(adsbygoogle = window.adsbygoogle || []).push({}); 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

[tex]E[/tex] [tex]\frac{mx²}{mi}[/tex] (Center of Mass)

[tex]E \tau[/tex] = Ia (Torque)

K_{R}= [tex]\frac{1}{2}[/tex] 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 ><;.

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# Homework Help: Parallel Axis Theorem/Center of Mass

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