}}^2\\ \text{M.I. of solid sphere about an axis through\hspace{0.17em}its\hspace{0.17em}centre,}{\mathrm{I}}_2=\frac{2}{5}{\mathrm{MR}}^2\\ \text{From the relation,\hspace{0.17em}}\mathrm{\tau }=\mathrm{I\alpha }\\ \text{Here,\hspace{0.17em}}\mathrm{\alpha },\mathrm{\tau }\text{\hspace{0.17em}and\hspace{0.17em}I\hspace{0.17em}represent the\hspace{0.17em}angular acceleration,\hspace{0.17em}torque\hspace{0.17em}and\hspace{0.17em}moment of inertia\hspace{0.17em}respectively.}\\ \text{Torque\hspace{0.17em}for the hollow cylinder,}{\mathrm{\tau }}_1={\mathrm{I}}_1{\mathrm{\alpha }}_1\\ \text{Torque\hspace{0.17em}for the solid sphere,}{\mathrm{\tau }}_2={\mathrm{I}}_2{\mathrm{\alpha }}_2\\ \text{Since an equal torque is exerted\hspace{0.17em}on both the bodies,}\\ \therefore {\mathrm{\tau }}_1={\mathrm{\tau }}_2\\ \therefore \frac{{\mathrm{\alpha }}_2}{{\mathrm{\alpha }}_1}=\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}\\ =\frac{{\mathrm{mr}}^2}{\frac{2}{5}{\mathrm{mr}}^2}\\ =\frac{5}{2}\\ \therefore {\mathrm{\alpha }}_2>{\mathrm{\alpha }}_1\to \left(\mathrm{i}\right)\\ \text{From the relation,\hspace{0.17em}}\mathrm{\omega }\text{=}{\mathrm{\omega }}_0+\mathrm{\alpha t}\\ \text{Here,\hspace{0.17em}}{\mathrm{\omega }}_0=\text{Initial angular velocity}\\ \text{t}=\text{Time of rotation}\\ \mathrm{\omega }=\text{Final angular velocity}\\ \text{For equal values\hspace{0.17em}of\hspace{0.17em}}{\mathrm{\omega }}_0\text{and t, we obtain:}\\ \mathrm{\omega }\propto \mathrm{\alpha }\to \left(\text{ii}\right)\\ \text{From equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{,\hspace{0.17em}we\hspace{0.17em}obtain:}\\ {\mathrm{\omega }}_2>{\mathrm{\omega }}_1\\ \therefore \text{The angular velocity of the solid sphere will be more than the angular} \\ \text{velocity of the hollow\hspace{0.17em}cylinder.}\end{array} \end{gathered}$$

Q.12 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^–1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder?
What is the magnitude of angular momentum of the cylinder about its axis?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of the cylinder, m}=\text{2}0\text{kg\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Angular speed\hspace{0.17em}of the cylinder,}\mathrm{\omega }=\text{1}00{\text{rad s}}^{–\text{1}}\\ \text{Radius of the cylinder, r}=0.\text{25 m}\\ \text{M.I. of the solid cylinder\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:\hspace{0.17em}}\\ \text{I=}\frac{{\mathrm{mr}}^2}{2}\\ =\frac{1}{2}\times 20\times {(0.\text{25})}^2\\ =0.625{\mathrm{kgm}}^2\\ \therefore \text{Kinetic energy\hspace{0.17em}of\hspace{0.17em}rotation=}\frac{1}{2}\text{I}{\mathrm{\omega }}^2\\ \text{=}\frac{1}{2}\times 0.625\times {\left(100\right)}^2\\ =3125\mathrm{J}\\ \therefore \text{Angular momentum\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}cylinder, L}=\text{I}\mathrm{\omega }\\ =\text{0.625}\times \text{1}00\mathrm{Js}\\ =\text{62}.\text{5 Js}\end{array}$

Q.13 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}initial angular velocity\hspace{0.17em}of\hspace{0.17em}child,}{\mathrm{\omega }}_{\text{1}}=\text{4}0{\text{rev min}}^{\text{-1}}\\ \text{Let\hspace{0.17em}final angular velocity of\hspace{0.17em}child}={\mathrm{\omega }}_{\text{2}}\\ {\text{Let\hspace{0.17em}I}}_{\text{1}}{\text{\hspace{0.17em}and\hspace{0.17em}I}}_{\text{2}}\text{\hspace{0.17em}be\hspace{0.17em}initial\hspace{0.17em}and\hspace{0.17em}final\hspace{0.17em}M.I.\hspace{0.17em}of\hspace{0.17em}child\hspace{0.17em}respectively.}\\ {\text{Initial\hspace{0.17em}and\hspace{0.17em}final\hspace{0.17em}M.I. are related as,\hspace{0.17em}I}}_{\text{2}}=\frac{2}{5}{\text{I}}_{\text{1}}\\ \text{As no external force acts on the child,}\\ \therefore \text{Angular momentum, L=\hspace{0.17em}constant.}\\ \therefore \text{For the two situations, we have:}\\ {\text{\hspace{0.17em}I}}_{\text{2}}{\mathrm{\omega }}_{\text{2}}={\text{I}}_{\text{1}}{\mathrm{\omega }}_{\text{1}}\\ \therefore {\mathrm{\omega }}_{\text{2}}=\frac{{\text{I}}_{\text{1}}}{{\text{I}}_{\text{2}}}{\mathrm{\omega }}_{\text{1}}\\ =\frac{{\text{I}}_{\text{1}}}{\frac{2}{5}{\text{I}}_{\text{1}}}\times 40\\ =\frac{5}{2}\times 40\\ =100{\mathrm{revmin}}^{-1}\\ \left(\text{b}\right){\text{Here,\hspace{0.17em}final K.E.\hspace{0.17em}of rotation\hspace{0.17em}of\hspace{0.17em}child, E}}_{\text{F}}=\frac{1}{2}{\text{I}}_{\text{2}}{{\mathrm{\omega }}_{\text{2}}}^2\\ {\text{Initial K.E.\hspace{0.17em}of rotation\hspace{0.17em}of\hspace{0.17em}child, E}}_{\text{I}}=\frac{1}{2}{\text{I}}_1{{\mathrm{\omega }}_1}^2\\ \therefore \frac{{\text{E}}_{\text{F}}}{{\text{E}}_{\text{I}}}=\frac{\frac{1}{2}{\text{I}}_{\text{2}}{{\mathrm{\omega }}_{\text{2}}}^2}{\frac{1}{2}{\text{I}}_1{{\mathrm{\omega }}_1}^2}\\ =\frac{2}{5}\frac{{\text{I}}_1}{{\text{I}}_1}\frac{{\left(100\right)}^2}{{\left(40\right)}^2}\\ =\frac{5}{2}\\ =2.5\\ \therefore {\text{E}}_{\text{F}}=2.5{\text{E}}_{\text{I}}\\ \therefore \text{\hspace{0.17em}K.E.\hspace{0.17em}of rotation\hspace{0.17em}of\hspace{0.17em}child\hspace{0.17em}increases.This\hspace{0.17em}is\hspace{0.17em}due\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}internal\hspace{0.17em}energy\hspace{0.17em}spent\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}child\hspace{0.17em}in\hspace{0.17em}folding\hspace{0.17em}his\hspace{0.17em}hands.}\end{array}$

Q.14 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of the hollow cylinder, m}=\text{3 kg\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Radius of the hollow cylinder, r}=\text{4}0\text{cm}\\ =0.\text{4 m}\\ \text{Force\hspace{0.17em}exerted, F}=\text{3}0\text{N}\\ \text{The moment of inertia of the hollow cylinder about its axis\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:\hspace{0.17em}I}={\text{mr}}^{\text{2}}\\ =\text{3}\times {(0.\text{4})}^{\text{2}}\\ =0.{\text{48 kg m}}^{\text{2}}\\ \text{Torque\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \mathrm{\tau }=\mathrm{F}\times \mathrm{r}\\ =\text{3}0\times 0.\text{4}\\ =\text{12 Nm}\\ \text{If\hspace{0.17em}}\mathrm{\alpha }\text{\hspace{0.17em}is\hspace{0.17em}the angular acceleration\hspace{0.17em}produced, torque is}\\ \text{given as:\hspace{0.17em}}\mathrm{\tau }=\text{I}\mathrm{\alpha }\\ \therefore \mathrm{\alpha }=\frac{\mathrm{\tau }}{\text{I}}\\ =\frac{\text{12}}{0.\text{4}8}\\ =25{\mathrm{rads}}^{-2}\\ \therefore \text{Linear acceleration,\hspace{0.17em}a}=\text{r}\mathrm{\alpha }\\ =0.\text{4}\times \text{25}\\ =\text{1}0{\text{ms}}^{–\text{2}}\end{array}$

