# Pivoting Rod and Ball - Rotational Dynamics

1. Nov 24, 2007

### DelPopolo

1. The problem statement, all variables and given/known data
A cylindrical rod 26.0 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 7.80 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?

2. Relevant equations
KE=1/2I$$omega$$^2
I=(2mr^2)/5
I=(2mL^2)/3

3. The attempt at a solution

Moment of inertia=I of rod + I of ball?
I=(2*m*(r+L)^2)/5)+(2mL^2)/3
I'm lost as to finding $$\omega$$ so I can find KE.

Please point me in the right direction.
Thank you

Last edited: Nov 24, 2007
2. Nov 24, 2007

### azatkgz

Moment of Inertia of a rod about pivot is

$$I_r=\frac{mL^2}{3}$$

And if you take level of pivet as zero level,after rod falls through 90°,total potential energy will be zero

so we have

$$PE_{initial}=\frac{1}{2}I_{total}\omega^2$$

3. Nov 24, 2007

### DelPopolo

I=(2*m*(r+L)^2)/5)+(mL^2)/3
I=.0985

so

KE=1/2(.0985)$$\omega$$^2

I need to find omega so I can find kinetic E. Correct?

How does one find omega? Obviously gravity affects the speed but gravity isnt the same over the entire rotation so how does that work out?

Last edited: Nov 24, 2007
4. Nov 25, 2007

### azatkgz

What's the potential energy of this system,when it's at rest?