Uniform Circular Motion Formula
Uniform Circular Motion Formula
A change in velocity direction is required for an object to travel in a curved circular path. It is because the tangent will provide the direction at each point along the circular path. The acceleration that results from a change in velocity will not be in the same direction as the velocity. Therefore, an acceleration that is always perpendicular to the velocity is required for an object to move in a circular path. The circular motion could either be uniform or non-uniform.
Uniform Circular Motion
Concept of Uniform Circular Motion:
The magnitude of the velocity will always be constant in a uniform circular motion. But from every point, the velocity’s direction will vary steadily. It implies that the object will travel in a circle. And the object will finish each of its numerous trips around the path in exactly the same amount of time.
The motion along a curved path can be described as circular. Uniform circular motion is the movement of any object along a circular path that travels the same distance around the circumference in the same amount of time. Any such motion has a constant speed and a continuously changing direction.
In a uniform circular motion, the tangential speed will be constant at every point along the circumference. Every point along the circumference of this tangential velocity vector is tangent.
Additionally, the acceleration vector is always pointed in the direction of the centre of the circle the object’s motion creates. Due to its distance from the center, this acceleration is either referred to as “radial acceleration” or “centripetal acceleration,” which denotes that it is “center seeking.”
The formula for Uniform Circular Motion
Numerous Indian educational boards include the Uniform Circular Motion Formula in their physics curricula. It is also covered in several Physics classes’ curricula. You can learn the Uniform Circular Motion Formula with the aid of textbooks. The recommended textbooks are the easiest for students to use as learning resources for the uniform circular motion formula.
Numerous questions from the Uniform Circular Motion Formula that may come up on exams and tests are also included in the textbook. As a result, textbooks become students’ preferred method of learning. However, some of the textbook questions might be difficult and perplexing. This does not imply that students should feel constrained by the answers to these questions. To supplement their education and prepare for exams, students may choose to learn about the Uniform Circular Motion Formula from other sources.
On the Extramarks platform, students can practise answering numerous questions. For students who don’t want to learn about Uniform Circular Motion Formula exclusively from textbooks, the platform provides a variety of learning options. They will therefore receive additional learning support as they get ready for their exams. Students who want to understand the Uniform Circular Motion Formula and its use can access the solved question papers. Understanding all of the aspects of the Uniform Circular Motion Formula topic can be difficult. Students should comprehend the material in the textbook in order to fully understand the Uniform Circular Motion Formula.
In higher classes, however, relying solely on the textbooks is insufficient. To continue learning, students must use a variety of sources. The Extramarks platform can be used to learn the Uniform Circular Motion Formula and other concepts. Students have access to all the necessary learning modules based on the Uniform Circular Motion Formula through this platform. Students gain confidence as a result, and it also helps them prepare the Uniform Circular Motion Formula for exams and tests. The combination of the textbooks and learning materials from the Extramarks make for a good and complete source of learning for students. Even teachers can refer to the curriculum on the website.
Solved Examples
Q.1: A player is moving with a constant tangential speed of 50 m per second. He takes one lap around a circular track in 40 seconds. Calculate the magnitude of the acceleration of the player.
Solution:
Given parameters:
The magnitude of Velocity, V = 50 m per second
Time period, T = 40 seconds.
We know that,
V=2πRT
Therefore,R=T×V2π
Putting values,
R=40×502×3.14
R = 31.84
Now,
arad=4π2RT2putR=T×V2πwe get,
= 4π2RT2
= 4π2×T×V2πT2
= 2×π×VT
= 2×3.14×5040
= 7.86
Acceleration will be 7.86ms−2