Q.15 To maintain a rotor at a uniform angular speed of 200 rad s^–1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

Ans.

\begin{array}{l}\text{Here,}\text{angular speed of the rotor,}\omega =\text{2}00{\text{rads}}^{\text{-1}}\\ \text{Required}\text{torque,}\tau =\text{18}0\text{Nm}\\ \text{The power of the rotor}\left(\text{P}\right)\text{is related to torque and}\\ \text{angular speed as,}\text{P}=\tau \omega \\ \therefore P=\text{18}0\times \text{2}00=\text{36}\times \text{1}{0}^{\text{3}}=\text{36 kW}\\ \therefore \text{The}\text{power required by the engine = 36 kW}\end{array}

Q.16 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Ans.

$$\begin{gathered} \begin{array}{l}\text{Let\hspace{0.17em}mass per unit area of the disc}=\text{m}\\ \text{Let\hspace{0.17em}radius of the original disc}=\text{R}\\ \therefore \text{Mass of the original disc, M}=\mathrm{\pi }{\text{R}}^{\text{2}}\mathrm{m}\\ \text{The disc with the removed portion is shown in the given figure.}\\ \text{Here,\hspace{0.17em}radius of the smaller disc}=\frac{\mathrm{R}}{2}\\ \therefore \text{Mass of the smaller disc, M’}=\mathrm{\pi }{\left(\frac{\mathrm{R}}{2}\right)}^2\mathrm{m}\\ =\frac{1}{4}\mathrm{\pi }{\text{R}}^{\text{2}}\mathrm{m}\\ =\frac{\mathrm{M}}{4}\\ \text{Let O and O¢\hspace{0.17em}be the respective centres of the original disc and the} \\ \text{disc removed from\hspace{0.17em}the original.}\\ \text{According\hspace{0.17em}to the definition of the centre of mass, the centre of mass} \\ \text{of the\hspace{0.17em}original disc is supposed to be concentrated at O,}\\ \text{while that of the smaller disc is supposed to be concentrated at O¢.}\\ \text{Given,\hspace{0.17em}OO}‘=\frac{\mathrm{R}}{2}\\ \text{After the smaller disc has been\hspace{0.17em}removed from the original, the remaining} \\ \text{portion is\hspace{0.17em}assumed to be a system of two masses.}\\ \text{These\hspace{0.17em}masses are:}\\ \text{M}\left(\text{concentrated at O}\right)\text{, and}–\text{M}‘(=\frac{\mathrm{M}}{4})\text{concentrated at O¢}\\ \text{(The negative sign shows that this portion has been cut\hspace{0.17em}off from the original\hspace{0.17em}disc.)}\\ \text{Let distance of centre of mass of the remaining portion = x}\\ \text{Relation between the centres of masses of two\hspace{0.17em}masses is given as:}\\ \mathrm{x}=\frac{{\mathrm{m}}_1{\mathrm{r}}_1+{\mathrm{m}}_2{\mathrm{r}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ \text{Using\hspace{0.17em}the\hspace{0.17em}above\hspace{0.17em}relation\hspace{0.17em}for the given system, we have,}\\ \mathrm{x}=\frac{\mathrm{M}\times 0-\mathrm{M}‘\times \left(\frac{\mathrm{R}}{2}\right)}{\mathrm{M}+(-\mathrm{M}‘)}\\ =\frac{-\frac{\mathrm{M}}{4}\times \frac{\mathrm{R}}{2}}{\mathrm{M}-\frac{\mathrm{M}}{4}}\\ =\frac{-\mathrm{MR}}{8}\times \frac{4}{3\mathrm{M}}\\ =-\frac{\mathrm{R}}{6}\\ \text{The negative sign shows that the centre of mass of\hspace{0.17em}disc\hspace{0.17em}gets displaced} \\ \text{toward the left of\hspace{0.17em}point O.}\end{array} \end{gathered}$$

Q.17 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Ans.

$\begin{array}{l}\text{Let,\hspace{0.17em}mass of the meter stick}=\text{m}’\\ \text{Here,\hspace{0.17em}mass of each coin,\hspace{0.17em}m}=\text{5 g}\\ \text{As the coins are placed 12 cm away from the end P, the centre of mass gets\hspace{0.17em}displaced by 5 cm from point R towards the end P.}\\ \text{The centre of mass is positioned at the\hspace{0.17em}seperation of 45 cm from point P.}\\ \text{For\hspace{0.17em}the\hspace{0.17em}equilibrium\hspace{0.17em}state\hspace{0.17em}about\hspace{0.17em}the\hspace{0.17em}45\hspace{0.17em}cm\hspace{0.17em}mark,\hspace{0.17em}we\hspace{0.17em}have:}\\ 10\times \mathrm{g}(45-12)\\ =\mathrm{m}‘\mathrm{g}(50-45)\\ \therefore \mathrm{m}‘=\frac{10\times 33}{5}\\ =66\mathrm{g}\\ \therefore \text{Mass of the metre stick,\hspace{0.17em}m’ =66 g}.\end{array}$

Q.18 A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
(a) Will it reach the bottom with the same speed in each case?
(b) Will it take longer to roll down one plane than the other?
(c) If so, which one and why?

Ans.

$\begin{array}{l}\text{Let,\hspace{0.17em}mass of the sphere}=\text{m}\\ \text{Let\hspace{0.17em}height of the inclined\hspace{0.17em}planes}=\text{h}\\ \text{Let\hspace{0.17em}velocity of the sphere at the bottom of the plane}=\text{v}\\ \text{At the top of the\hspace{0.17em}inclined plane,}\\ \text{Total energy of the sphere=Potential energy}=\text{mgh}\\ \text{According\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}principle\hspace{0.17em}of\hspace{0.17em}conservation\hspace{0.17em}of\hspace{0.17em}energy,\hspace{0.17em}we\hspace{0.17em}have:}\\ \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2=\text{mgh}\\ \text{In\hspace{0.17em}case\hspace{0.17em}of\hspace{0.17em}a\hspace{0.17em}solid\hspace{0.17em}sphere,\hspace{0.17em}M.I.\hspace{0.17em}about\hspace{0.17em}its\hspace{0.17em}centre\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \mathrm{I}=\frac{2}{5}{\mathrm{mr}}^2\\ \therefore \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left(\frac{2}{5}{\mathrm{mr}}^2\right){\mathrm{\omega }}^2=\text{mgh}\\ \text{Linear\hspace{0.17em}and\hspace{0.17em}angular\hspace{0.17em}velocity\hspace{0.17em}are\hspace{0.17em}related\hspace{0.17em}as:}\\ \mathrm{r\omega }=\mathrm{v}\\ \therefore \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{5}{\mathrm{mv}}^2=\text{mgh}\\ \therefore \mathrm{v}=\sqrt{\frac{10}{7}\mathrm{gh}}\\ \text{Since\hspace{0.17em}h\hspace{0.17em}is\hspace{0.17em}same\hspace{0.17em}in\hspace{0.17em}each\hspace{0.17em}case,\hspace{0.17em}v\hspace{0.17em}must\hspace{0.17em}also\hspace{0.17em}be\hspace{0.17em}same.\hspace{0.17em}}\\ \therefore \text{Sphere\hspace{0.17em}will\hspace{0.17em}reach\hspace{0.17em}the\hspace{0.17em}bottom\hspace{0.17em}with\hspace{0.17em}same\hspace{0.17em}speed\hspace{0.17em}in\hspace{0.17em}each\hspace{0.17em}case.}\end{array}$

Q.19 A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}radius of the hoop, r}=\text{2 m}\\ \text{Mass of the hoop, m}=\text{1}00\text{kg}\\ \text{Velocity of the hoop, v}=\text{2}0{\text{cms}}^{\text{-1}}=0.{\text{2 ms}}^{\text{-1}}\\ \text{Moment of inertia of the hoop about its centre,\hspace{0.17em}I}={\text{mr}}^{\text{2}}\\ \left(\text{Total energy of the hoop = Translational KE + Rotational KE}\right)\\ \text{=}\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2\\ \text{=}\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left({\text{mr}}^{\text{2}}\right){\mathrm{\omega }}^2(\because \mathrm{I}={\mathrm{mr}}^2)\\ =\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{mv}}^2(\because \mathrm{r\omega }=\mathrm{v})\\ ={\mathrm{mv}}^2\\ \text{Work required to be\hspace{0.17em}done\hspace{0.17em}to\hspace{0.17em}stop the hoop = Total energy of\hspace{0.17em}the hoop.}\\ \therefore \text{Required work, W}={\text{mv}}^{\text{2}}\\ =\text{1}00\times {(0.\text{2})}^{\text{2}}=\text{4 J}\end{array}$

