(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The rigid body assembly is made from a 15 Kg disk, an 8 Kg slender rod, and a small 4 Kg sphere. The radius of the disk is R = 0.4 m and the length of the rod is L = 1.2 m. The pivot O is at the center of the disk. The assembly is released from rest at the horizontal position shown. Find the angular velocity of the assembly when it rotates 90° down to the vertical position.

http://img269.imageshack.us/img269/4544/problemtp.png [Broken]

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2. Relevant equations

COE: K1+UG1= K2+UG2

Inertia equations for particle, rod, and disk.

3. The attempt at a solution

I Found Inertia for disk, rod, and sphere seperatedly, with I disk = 1.2, I rod = 1.28 , I sphere= 3.2

Then I use reference point as the horizontal line it started, therefore UG1 are all zero. So my energy equation look like:

0= K2+ UG2

0= 1/2 (I)w^2 + UG2rod + UG2sphere

0= 1/2 (5.68) w^2 + (-0.2) (8) (g) + (-0.8) (4) (g)

I got 4.07 for w, angular velocity.

The answer is 4.32. I dont see what I am doing wrong. I did Irod= Icm + md^2 for rod inertia since its parallel axis theorem.

Thanks

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# Homework Help: Rigid body dynamics question

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