Answer to Question #157300 in Classical Mechanics for mehrab hossain

Question

A particle of mass m is attached to the end of a string and moves in a circle of radius of

radius 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) According to law of conservation of angular momentum:

$$I_0\omega_0=I\omega,\\ mr_0^2\omega_0=mr^2\omega,\\ r_0^2\omega_0=r^2\omega,\\\space\\ \omega(r)=\omega_0\cdot\bigg(\frac{r_0}{r}\bigg)^2.$$

As we see, as the radius decreases, the angular velocity increases.

b) The work is force times distance integrated. The force changes because the centripetal acceleration changes:

$$a(r)=\omega^2r=\frac{\omega_0^2r_0^4}{r^3},\\\space\\ F(r)=ma=\frac{m\omega_0^2r_0^4}{r^3}.$$

The work, thus, is

$$W=\int^{r_0}_{r_0/2}F(r)\text{d}r=m\omega_0^2r_0^4\int^{r_0}_{r_0/2}\frac{\text{d}r}{r^3}=\\\space\\ =m\omega_0^2r_0^4\cdot\frac{3}{2r_0^2}=\frac32m\omega_0^2r_0^2.$$