Q.20 The oxygen molecule has a mass of 5.30 × 10^–26 kg and a moment of inertia of 1.94×10^–46 kg m² about an axis through its centre perpendicular to the lines joining the two atoms.
Suppose the mean speed of such a molecule in a gas is 500 ms^-1 and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of the oxygen molecule, m}=\text{5}.\text{3}0\times \text{1}{0}^{–\text{26}}\text{kg}\\ \text{Moment of inertia\hspace{0.17em}of\hspace{0.17em}oxygen\hspace{0.17em}molecule, I}=\text{1}.\text{94}\times \text{1}{0}^{–\text{46}}{\text{kg m}}^{\text{2}}\\ \text{Velocity of the oxygen molecule, v}=\text{5}00{\text{ms}}^{\text{-1}}\\ \text{Let distance\hspace{0.17em}of\hspace{0.17em}separation between the two atoms of the oxygen\hspace{0.17em}molecule}=\text{2r}\\ \text{Mass of each oxygen atom}=\frac{\mathrm{m}}{2}\\ \therefore \text{Moment of inertia I\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as}:\\ \mathrm{I}=\frac{\mathrm{m}}{2}{\mathrm{r}}^2+\frac{\mathrm{m}}{2}{\mathrm{r}}^2\\ ={\mathrm{mr}}^2\to \text{(i)}\\ \therefore \mathrm{r}=\sqrt{\frac{\mathrm{I}}{\mathrm{m}}}\\ =\sqrt{\frac{\text{1}.\text{94}\times \text{1}{0}^{–\text{46}}}{\text{5}.\text{3}0\times \text{1}{0}^{–\text{26}}}}\\ =0.61\times \text{1}{0}^{–10}\mathrm{m}\\ \text{Given,\hspace{0.17em}K.E.\hspace{0.17em}of\hspace{0.17em}rotation=}\frac{2}{3}\times \text{K.E.\hspace{0.17em}of\hspace{0.17em}translation}\\ \therefore \frac{1}{2}{\mathrm{I\omega }}^2=\frac{2}{3}\times \frac{1}{2}{\mathrm{mv}}^2\\ \frac{1}{2}\left({\mathrm{mr}}^2\right){\mathrm{\omega }}^2=\frac{1}{3}{\mathrm{mv}}^2(\because \mathrm{I}={\mathrm{mr}}^2)\\ \therefore \mathrm{\omega }=\sqrt{\frac{2}{3}}\frac{\mathrm{v}}{\mathrm{r}}\\ =\sqrt{\frac{2}{3}}\times \frac{500}{0.61\times {10}^{-10}}\\ =6.75\times {10}^{12}{\mathrm{rads}}^{-1}\end{array}$

Q.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?

Ans.

$$\begin{gathered} \begin{array}{l}\text{A solid cylinder rolling up an inclination is shown in the given figure.}\\ \text{Here,\hspace{0.17em}Angle of inclination,}\mathrm{\theta }=\text{3}0\mathrm{°}\\ \text{Initial velocity of the solid cylinder, v}={\text{5 ms}}^{\text{-1}}\\ \text{Height attained by the cylinder}=\text{h}\\ \left(\text{a}\right)\text{Energy of the cylinder at point P:}\\ {\text{K.E.}}_{\mathrm{Rot}}={\text{K.E.}}_{\mathrm{Trans}}\\ \text{Energy of the cylinder at point Q=mgh According\hspace{0.17em}to the law of conservation of energy, we have:}\\ \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2=\mathrm{mgh}\to \text{(i)}\\ \text{Moment of inertia of the solid cylinder,}\mathrm{I}=\frac{1}{2}{\mathrm{mr}}^2\to \text{(ii)}\\ \text{Substituting\hspace{0.17em}equation\hspace{0.17em}(ii)\hspace{0.17em}in\hspace{0.17em}equation\hspace{0.17em}(i),\hspace{0.17em}we\hspace{0.17em}obtain:}\\ \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left(\frac{1}{2}{\mathrm{mr}}^2\right){\mathrm{\omega }}^2=\mathrm{mgh}\to \text{(iii)}\\ \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{4}{\mathrm{mv}}^2=\mathrm{mgh}(\because \mathrm{r\omega }=\mathrm{v})\\ \frac{3}{4}{\mathrm{mv}}^2=\mathrm{mgh}\\ \therefore \mathrm{h}=\frac{3{\mathrm{v}}^2}{4\mathrm{g}}\\ =\frac{3\times {\left(5\right)}^2}{4\times 9.8}\\ =1.913\mathrm{m}\\ \text{Let\hspace{0.17em}s\hspace{0.17em}be\hspace{0.17em}the\hspace{0.17em}distance\hspace{0.17em}up\hspace{0.17em}the\hspace{0.17em}inclined\hspace{0.17em}plane, then\hspace{0.17em}using the\hspace{0.17em}relation,}\\ \mathrm{sin\theta }=\frac{\mathrm{h}}{\mathrm{s}}\\ \therefore \mathrm{s}=\frac{\mathrm{h}}{\mathrm{sin\theta }}\\ =\frac{1.913}{\sin {30}^{\mathrm{o}}}\\ =3.826\mathrm{m}\\ \text{In}\mathrm{\Delta PQR}:\\ \mathrm{sin\theta }=\frac{\mathrm{QR}}{\mathrm{PQ}}\\ \sin {30}^0=\frac{\mathrm{h}}{\mathrm{PQ}}\\ \mathrm{PQ}=\frac{1.91}{0.5}\\ =3.82\mathrm{m}\\ \therefore \text{The cylinder will move 3.82 m up the inclined plane.}\\ \left(\text{b}\right)\text{For radius of gyration K, the velocity of the cylinder at the instance it rolls\hspace{0.17em}back to} \\ \text{the bottom is given as:}\\ \mathrm{v}=\left(\frac{2\mathrm{gh}}{1+\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}}\right)\\ \text{In\hspace{0.17em}case\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}solid\hspace{0.17em}cylinder,\hspace{0.17em}}{\mathrm{K}}^2=\frac{{\mathrm{R}}^2}{2}\end{array} \end{gathered}$$ \begin{array}{l}\therefore v={\left(\frac{2gAB\sin \theta }{1+\frac{1}{2}}\right)}^{\frac{1}{2}}\\ ={\left(\frac{4}{3}gAB\sin \theta \right)}^{\frac{1}{2}}\\ \text{The time taken to come}\text{back to the bottom is}\text{given}\text{as:}\\ t=\frac{AB}{v}\\ =\frac{AB}{{\left(\frac{4}{3}gAB\sin \theta \right)}^{\frac{1}{2}}}\\ ={\left(\frac{3AB}{4g\sin \theta }\right)}^{\frac{1}{2}}\\ \therefore \text{Total time taken by the cylinder to come}\text{back to the bottom =}\left(\text{2}\times 0.\text{764}\right)\text{s}\\ \text{=1}.\text{53 s}\text{.}\end{array}

Q.22  As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 ms^-2)(Hint: Consider the equilibrium of each side of the ladder separately.)

Ans.

The situation of the problem can be shown in the given figure.

