Question

A mass m moves in a circle on a smooth horizontal plane with velocity $ v_0 $ at a radius $ R_0. $ The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius $ \frac{R_0}{2}. $ The final value of the kinetic energy is

• A. $ 2 m v_0^2 $
• B. $ \frac{1}{2} m v_0^2 $
• C. $ m v_0^2 $
• D. $ \frac {1}{4} m v_0^2 $

Answer 1

The Correct Option is A

To solve this problem, let's analyze the situation carefully and use the conservation of angular momentum and the kinetic energy formula.

Initially, the mass $ m $ moves in a circle of radius $ R_0 $ with velocity $ v_0 $. Hence, the initial angular momentum is given by:

L_i = m \cdot v_0 \cdot R_0

When the radius reduces to $ \frac{R_0}{2} $, due to the conservation of angular momentum (since there is no external torque), the final angular momentum $ L_f $ should equal the initial angular momentum $ L_i $:

L_f = m \cdot v_f \cdot \frac{R_0}{2}

Equating the two:

m \cdot v_0 \cdot R_0 = m \cdot v_f \cdot \frac{R_0}{2}

Solving for $ v_f $:

v_f = 2 \cdot v_0

Next, let's calculate the final kinetic energy. Initially, the kinetic energy $ K_i $ is:

K_i = \frac{1}{2} m v_0^2

The final kinetic energy $ K_f $ when the radius is $ \frac{R_0}{2} $ is given by:

K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (2 v_0)^2 = \frac{1}{2} m \cdot 4v_0^2 = 2m v_0^2

Therefore, the final value of the kinetic energy is $ 2 m v_0^2 $.

Thus, the correct answer is: $ 2 m v_0^2 $.