Question

A body of mass 500 gram is rotating in a vertical circle of radius 1 m. What is the difference in its kinetic energies at the top and the bottom of the circle?

• A. 4.9 J
• B. 19.8 J
• C. 2.8 J
• D. 9.8 J

Hint:

In vertical circular motion, kinetic energy is highest at the bottom and lowest at the top. The difference equals \( 2mgr \).

Answer 1

The Correct Option is D

Step 1: Given mass \( m = 500 \, \text{g} = 0.
5 \, \text{kg} \), radius \( r = 1 \, \text{m} \), and gravitational acceleration \( g = 9.
8 \, \text{m/s}^2 \).
Step 2: The difference in kinetic energy between the bottom and the top of the vertical circle is equal to the change in potential energy: \[ \Delta KE = m g (2r) = 0.
5 \times 9.
8 \times 2 = 9.
8 \, \text{J} \]