\begin{array}{l}\text{Here,}{\text{N}}_{\text{B}}=\text{Force applied on the ladder by the floor at}\text{point B}\\ {\text{N}}_{\text{C}}=\text{Force applied on the ladder by the floor at}\text{point C}\\ \text{T}=\text{Tension produced}\text{in the rope BA}=\text{CA}=\text{1}.\text{6 m}\\ \text{DE}=0.\text{5 m}\\ \text{BF}=\text{1}.\text{2 m}\\ \text{Mass of the weight, m}=\text{4}0\text{kg}\\ \text{Draw a normal from A on the floor BC,}\text{which intersects DE at the}\text{mid}\text{–}\text{point H}\text{.}\\ \text{Since,}\Delta \text{ABI and}\Delta \text{AIC are similar}\\ \therefore \text{BI}=\text{IC}\\ \therefore \text{I is the mid – point of BC}\text{.}\\ \text{DE}\left|\right|\text{BC}\\ \text{BC}=\text{2}\times \text{DE}\\ =\text{1 m}\\ \text{AF}=\text{BA}–\text{BF}\\ =0.\text{4 m}\dots \left(\text{i}\right)\\ \text{Since}\text{D is the mid – point of AB}\\ \therefore \text{AD}=\frac{\text{1}}{\text{2}}\times \text{BA}\\ =0.8m\to \text{(ii)}\\ \text{From equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we obtain:}\\ \text{FE}=0.\text{4 m}\\ \therefore \text{F is the mid – point of AD}\text{.}\\ \text{FG || DH and F is the middle}\text{point of AD}\text{.}\\ \therefore \text{G will also be the middle}\text{point of AH}\text{.}\\ \text{As}\Delta \text{AFG and}\Delta \text{ADH are similar}\end{array} $\begin{array}{l}\therefore \frac{\text{FG}}{\text{DH}}\text{=}\frac{\text{AF}}{\text{AD}}\\ \frac{\text{FG}}{\text{DH}}=\frac{0.4}{0.8}\\ =\frac{1}{2}\\ \text{FG}=\frac{1}{2}\text{DH}\\ =\frac{1}{2}\times 0.25\\ =0.125\mathrm{m}\\ \text{In}\mathrm{\Delta }\text{ADH}:\\ \text{AH}=\sqrt{{\text{AD}}^{\text{2}}{\text{-DH}}^{\text{2}}}\\ =\sqrt{{(0.8)}^2-{(0.25)}^2}\\ =0.76\mathrm{m}\\ \text{In\hspace{0.17em}case\hspace{0.17em}of\hspace{0.17em}translational equilibrium of the ladder, the upward force must be same\hspace{0.17em}as\hspace{0.17em}that\hspace{0.17em}of the\hspace{0.17em}downward\hspace{0.17em}force.}\\ \therefore {\text{N}}_{\text{C}}+{\text{N}}_{\text{B}}=\text{mg}\\ =\text{392 N}\dots \left(\text{iii}\right)\\ \text{In\hspace{0.17em}case\hspace{0.17em}of rotational equilibrium of the ladder, the total moment about A \hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ -{\text{N}}_{\text{B}}\times \mathrm{BI}+\mathrm{mg}\times \mathrm{FG}+{\text{N}}_{\text{C}}\times \mathrm{CI}+\mathrm{T}\times \mathrm{AG}-\mathrm{T}\times \mathrm{AG}=0\\ -{\text{N}}_{\text{B}}\times 0.5+40\times 9.8\times 0.125+{\text{N}}_{\text{C}}\times (0.5)=0\\ ({\text{N}}_{\text{C}}-{\text{N}}_{\text{B}})\times 0.5=49\\ {\text{N}}_{\text{C}}-{\text{N}}_{\text{B}}=98\mathrm{N}...\text{(iv)}\\ \text{By\hspace{0.17em}adding\hspace{0.17em}equation\hspace{0.17em}(iii)\hspace{0.17em}and\hspace{0.17em}(iv),\hspace{0.17em}we\hspace{0.17em}obtain:}\\ {\text{N}}_{\text{C}}=245\mathrm{N}\\ {\text{N}}_{\text{B}}=147\mathrm{N}\\ \text{In\hspace{0.17em}case\hspace{0.17em}of\hspace{0.17em}rotational\hspace{0.17em}equilibrium\hspace{0.17em}of\hspace{0.17em}side\hspace{0.17em}AB,\hspace{0.17em}assume\hspace{0.17em}the moment\hspace{0.17em}about\hspace{0.17em}A.}\\ -{\text{N}}_{\text{B}}\times \mathrm{BI}+\mathrm{mg}\times \mathrm{FG}+\mathrm{T}\times \mathrm{AG}=0\\ -245\times 0.5+40+9.8\times 0.125+\mathrm{T}\times 0.76=0\\ 0.76\mathrm{T}=122.5-49\\ \therefore \mathrm{T}=96.7\mathrm{N}\end{array}$

Q.23 A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m².
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Here,\hspace{0.17em}M.I. of the man-platform system,I}=\text{7}.{\text{6 kgm}}^{\text{2}}\\ \text{when the man stretches his hands to a distance of 9}0{\text{cm,\hspace{0.17em}M.I.,\hspace{0.17em}I}}_{\mathrm{a}}\text{=2}\times {\text{mr}}^{\text{2}}\\ =\text{2}\times \text{5}\times {(0.\text{9})}^{\text{2}}\\ =\text{8}.{\text{1 kgm}}^{\text{2}}\\ \text{Initial M.I. of the system,\hspace{0.17em}}{\mathrm{I}}_1=\mathrm{I}+{\text{I}}_{\mathrm{a}}\\ \text{=7}.\text{6 +8}.\text{1}\\ {\text{=15.7\hspace{0.17em}kgm}}^{\text{2}}\\ \text{Angular speed,\hspace{0.17em}}{\mathrm{\omega }}_1=30\mathrm{rpm}\\ \text{Angular momentum,\hspace{0.17em}}{\mathrm{L}}_1={\mathrm{I}}_1{\mathrm{\omega }}_1\\ =\text{15.7}\times 30\\ =471{\text{\hspace{0.17em}kgm}}^{\text{2}}{\mathrm{s}}^{-1}\\ {\text{Moment of inertia when the man bends his hands to a distance of 20 cm,\hspace{0.17em}I}}_{\text{b}}=\text{2}\times {\text{mr}}^{\text{2}}\\ =\text{2}\times \text{5}\times {(0.2)}^{\text{2}}\\ =0.{\text{4 kgm}}^{\text{2}}\\ \text{Final moment of inertia,}{\mathrm{I}}_2=\mathrm{I}+{\text{I}}_{\text{b}}\\ =\text{7}.\text{6 +}0.\text{4}\\ =8{\mathrm{kgm}}^2\\ \text{Let,\hspace{0.17em}final angular speed}={\mathrm{\omega }}_2\\ \text{Final angular momentum,\hspace{0.17em}}{\mathrm{L}}_2={\mathrm{I}}_2{\mathrm{\omega }}_2\\ =8{\mathrm{\omega }}_2\dots \left(\text{ii}\right)\\ \text{According\hspace{0.17em}to the law\hspace{0.17em}of\hspace{0.17em}conservation of angular momentum, we have:}\\ \text{Final\hspace{0.17em}angular\hspace{0.17em}momentum = Initial\hspace{0.17em}angular\hspace{0.17em}momentum}\\ {\mathrm{I}}_2{\mathrm{\omega }}_2={\mathrm{I}}_1{\mathrm{\omega }}_1\\ \therefore {\mathrm{\omega }}_2=\frac{471}{8}\\ =\text{58.875 rev/min}\\ \left(\text{b}\right)\text{No,\hspace{0.17em}K.E. is not conserved in this\hspace{0.17em}process.\hspace{0.17em}The\hspace{0.17em}fact\hspace{0.17em}is\hspace{0.17em}that,\hspace{0.17em}as\hspace{0.17em}the M.I.\hspace{0.17em}decreases,\hspace{0.17em}the K.E. increases.}\\ \text{The work done by the man to bend his hands\hspace{0.17em}closer\hspace{0.17em}to\hspace{0.17em}his body provides the additional K.E.}\end{array}$

Q.24 A bullet of mass 10 g and speed 500 ms^-1 is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.. (Hint: The moment of inertia of the door about the vertical axis at one end is ML²/3.)

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of the bullet, m}=\text{1}0\text{g}\\ =\text{1}0\times \text{1}{0}^{–\text{3}}\text{\hspace{0.17em}kg}\\ \text{Speed of the bullet, v}=\text{5}00{\text{ms}}^{\text{-1}}\\ \text{Thickness of the door, L}=\text{1 m}\\ \text{Radius of door, r}=\frac{1}{2}\mathrm{m}\\ \text{Mass of door, M}=\text{12 kg}\\ \text{Angular momentum provided by the bullet on the door,\hspace{0.17em}}\mathrm{\alpha }=\text{mvr}\\ \text{=}(\text{1}0\times \text{1}{0}^{–\text{3}})\times 500\times \frac{1}{2}\\ {\text{=2.5\hspace{0.17em}kgm}}^{\text{2}}{\text{s}}^{\text{-1}}\\ \text{M.I. of the door,\hspace{0.17em}I=}\frac{{\mathrm{ML}}^2}{3}\\ =\frac{12\times 1^2}{3}\\ =4{\mathrm{kgm}}^2\\ \text{Since,\hspace{0.17em}angular\hspace{0.17em}momentum\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as}:\mathrm{L}=\mathrm{I\omega }\\ \therefore \mathrm{\omega }=\frac{\mathrm{L}}{\mathrm{I}}\\ =\frac{2.5}{4}\\ =0.625{\mathrm{rads}}^{-1}\\ \therefore \text{Angular\hspace{0.17em}speed\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}door,\hspace{0.17em}}\mathrm{\omega }=0.625{\mathrm{rads}}^{-1}\end{array}$

Q.25 Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident.
(a) What is the angular speed of the two-disc system?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω₁ ≠ ω₂.

Ans.

