Answer to Question #197199 in Physics for gibson

Question

A hoop of mass 0.1kg and radius 0.25m is rotating in a horizontal plane with angular

momentum of 4.0kgm2/s. A lump of clay of mass 0.2kg is placed (gently) on the hoop.

Show that the angular velocity of the hoop reduces

Answer 1

Let's first find the angular velocity of the hoop before the lump of clay was placed on it:

$$L_i=I_{hoop}\omega_i=m_{hoop}r^2\omega_i,$$ $$\omega_i=\dfrac{L_i}{m_{hoop}r^2}=\dfrac{4\ \dfrac{kg\cdot m^2}{s}}{0.1\ kg\cdot(0.25\ m)^2}=640\ \dfrac{rad}{s}.$$

According to the law of conservation of angular momentum, we have:

$$L_i=L_f=(I_{hoop}+I_{clay})\omega_f=(m_{hoop}r^2+m_{clay}r^2)\omega_f,$$ $$\omega_f=\dfrac{L_i}{(m_{hoop}r^2+m_{clay}r^2)},$$ $$\omega_f=\dfrac{4\ \dfrac{kg\cdot m^2}{s}}{(0.1\ kg\cdot(0.25\ m)^2+0.2\ kg\cdot(0.25\ m)^2)}=213\ \dfrac{rad}{s}.$$

As we can see from calculations, the angular velocity of the hoop reduces.