Answer to Question #158056 in Classical Mechanics for Light Fury

Question

A particle of mass m is attached to the end of a string and moves in a circle of radius of r on a frictionless horizontal table. The string passes through a frictionless hole in the table and, initially, the other end id fixed.

a) if the string is pulled so that the radius of the circular orbit decreases, how does the angular velocity change if it is ω₀ when r = r₀?

b) what work is done when the particle is pulled slowly in from a radius r₀ to a radius r₀/2?

Answer 1

(a) The law of conservation of an angular momentum says

$$I_0\omega_0=I\omega\\ mr_0^2\omega_0=mr^2\omega\\ r_0^2\omega_0=r^2\omega$$

If the radius of the circular orbit decreases, then angular velosity increases.

(b) The work done

$$W=\Delta K=\frac{I\omega^2}{2}-\frac{I_0\omega_0^2}{2}$$

$$W=\frac{m(r_0/2)^2(4\omega_0)^2}{2}-\frac{m r_0 ^2 \omega_0 ^2}{2}=\frac{3m r_0 ^2 \omega_0 ^2}{2}$$