$\begin{array}{l}\left(\text{a}\right){\text{Here,\hspace{0.33em}two\hspace{0.33em}discs\hspace{0.33em}of\hspace{0.33em}moments\hspace{0.33em}of\hspace{0.33em}inertia\hspace{0.33em}I}}_1{\text{\hspace{0.33em}and\hspace{0.33em}I}}_2\text{\hspace{0.33em}about\hspace{0.33em}their\hspace{0.33em}respective}\\ \text{axes\hspace{0.33em}are\hspace{0.33em}brought\hspace{0.33em}into\hspace{0.33em}contact.}\\ {\text{Total\hspace{0.33em}initial\hspace{0.33em}angular\hspace{0.33em}momentum\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}two\hspace{0.33em}discs,\hspace{0.33em}L}}_1={\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_2{\mathrm{\omega }}_2\\ \text{As the two discs are joined, their M.I. get added up.}\\ \text{M.I. of the system of two discs\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:\hspace{0.33em}}\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2\\ \text{Let the angular speed of the system=}\mathrm{\omega }\\ \text{Total final angular momentum\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}system,}\\ {\mathrm{L}}_2=\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)\mathrm{\omega }\\ \text{According\hspace{0.33em}to the law of conservation of angular momentum,\hspace{0.33em}we have:}\\ \text{Final\hspace{0.33em}angular momentum = Initial\hspace{0.33em}angular\hspace{0.33em}momentum\hspace{0.33em}\hspace{0.33em}}\\ \left({\mathrm{I}}_1+{\mathrm{I}}_2\right)\mathrm{\omega }={\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_2{\mathrm{\omega }}_2\\ \therefore \mathrm{\omega }=\frac{{\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_2{\mathrm{\omega }}_2}{{\mathrm{I}}_1+{\mathrm{I}}_2}\to \text{(i)}\\ \left(\text{b}\right)\text{Let,\hspace{0.33em}kinetic energy of disc I be}\frac{1}{2}{\mathrm{I}}_1{{\mathrm{\omega }}_1}^2\\ \text{Kinetic energy of disc II be}\frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}{{\text{ω}}_{\text{2}}}^{\text{2}}\\ {\text{Total initial kinetic energy,E}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{I}}_1{{\mathrm{\omega }}_1}^2+\frac{1}{2}{\mathrm{I}}_2{{\mathrm{\omega }}_2}^2\\ \text{As the discs are joined\hspace{0.33em}together, their M.I. get added up.}\\ \text{Moment of inertia of the system,\hspace{0.33em}}\mathrm{I}=\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)\\ \text{Let\hspace{0.33em}angular speed of the\hspace{0.33em}system}=\mathrm{\omega }\\ {\text{Final kinetic energy, E}}_{\text{f}}=\frac{1}{2}\left({\mathrm{I}}_1+{\mathrm{I}}_2\right){\mathrm{\omega }}^2\\ \text{Using\hspace{0.33em}equation\hspace{0.33em}(i),\hspace{0.33em}we\hspace{0.33em}get}\\ {\text{E}}_{\text{f}}=\frac{1}{2}\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)\frac{{\left({\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_2{\mathrm{\omega }}_2\right)}^2}{{\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)}^2}=\frac{{\left({\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_2{\mathrm{\omega }}_2\right)}^2}{2\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)}\\ {\text{Now,\hspace{0.33em}E}}_{\mathrm{i}}-{\text{E}}_{\text{f}}=\frac{1}{2}{\mathrm{I}}_1{{\mathrm{\omega }}_1}^2+\frac{1}{2}{\mathrm{I}}_2{{\mathrm{\omega }}_2}^2-\frac{{\left({\mathrm{I}}_1{\mathrm{\omega }}_1+{\mathrm{I}}_2{\mathrm{\omega }}_2\right)}^2}{2\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)}\\ {\text{Thus,\hspace{0.33em}E}}_{\mathrm{i}}-{\text{E}}_{\text{f}}=\frac{{\mathrm{I}}_1{\mathrm{I}}_2{\left({\mathrm{\omega }}_1-{\mathrm{\omega }}_2\right)}^2}{2\left({\mathrm{I}}_1+{\mathrm{I}}_2\right)}\\ \therefore {\text{E}}_{\mathrm{i}}-{\text{E}}_{\text{f}}>0\\ {\text{Therefore,\hspace{0.33em}E}}_{\text{f}}<{\text{E}}_{\mathrm{i}}\\ \text{The\hspace{0.33em}loss\hspace{0.33em}of\hspace{0.33em}K.E.\hspace{0.33em}takes\hspace{0.33em}place\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}process\hspace{0.33em}due\hspace{0.33em}to the frictional force}\\ \text{that comes into play when the\hspace{0.33em}two discs come in contact with each other.}\end{array}$

Q.26 (a) Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x² + y²).
(b) Prove the theorem of parallel axes. (Hint: If the centre of mass is chosen to be the origin).

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{According\hspace{0.33em}to\hspace{0.33em}the theorem of perpendicular axes, the MI of a planar\hspace{0.33em}body}\\ \text{about an axis perpendicular to its plane is equalto the sum of its\hspace{0.33em}M.I.\hspace{0.33em}about}\\ \text{two perpendicular axes concurrent with perpendicular axis and lying in the}\\ \text{plane of the body.\hspace{0.33em}A physical body with centre O and a point mass m,\hspace{0.33em}in}\\ \text{the x}–\text{y plane at}\left(\text{x},\text{y}\right)\text{is shown\hspace{0.33em}in the given figure.}\\ \text{Here,\hspace{0.33em}M.I. about x}-\text{axis},{\text{I}}_{\text{x}}={\text{mx}}^{\text{2}}\\ \text{M.I. about y}-\text{axis},{\text{I}}_{\text{y}}={\text{my}}^{\text{2}}\\ \text{M.I. about z}-\text{axis},{\text{I}}_{\text{z}}=\mathrm{m}{\left(\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\right)}^2\\ ={\text{I}}_{\text{x}}+{\text{I}}_{\text{y}}\\ ={\text{mx}}^{\text{2}}+{\text{my}}^{\text{2}}\\ =\text{m}({\text{x}}^{\text{2}}+{\text{y}}^{\text{2}})\\ =\mathrm{m}{\left(\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\right)}^2\\ {\text{I}}_{\text{x}}+{\text{I}}_{\text{y}}={\text{I}}_{\text{z}}\\ \text{Hence,\hspace{0.33em}theorem\hspace{0.33em}is\hspace{0.33em}proved.}\\ \left(\text{b}\right)\text{According\hspace{0.33em}to\hspace{0.33em}the theorem of parallel axes,\hspace{0.33em}the MI of a\hspace{0.33em}body about\hspace{0.33em}any}\\ \text{axis is equal to the sum of the MI of the body abouta parallel\hspace{0.33em}axis passing}\\ \text{through its CM and the product of its mass and the square of\hspace{0.33em}the distance}\\ \text{between the two parallel axes. Suppose a rigid body is made up of n particles,}\\ {\text{having masses m}}_{\text{1}},{\text{m}}_{\text{2}},{\text{m}}_{\text{3}},\dots ,{\text{m}}_{\text{n}},{\text{at\hspace{0.33em}normal\hspace{0.33em}distances r}}_{\text{1}},{\text{r}}_{\text{2}},{\text{r}}_{\text{3}},\dots ,{\text{r}}_{\text{n}}\\ \text{respectively from the centre of mass of the\hspace{0.33em}rigid body.}\end{array}$

Q.27

$\begin{array}{l}\text{Prove the result that the velocity v of translation of a rolling body}\\ \text{(like a ring, disc, cylinder or sphere )\hspace{0.33em}at the bottom of an inclined}\\ \text{plane of a\hspace{0.33em}height h is given by\hspace{0.33em}}{\mathrm{v}}^2=\frac{2\mathrm{gh}}{\left(1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}\right)}.\text{Using dynamical consideration}\\ \text{(i.e. by consideration of forces and torques). Note k is the radius of gyration}\\ \text{of the\hspace{0.33em}body about its symmetry axis, and R is the radius of the body The body}\\ \text{starts from rest at\hspace{0.33em}the top of the plane.}\end{array}$

Ans.

