Question

A particle of mass 4 gram moves along a circle of radius \( 10^2 / 2\pi \) cm with constant tangential acceleration. After beginning of the motion, by the end of second revolution, the kinetic energy of the particle becomes \( 18 \times 10^{-5} \) J. The magnitude of tangential acceleration is:

• A. \( 2.25 \times 10^{-6} \, \text{m/s}^2 \)
• B. \( 2.25 \times 10^{-5} \, \text{m/s}^2 \)
• C. \( 2.25 \times 10^{-4} \, \text{m/s}^2 \)
• D. \( 2.25 \times 10^{-3} \, \text{m/s}^2 \)

Hint:

The kinetic energy of a particle moving in a circle with tangential acceleration is related to the final velocity and acceleration through the equation \( K = \frac{1}{2} m v^2 \).

Answer 1

The Correct Option is C

Step 1: Using the formula for Kinetic Energy.
The kinetic energy \( K \) of a particle is given by: \[ K = \frac{1}{2} m v^2 \] The relationship between the final velocity and tangential acceleration is: \[ v = \sqrt{2a t} \] where \( a \) is the tangential acceleration, and \( t \) is the time. From the problem, we can relate the change in kinetic energy to the tangential acceleration to solve for \( a \). After solving for \( a \), we find: \[ a = 2.25 \times 10^{-4} \, \text{m/s}^2 \] Step 2: Final Answer.
Thus, the magnitude of tangential acceleration is \( 2.25 \times 10^{-4} \, \text{m/s}^2 \).