Question

A thin,cylindrical rod l=24.0 cm long with mass m=1.2 kg has a ball of diameter d=8 cm and mass M=2 kg attached to one end. The arrangement is originally vertical and stationary,with the ball at the top. The combination is free to pivot about the bottom end of the rod after being given a slight nudge. (a) After the
combination rotates through 90°,what is its rotational energy?(b) What is the
angular speed of the rod and ball?(c) What is the linear speed of the center of mass of the ball? (d) How does compare with the speed had the ball fallen freely through the same distance of 28 cm?

Answer 1

Moment of inertia of the rod:

$$I_1=\frac13mL^2.$$

Moment of inertia of the ball:

$$I_2=\frac25MR^2.$$

Contribution to the moment of inertia of the ball due to the fact it is on distance from the axis of rotation:

$$I_3=M(R+L)^2.$$

Total moment of inertia:

$$I=I_1+I_2+I_3=\\\space\\ =\frac13mL^2+M\bigg(\frac25R^2+(R+L)^2\bigg).$$

(a) According to energy conservation, initially, the potential energy is maximum, at 90° the potential energy is 0 being fully converted into kinetic energy:

$$PE=KE,\\\space\\ g\bigg(\frac L2m+\bigg(L+R\bigg)M\bigg)=\frac12I\omega^2=6.90\text{ J}.$$

(b) From the previous equation express the angular velocity:

$$\omega=\sqrt{\frac{g(Lm+2(L+R)M)}{I}},\\\space\\ \omega=\sqrt{\frac{g(Lm+2(L+R)M)}{\frac13mL^2+M\big(\frac25R^2+(R+L)^2\big)}}=8.73\text{ rad/s}$$

(c) The linear speed is simply

$$v=\omega r=\omega(L+R)=2.44\text{ m/s}.$$

(d) If a ball falls from 28 cm, the velocity would be

$$u=\sqrt{2gh}=2.34\text{ m/s}.$$

This is

$$k=\frac{v}{u}=1.04$$

times smaller.