Question

A mass m moves in a circle on a smooth horizontal plane with velocity v₀ at a radius R₀. 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. 1/2 mv₀²
• B. mv₀²
• C. 1/4 mv₀²
• D. 2 mv₀²

Answer 1

The correct option is D

2 mv₀²

When a mass moves in a circle of radius $R_0$ with velocity $v_0$, its kinetic energy is given by

$K{E}_1=\frac{1}{2}m{v}_0^2\dots .\left(1\right)$

The centripetal force required for circular motion is

$F_c=\frac{m{v}_0^2}{R_0}\dots .\left(2\right)$

The tension in the string is gradually increased and the radius of the circle decreased to $\frac{R_0}{2}$.

When the radius of the circle is $R(R_0>R>\frac{R_0}{2})$ the tension in the string is the same as the centripetal force.

$T=F_c=\frac{m{v}^2}{R}=\frac{L^2}{m{R}^3}...\left(3\right)$

where L = mRv is the angular momentum which is conserved.

Work done in reducing the radius of the circle from $R_0$ to $\frac{R_0}{2}$ is

$W=-{\int }_{R_0}^{\frac{R_0}{2}}{F}_{c}dR={\int }_{R_0}^{\frac{R_0}{2}}\frac{L_{2}dR}{m{R}^3}=-\frac{L^2}{m}{\int }_{R_0}^{\frac{R_0}{2}}\frac{dR}{R^3}=-\frac{L^2}{m}{[-\frac{1}{2R^2}]}_{R_0}^{\frac{R_0}{2}}=-\frac{L^2}{2m}{\left[\frac{1}{R^2}\right]}_{R_0}^{\frac{R_0}{2}}=\frac{L^2}{2m}[\frac{4}{R_0^2}-\frac{1}{R_0^2}]=\frac{L^2}{2m}\frac{3}{R_0^2}=\frac{m^2{v}_0^2{R}_0^2}{2m}\frac{3}{R_0^2}=\frac{3}{2}m{v}_0^2$

Total kinetic energy = Initial kinetic energy + Work done

$=\frac{1}{2}m{v}_0^2+\frac{3}{2}m{v}_0^2=2m{v}_0^2$