$\begin{array}{l}\text{A body rolling on an inclined plane of height h,is shown in the given figure.}\\ \text{Here,\hspace{0.17em}radius of the body=R}\\ \text{Height of the inclined plane}=\text{h}\\ \text{Acceleration due to gravity}=\text{g}\\ \text{Translational velocity of the body}=\text{v}\\ \text{Radius of gyration of the body}=\text{k}\\ \text{Let\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}body\hspace{0.17em}=\hspace{0.17em}m}\\ {\text{Total energy at the top of the\hspace{0.17em}plane, E}}_{\text{1}}=\text{mgh}\\ {\text{M.I.\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}rolling\hspace{0.17em}body\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:\hspace{0.17em}I\hspace{0.17em}=\hspace{0.17em}mk}}^{\text{2}}\\ \text{According\hspace{0.17em}to the principle of conservation of energy:}\\ \text{K.E.\hspace{0.17em}of\hspace{0.17em}translation + K.E.\hspace{0.17em}of\hspace{0.17em}rotation}=\text{P.E.\hspace{0.17em}at\hspace{0.17em}the\hspace{0.17em}top}\\ \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2=\mathrm{mgh}\\ \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left({\mathrm{mk}}^2\right){\mathrm{\omega }}^2=\mathrm{mgh}(\because \mathrm{I}={\mathrm{mk}}^2)\\ \text{Since,\hspace{0.17em}}\mathrm{\omega }=\frac{\mathrm{v}}{\mathrm{R}}\\ \therefore \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\mathrm{m}\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}{\mathrm{v}}^2=\mathrm{mgh}\\ \therefore \frac{1}{2}{\mathrm{mv}}^2(1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2})=\mathrm{mgh}\\ \therefore {\mathrm{v}}^2=\frac{2\mathrm{gh}}{(1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2})}\\ \text{Hence, proved.}\end{array}$

Q.28 A disc rotating about its axis with angular speed ω_o is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}angular speed of the disc}={\mathrm{\omega }}_{\text{o}}\\ \text{Radius of the disc}=\text{R}\\ \text{From the relation for linear velocity, v}={\mathrm{\omega }}_{\text{o}}\text{R For point A:}\\ {\text{v}}_{\text{A}}=\text{R}{\mathrm{\omega }}_{\text{o}};\text{\hspace{0.17em}tangential\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}right For point B:}\\ {\text{v}}_{\text{B}}=\text{R}{\mathrm{\omega }}_{\text{o}};\text{\hspace{0.17em}tangential\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}left For point C:}\\ {\mathrm{v}}_{\mathrm{C}}=\left(\frac{\mathrm{R}}{2}\right){\mathrm{\omega }}_{\text{o}};\text{\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}direction\hspace{0.17em}of\hspace{0.17em}}{\mathrm{v}}_{\mathrm{A}}\\ \text{The directions of motion of points A, B, and C on the disc are shown in the given\hspace{0.17em}figure As the disc is placed on a frictionless table,}\\ \text{it will not\hspace{0.17em}roll\hspace{0.17em}at\hspace{0.17em}all as friction is essential for the rolling of a body.}\end{array}$

Q.29 Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?

Ans.

In order to roll a disc, we need a torque tangential to the disc; provided only by frictional force in this case.
(a) Frictional force acts in the direction opposite to the direction of velocity at point B. The linear velocity at point B acts along tangentially leftward. Thus, force of friction will act tangentially rightward and the sense of frictional torque will be in normal direction to the plane of the disc and outwards before the start.
(b) Since force of friction acts opposite velocity at point B, perfect rolling will start when the velocity at that point becomes zero. The force of friction will also become zero at this stage.

Q.30 A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s^-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μ_k = 0.2.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.33em}radii of the ring and the disc, r}=\text{1}0\text{cm}\\ \text{=}0.\text{1 m}\\ \text{Initial angular speed,}{\mathrm{\omega }}_0=\text{1}0\mathrm{\pi }{\text{rads}}^{–\text{1}}\\ \text{Coefficient of kinetic friction,}{\mathrm{\mu }}_{\text{k}}=0.\text{2}\\ \text{Initial velocity of\hspace{0.33em}ring\hspace{0.33em}and\hspace{0.33em}disc, u=}0\\ \text{Frictional force\hspace{0.33em}gives\hspace{0.33em}rise\hspace{0.33em}to\hspace{0.33em}motion of the two objects.}\\ \text{According\hspace{0.33em}to Newton’s second law of\hspace{0.33em}motion:}\\ \text{Force\hspace{0.33em}of\hspace{0.33em}friction, f}=\text{ma}\\ \therefore {\mathrm{\mu }}_{\text{k}}\text{mg}=\text{ma}\\ \text{Here,\hspace{0.33em}a=Acceleration produced}\\ \text{m}=\text{Mass}\\ \therefore \text{a}=\mathrm{\mu }\text{kg}\dots \left(\text{i}\right)\\ \text{According\hspace{0.33em}to the first equation of motion,}\\ \text{v}=\text{u}+\text{at}\\ =0+{\mathrm{\mu }}_{\text{k}}\text{gt}\\ \therefore \mathrm{v}={\mathrm{\mu }}_{\text{k}}\text{gt}\dots \left(\text{ii}\right)\\ \text{Torque due\hspace{0.33em}to\hspace{0.33em}frictional\hspace{0.33em}force\hspace{0.33em}will act perpendicularly outwards\hspace{0.33em}and give}\\ \text{rise\hspace{0.33em}to retardation in the initial angular speed\hspace{0.33em}}{\mathrm{\omega }}_{\text{o}}.\\ \text{Torque\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by\hspace{0.33em}the\hspace{0.33em}relation,}\mathrm{\tau }=–\text{I}\mathrm{\alpha }\\ \text{Here,\hspace{0.33em}}\mathrm{\alpha }=\text{Angular acceleration}\\ \therefore {\mathrm{\mu }}_{\mathrm{k}}\text{mgr}=–\text{I}\mathrm{\alpha }\\ \therefore \mathrm{\alpha }=\frac{-{\mathrm{\mu }}_{\mathrm{k}}\text{mgr}}{\mathrm{I}}\to \text{(iii)}\\ \text{Applying the first equation of rotational motion,\hspace{0.33em}we\hspace{0.33em}have:\hspace{0.33em}}\mathrm{\omega }={\mathrm{\omega }}_{\mathrm{o}}+\mathrm{\alpha t}\\ \therefore \mathrm{\omega }={\mathrm{\omega }}_{\mathrm{o}}+\frac{-{\mathrm{\mu }}_{\mathrm{k}}\text{mgr}}{\mathrm{I}}\mathrm{t}\to \left(\mathrm{iv}\right)\\ \text{Rolling begins when linear velocity, v}=\text{r}\mathrm{\omega }\\ \therefore \text{v}=\text{r}\left({\mathrm{\omega }}_{\mathrm{o}}-\frac{{\mathrm{\mu }}_{\mathrm{k}}\text{mgrt}}{\mathrm{I}}\right)\to \left(\mathrm{v}\right)\\ \text{Equating equations}\left(\text{ii}\right)\text{and}\left(\text{v}\right)\text{, we obtain:}\\ {\mathrm{\mu }}_{\mathrm{k}}\text{g}\mathrm{t}=\text{r}\left({\mathrm{\omega }}_{\mathrm{o}}-\frac{{\mathrm{\mu }}_{\mathrm{k}}\text{mgrt}}{\mathrm{I}}\right)\\ \therefore {\mathrm{\mu }}_{\mathrm{k}}\text{g}\mathrm{t}=\text{r}{\mathrm{\omega }}_{\mathrm{o}}-\frac{{\mathrm{\mu }}_{\mathrm{k}}{\text{mgr}}^{\text{2}}\text{t}}{\mathrm{I}}\to \left(\mathrm{vi}\right)\\ \text{In\hspace{0.33em}case\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}ring:\hspace{0.33em}}\\ \mathrm{I}={\mathrm{mr}}^2\\ \therefore {\mathrm{\mu }}_{\mathrm{k}}\text{g}\mathrm{t}=\text{r}{\mathrm{\omega }}_{\mathrm{o}}-\frac{{\mathrm{\mu }}_{\mathrm{k}}{\text{mgr}}^{\text{2}}\text{t}}{{\mathrm{mr}}^{{}^2}}\end{array}$

Q.31 A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µs = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?

Ans.

\begin{array}{l}\text{The several forces acting on the cylinder are shown in the given figure}\text{.}\\ \text{Acceleration of the cylinder}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{a=}\frac{mg\sin \theta }{m+\frac{I}{r^2}}\\ =\frac{mg\sin \theta }{m+\frac{1}{2}\frac{m{r}^2}{r^2}}\\ =\frac{2}{3}g\sin {30}^o\\ =\frac{2}{3}\times 9.8\times 0.5\\ =3.27m{s}^{-2}\\ \left(\text{a}\right)\text{According}\text{to Newton’s second law of motion, we}\text{have:}{\text{f}}_{\text{net}}=\text{ma}\\ \therefore \text{mgsin}{30}^o-f=ma\\ \therefore f=\text{mgsin}{30}^o-ma\\ =10\times 9.8\times 0.5-10\times 3.27\\ =49-32.7\\ =16.3N\\ \left(\text{b}\right)\text{At}\text{the}\text{time}\text{of rolling, the instantaneous point of contact with the plane comes to rest}\text{.}\\ \therefore \text{Work done against the}\text{force of}\text{friction = 0}\\ \left(\text{c}\right)\text{In}\text{case}\text{of rolling without slipping, we have:}\\ \mu =\frac{1}{3}tan\theta \\ \therefore tan\theta =3\mu \\ =3\times 0.25\\ =0.75\\ \therefore \theta =ta{n}^{-1}\left(0.75\right)\\ ={36.87}^o\end{array}

