Parallel Axis Theorem/Center of Mass

1. Jan 12, 2009

Omniscient

1. The problem statement, all variables and given/known data
A cylindrical rod 24.0 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball with a diameter of 8.00 cm and a mass of 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the bottom end of the rod. (a) After it falls through 90°, what is its rotational kinetic energy? (b) What is the angular speed of the rod and ball? (c) What is the linear speed of the ball? (d) How does this compare with the speed if the ball had fallen freely through the same distance of 28 cm?

2. Relevant equations
$$E$$ $$\frac{mx²}{mi}$$ (Center of Mass)

$$E \tau$$ = Ia (Torque)

KR = $$\frac{1}{2}$$ I w 2 (Rotational Kinetic Energy)

3. The attempt at a solution

Alright. So I have the main idea that you need to get the center of mass (which I'm at a loss to get -> I know you've gotta integrate with respect to dm, and probably using... volume? As a u substitute?). And then from there, you can get the potential energy, and set it equal to the rotational kinetic energy at 90°. But the thing is, once the CM of is calculated, how can you get the moment of inertia? It doesn't go through the center of mass, so parallel axis theorem needs to be used.... Somewhere? Once (a) is figured out, it's pretty much plug and chug with standard rotational kinematics and their analogs ><;.

Last edited: Jan 12, 2009
2. Jan 13, 2009

chrisk

Use the parallel axis theorem. First, find the center of mass of the system. This can be done by locating the center of mass of the rod and the ball separately; symmetry gives these locations to be at the geometric center of each. These locations can be treated as point masses. Then find the center of mass of these two point masses.

CM = (XrMr + XbMb)/(Mr + Mb)

where Xr is the center of mass coordinate of the rod, Xb the center of mass coordinate of the ball, Mr the mass of the rod, and Mb the mass of the ball.

No integration is required. The system center of mass is now the axis of rotation to utilize the parallel axis theorem, so find the moments of inertia for the rod alone and the ball alone about the center of mass axis (integation is required for the rod but not for the ball because it can be treated as a point mass at the center of gravity). Add these two moments to find the total moment of inertia of the system about the center of mass. The axis of rotation in the problem is at the end of the rod. Use the parallel axis theorem for this to give the moment of inertia about the given axis of rotation. Once the moment of inertia is known, you can figure the rest out using the rotational kinematic equations and energy equations.