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Kelvi

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Kelvi

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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

medium

Work, Energy and Power

2015

physics

$\frac{1}{4} m v_0^2$

$2 m v_0^2$

Explanation

To solve this problem, we will use the principle of conservation of angular momentum. $\\$ Given: $\\$ • Initial velocity $v_0$ at radius $R_0. \\$

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$\frac{1}{2} m v_0^2$

$m v_0^2$

• Final radius

R = \frac{R_0}{2}. \\

The angular momentum is conserved, so:

\\ m v_0 R_0 = m v R \\

Substitute the final radius:

\\ m v_0 R_0 = m v \left(\frac{R_0}{2}\right) \\

Simplify to find the final velocity

v: \\ v_0 R_0 = v \frac{R_0}{2} \\ v = 2 v_0 \\

The initial kinetic energy

K_i

is:

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

The final kinetic energy

K_f

is:

\\ K_f = \frac{1}{2} m v^2 \\

Substitute the final velocity:

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

Therefore, the final value of the kinetic energy is

2 m v_0^2. \\

This corresponds to Option 2.