Q.32

$\begin{array}{l}\text{Separation of Motion of a system of particles into\hspace{0.33em}motion of the}\\ \text{centre of mass and motion about the centre of mass:}\\ \left(\text{a}\right)\text{Show pi}={\text{p’}}_{\text{i}}{\text{+ m}}_{\text{i}}{\text{V\hspace{0.33em}Where p}}_{\text{i}}{\text{\hspace{0.33em}is the momentum of the i}}^{\text{th}}\text{particle}\\ \text{(of mass mi)\hspace{0.33em} and p}{‘}_{\text{i}}={\text{m}}_{\text{i}}{\text{v’}}_{\text{i}}\text{. Note v}{‘}_{\text{i}}{\text{\hspace{0.33em}is the velocity of the i}}^{\text{th}}\text{\hspace{0.33em}particle}\\ \text{relative to the centre of mass.\hspace{0.33em}Also, prove using the definition of\hspace{0.33em}the}\\ \text{centre of mass}\sum _{}^{}\text{p}{‘}_{\text{i}}=\text{0}\\ \left(\text{b}\right)\text{Show}\mathrm{K}={\mathrm{K}}^{‘}\text{\hspace{0.17em}+}\frac{\text{1}}{\text{2}}{\text{MV}}^{\text{2}}\text{\hspace{0.33em}Where K is the total kinetic energy of the}\\ \text{system\hspace{0.33em}of particles,}{\mathrm{K}}^{‘}\text{\hspace{0.33em}is the\hspace{0.33em}total kinetic\hspace{0.33em}energy of the system when}\\ \text{the particle velocities are taken\hspace{0.33em}with respect to the centre of mass}\\ \text{and}\frac{{\text{MV}}^{\text{2}}}{\text{2}}\text{is the kinetic energy of the translation of the system as a\hspace{0.33em}}\\ \text{whole (i.e. of the centre of mass motion of the\hspace{0.33em}system).The result}\\ \text{has been used in\hspace{0.33em}Sec. 7.14.}\\ \left(\text{c}\right)\text{Show L}=\text{L}‘+\text{R}\times \text{MV Where L}‘=\sum \text{r}{’}_{\text{i}}\times \text{p}{’}_{\text{i}}\text{is the angular momentum}\\ \text{of the system about the centre of\hspace{0.33em}mass with velocities taken relative to the}\\ \text{centre of mass. Remember r}{’}_{\text{i}}={\text{r}}_i–\text{R};\text{rest of the notation is the standard}\\ \text{notation used in the chapter. Note L}‘\text{and MR × V can\hspace{0.33em}be said to be angular}\\ \text{momenta, respectively, about and of the centre of mass of the system of particles.}\\ \left(\text{d}\right)\text{Show}\frac{\mathrm{dL}‘}{\mathrm{dt}}=\sum \text{r}{’}_{\text{i}}\mathrm{x}\frac{\mathrm{dp}‘}{\mathrm{dt}}\text{Further, show that\hspace{0.33em}where}\frac{\mathrm{dL}‘}{\mathrm{dt}}=\mathrm{\tau }{‘}_{\text{ext}}\text{is the sum}\\ \text{of all external torques\hspace{0.33em}acting on the system about the centre of mass.}\\ \text{(Hint: Use the definition of centre of mass and Newton’s Third Law.}\\ \text{Assume the\hspace{0.33em}internal forces between any two particles act along the}\\ \text{line joining the particles.)}\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Let\hspace{0.17em}us\hspace{0.17em}consider a system of i moving particles.}\\ {\text{Here,\hspace{0.17em}mass of the i}}^{\text{th}}\text{particle}={\text{m}}_{\text{i}}\\ {\text{Velocity of the i}}^{\text{th}}\text{particle}={\text{v}}_{\text{i}}\\ \therefore {\text{Momentum of the i}}^{\text{th}}\text{particle,}{\overrightarrow{\mathrm{p}}}_{\text{i}}={\text{m}}_{\text{i}}{\overrightarrow{\mathrm{v}}}_{\text{i}}\\ \text{Velocity of the CM}=\overrightarrow{\mathrm{V}}\\ {\text{Velocity of the i}}^{\text{th}}\text{particle w.r.t. the CM of the system is\hspace{0.17em}given by\hspace{0.17em}the\hspace{0.17em}relation:}\\ \overrightarrow{\mathrm{v}}{’}_{\text{i}}={\overrightarrow{\mathrm{v}}}_{\text{i}}–\overrightarrow{\mathrm{V}}\dots \left(\text{1}\right)\\ \text{By\hspace{0.17em}multiplying equation\hspace{0.17em}}\left(\text{1}\right){\text{\hspace{0.17em}with\hspace{0.17em}m}}_{\text{i}}\text{, we obtain:}\\ {\text{m}}_{\text{i}}\overrightarrow{\mathrm{v}}{’}_{\text{i}}={\text{m}}_{\text{i}}{\overrightarrow{\mathrm{v}}}_{\text{i}}–{\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ \overrightarrow{\mathrm{p}}{’}_{\text{i}}={\overrightarrow{\mathrm{p}}}_{\text{i}}–{\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ \text{Here},\\ {\overrightarrow{\mathrm{p}}}_{\text{i}}’={\text{m}}_{\text{i}}{\overrightarrow{\mathrm{v}}}_{\text{i}}’\\ ={\text{Momentum of the i}}^{\text{th}}\text{particle w.r.t.\hspace{0.17em}the CM of the\hspace{0.17em}system}\\ \therefore {\overrightarrow{\mathrm{p}}}_{\text{i}}=\overrightarrow{\mathrm{p}}{’}_{\text{i}}+{\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ \text{From the relation:}\\ \overrightarrow{\mathrm{p}}{’}_{\text{i}}={\text{m}}_{\text{i}}{\overrightarrow{\mathrm{v}}}_{\text{i}}’\\ \text{Summation of momentum of all the particles\hspace{0.17em}w.r.t. the CM of the system\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \sum _{\mathrm{i}}\overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}=\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\overrightarrow{\mathrm{v}}{‘}_{\mathrm{i}}\\ =\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\frac{\mathrm{d}\overrightarrow{{\mathrm{r}}_{\mathrm{i}}^{‘}}}{\mathrm{dt}}\\ \mathrm{Here},\overrightarrow{{\mathrm{r}}_{\mathrm{i}}^{‘}}=\text{Position\hspace{0.17em}vector\hspace{0.17em}of\hspace{0.17em}}{\mathrm{i}}^{\mathrm{th}}\text{\hspace{0.17em}particle\hspace{0.17em}w.r.t.\hspace{0.17em}the\hspace{0.17em}CM}\\ \overrightarrow{\mathrm{v}}{‘}_{\mathrm{i}}=\frac{\mathrm{d}\overrightarrow{{\mathrm{r}}_{\mathrm{i}}^{‘}}}{\mathrm{dt}}\\ \text{According\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}definition\hspace{0.17em}of\hspace{0.17em}CM,\hspace{0.17em}we\hspace{0.17em}have}\\ \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\overrightarrow{\mathrm{r}}{‘}_{\mathrm{i}}=0\\ \therefore \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\frac{\mathrm{d}\overrightarrow{{\mathrm{r}}_{\mathrm{i}}^{‘}}}{\mathrm{dt}}=0\\ \sum _{\mathrm{i}}\overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}=0\\ \left(\text{b}\right){\text{As the relation for velocity of the i}}^{\text{th}}\text{particle\hspace{0.17em}is given\hspace{0.17em}as:}\\ {\overrightarrow{\mathrm{v}}}_{\text{i}}=\overrightarrow{\mathrm{v}}{’}_{\text{i}}+\overrightarrow{\mathrm{V}}\\ \therefore \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\overrightarrow{\mathrm{v}}}_{\mathrm{i}}=\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\overrightarrow{\mathrm{v}}{‘}_{\mathrm{i}}+\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\overrightarrow{\mathrm{V}}\dots \left(\text{2}\right)\\ \text{By\hspace{0.17em}taking the dot product of equation}\left(\text{2}\right)\text{with itself, we\hspace{0.17em}obtain:}\\ \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\overrightarrow{\mathrm{v}}}_{\mathrm{i}}.\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\overrightarrow{\mathrm{v}}}_{\mathrm{i}}=\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}(\overrightarrow{\mathrm{v}}{‘}_{\mathrm{i}}+\overrightarrow{\mathrm{V}}).\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}(\overrightarrow{\mathrm{v}}{‘}_{\mathrm{i}}+\overrightarrow{\mathrm{V}})\\ {\mathrm{M}}^2\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^2={\mathrm{M}}^2\sum _{\mathrm{i}}\mathrm{v}{‘}_{\mathrm{i}}^2+{\mathrm{M}}^2\sum _{\mathrm{i}}{\overrightarrow{\mathrm{v}}}_{\mathrm{i}}.\overrightarrow{\mathrm{v}}{‘}_{\mathrm{i}}+{\mathrm{M}}^2\sum _{\mathrm{i}}{\overrightarrow{\mathrm{v}‘}}_{\mathrm{i}}{\overrightarrow{\mathrm{v}}}_{\mathrm{i}}+{\mathrm{M}}^2{\mathrm{V}}^2\\ \text{In\hspace{0.17em}case\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}CM\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}system\hspace{0.17em}of\hspace{0.17em}particles,\hspace{0.17em}we\hspace{0.17em}have:}\\ \sum _{\mathrm{i}}{\overrightarrow{\mathrm{v}}}_{\mathrm{i}}.