# Calculating Rotational Kinetic Energy of a Sphere at the End of a Rod (HW 9 Question 5)

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A thin, cylindrical rod = 20.6 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge.  After the combination rotates through 90 degrees, what is its rotational kinetic energy?

Set the vertically standing to be initial position and horizontal to be the final position. Use energy conservation. Consider the change of potential energy: For the ball, its center of mass falls $l+\frac d 2$. For the rod, its center of mass falls $\frac l 2$. All these changes goes into the rotational kinetic energy.

Set y=0 at the horizontal level and use $U_1+K_{r1}+K{t1}=U_2+K_{r2}+K_{t2}$.

$$\therefore Mg(L+\frac d 2)+mg\frac {l}{2}=\frac 1 2 I_{total}\omega^2.$$

For parts 2 and 3, you will need to calculate $I_{total}$. Need to use parallel axis theorem to find the ball's moment of inertia. $$I_{total}=\frac 2 5 MR^2+M(l+\frac d 2)^2+\frac {1} {3} ml^2.$$

answered Oct 20, 2013 by (21,750 points)