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kpg0001
Mar 29, 2011, 03:49 PM
So there is a rod standing straight up that gets hit with a ball of clay that sticks to it.

m=mass of rod
m/2=mass of clay
v=speed of clay
L=length of rod
3L/4=point where clay hits rod
I=(mL^2)/3=rotational moment of inertia of the rod about its end
g=gravity
point 0=center of mass of rod

1)Angular momentum of clay about point O

2)Angular momentum of clay and rod about point O

3)Moment of inertia of clay and rod after collision

4)Angular speed of the rod and clay about point O when the rod is vertically downward

5)Change in gravitational potential energy of the rod and clay as the rod goes from vertically upward to vertically downward

I have the answers but I don't know how to get them. Any help is appreciated.

Unknown008
Mar 29, 2011, 10:31 PM
What is your attempt at the problem? I think you have all the necessary information/formulae to work this out.

kpg0001
Mar 29, 2011, 11:48 PM
For 1 and 2 I have (3/8)mvL but I am stuck on 3.

The answer to 3 is supposed to be (59/96)mL^2.

I am trying to use the parallel axis theorem I=I(center of mass)+Md^2.

d=distance from center of mass

but I don't believe I understand it correctly or if that is even the right way to go with it.

Again, help would be appreciated. I'm not looking for quick answers to homework(that's what cramster is for), I'm trying to understand the information.

kpg0001
Mar 30, 2011, 05:04 AM
Yeah I ended up figuring out 3 using the parallel axis theorem.

Unknown008
Mar 31, 2011, 07:14 AM
4. You have the moment of inertia of the rod and clay, and the angular momentum of the rod and clay. You should have a formula relating angular momentum, inertia and angular speed.

5. Now that you have the angular speed, find the amount of rotational kinetic energy. I think that from there you can get the potential energy.

But then, be aware that I never worked/learned about angular momentum or rotational kinetic energy.