\overrightarrow{\mathrm{v}}{‘}_{\mathrm{i}}=-\sum _{\mathrm{i}}{\overrightarrow{\mathrm{v}‘}}_{\mathrm{i}}{\overrightarrow{\mathrm{v}}}_{\mathrm{i}}\end{array}$ $\begin{array}{l}{\mathrm{M}}^2\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^2={\mathrm{M}}^2\sum _{\mathrm{i}}\mathrm{v}{‘}_{\mathrm{i}}^2+{\mathrm{M}}^2{\mathrm{V}}^2\\ \frac{1}{2}\mathrm{M}\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^2=\frac{1}{2}\mathrm{M}\sum _{\mathrm{i}}\mathrm{v}{‘}_{\mathrm{i}}^2+\frac{1}{2}{\mathrm{MV}}^2\\ \mathrm{K}=\mathrm{K}‘+\frac{1}{2}{\mathrm{MV}}^2\\ \text{Here,\hspace{0.17em}}\\ \mathrm{K}=\frac{1}{2}\mathrm{M}\sum _{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}^2\\ =\text{Total\hspace{0.17em}K.E.\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}system\hspace{0.17em}of\hspace{0.17em}particles}\\ \mathrm{K}‘=\frac{1}{2}\mathrm{M}\sum _{\mathrm{i}}\mathrm{v}{‘}_{\mathrm{i}}^2\\ =\text{Total\hspace{0.17em}K.E.\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}system\hspace{0.17em}of\hspace{0.17em}particles\hspace{0.17em}w.r.t. the\hspace{0.17em}CM}\\ \frac{1}{2}{\mathrm{MV}}^2=\text{K.E.\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}translation\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}system\hspace{0.17em}w.r.t.\hspace{0.17em}the\hspace{0.17em}CM}\\ \left(\text{c}\right){\text{Position vector of the i}}^{\text{th}}\text{particle w.r.t. origin}={\overrightarrow{\mathrm{r}}}_{\text{i}}\\ {\text{Position vector of the i}}^{\text{th}}\text{particle w.r.t. the CM}=\overrightarrow{\mathrm{r}}{’}_{\text{i}}\\ \text{Position vector of the CM w.r.t.\hspace{0.17em}origin}=\overrightarrow{\mathrm{R}}\\ \text{Given that,}\\ \overrightarrow{\mathrm{r}}{’}_{\text{i}}={\overrightarrow{\mathrm{r}}}_{\text{i}}–\overrightarrow{\mathrm{R}}\\ {\overrightarrow{\mathrm{r}}}_{\text{i}}=\overrightarrow{\mathrm{r}}{’}_{\text{i}}+\overrightarrow{\mathrm{R}}\\ \text{From part}\left(\text{a}\right)\text{,\hspace{0.17em}we\hspace{0.17em}obtain:}\\ {\overrightarrow{\mathrm{p}}}_{\text{i}}=\overrightarrow{\mathrm{p}}{’}_{\text{i}}+{\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ \text{Taking the cross product of the\hspace{0.17em}above equation by}{\overrightarrow{\mathrm{r}}}_{\text{i}},\\ \text{we obtain:}\\ \sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}}}_{\text{i}}\times {\overrightarrow{\mathrm{p}}}_{\text{i}}=\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}}}_{\text{i}}\times \overrightarrow{\mathrm{p}}{’}_{\text{i}}+\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}}}_{\text{i}}\times {\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ \overrightarrow{\mathrm{L}}=\sum _{\mathrm{i}}({\overrightarrow{\mathrm{r}‘}}_{\text{i}}+\overrightarrow{\mathrm{R}})\times \overrightarrow{\mathrm{p}}{’}_{\text{i}}+\sum _{\mathrm{i}}({\overrightarrow{\mathrm{r}‘}}_{\text{i}}+\overrightarrow{\mathrm{R}})\times {\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ =\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times \overrightarrow{\mathrm{p}}{’}_{\text{i}}+\sum _{\mathrm{i}}\overrightarrow{\mathrm{R}}\times \overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}+\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times {\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}+\sum _{\mathrm{i}}\overrightarrow{\mathrm{R}}\times {\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ =\overrightarrow{\mathrm{L}‘}+\sum _{\mathrm{i}}\overrightarrow{\mathrm{R}}\times \overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}+\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times {\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}+\sum _{\mathrm{i}}\overrightarrow{\mathrm{R}}\times {\text{m}}_{\text{i}}\overrightarrow{\mathrm{V}}\\ \text{Here,\hspace{0.17em}}\overrightarrow{\mathrm{R}}\times \sum _{\mathrm{i}}\overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}=0,\\ \left(\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\right)\times \mathrm{M}\overrightarrow{\mathrm{V}}=0,\\ \sum _{\mathrm{i}}{\text{m}}_{\text{i}}=\mathrm{M},\\ \therefore \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{L}‘}+\mathrm{R}\times \mathrm{M}\overrightarrow{\mathrm{V}}\\ \left(\text{d}\right)\text{We have,}\\ \overrightarrow{\mathrm{L}‘}=\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times \overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}\\ \frac{\mathrm{d}\overrightarrow{\mathrm{L}‘}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times \overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}})\end{array}$ $\begin{array}{l}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\right)\times \overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}+\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times \frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}\right)\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(\sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\right)\times \overrightarrow{{\mathrm{v}}_{\mathrm{i}}^{‘}}+\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times \frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}\right)\\ \text{Here,\hspace{0.17em}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}=\text{Position\hspace{0.17em}vector\hspace{0.17em}w.r.t.\hspace{0.17em}the\hspace{0.17em}CM\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}system\hspace{0.17em}}\\ \text{of\hspace{0.17em}particles.}\\ \therefore \sum _{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}=0\\ \therefore \frac{\mathrm{d}\overrightarrow{\mathrm{L}‘}}{\mathrm{dt}}=\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times \frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}\right)\\ \text{From the relation:}\\ \frac{\mathrm{d}\overrightarrow{\mathrm{L}‘}}{\mathrm{dt}}=\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times \frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{p}}_{\mathrm{i}}^{‘}}\right)\\ =\sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times {\mathrm{m}}_{\mathrm{i}}\frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{v}}_{\mathrm{i}}^{‘}}\right)\\ \text{Here,\hspace{0.17em}}\frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{v}}_{\mathrm{i}}^{‘}}\right)=\text{\hspace{0.17em}Rate\hspace{0.17em}of\hspace{0.17em}change\hspace{0.17em}of\hspace{0.17em}velocity\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}}{\mathrm{i}}^{\mathrm{th}}\text{particle\hspace{0.17em}w.r.t.\hspace{0.17em}the\hspace{0.17em}CM\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}system.}\\ \therefore \text{As\hspace{0.17em}per\hspace{0.17em}Newton’s\hspace{0.17em}third\hspace{0.17em}law\hspace{0.17em}of\hspace{0.17em}motion,\hspace{0.17em}we\hspace{0.17em}have:}\\ {\mathrm{m}}_{\mathrm{i}}\frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{v}}_{\mathrm{i}}^{‘}}\right)=\text{External\hspace{0.17em}force\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}}{\mathrm{i}}^{\mathrm{th}}\text{\hspace{0.17em}particle}=\sum _{\mathrm{i}}{\left(\overrightarrow{{\mathrm{\tau }}_{\mathrm{i}}‘}\right)}_{\mathrm{ext}}\\ \therefore \sum _{\mathrm{i}}{\overrightarrow{\mathrm{r}‘}}_{\text{i}}\times {\mathrm{m}}_{\mathrm{i}}\frac{\mathrm{d}}{\mathrm{dt}}\left(\overrightarrow{{\mathrm{v}}_{\mathrm{i}}^{‘}}\right)={\overrightarrow{\mathrm{\tau }‘}}_{\mathrm{ext}}\\ =\text{External\hspace{0.17em}torque\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}system}\\ \therefore \frac{\mathrm{d}\overrightarrow{\mathrm{L}‘}}{\mathrm{dt}}={\overrightarrow{\mathrm{\tau }‘}}_{\mathrm{ext}}\end{array}$