Solving Rotational Motion Problem Involving Mass and Length of Rod

[Poisonous]

I ran across this problem studying for my exam, and I'm not sure how to solve it:

Homework Statement

A thin rod of Mass M and length L is supported by a pivot a distance of L/4 from its left end. A second support a distance 3L/4 from the left end prevents it from falling. A clay ball of mass M/2 drops from a height H above the beam, strikes the left end and completely sticks to the rod. The beam swings upward to make an angle $$\Theta$$ with the horizontal.


a. Calculate the moment of Inertia, I_z, of the beam and clay ball together around the pivot point, after the clay sticks to the beam.


b. Calculate the angle $$\Theta$$ to which the beam rises. Express your answer in terms of $$\omega$$, V, and/or I_z.

The Attempt at a Solution

For a, I was thinking of finding the moment of Inertia at the new center of mass after the ball sticks, then using the parallel axis theorem to get I_z, but that seems too messy..

For b, I think there must be some way to do it with energy conservation, but I really don't know where to start.

Thanks for the help.

 

[tiny-tim]

Hi Poisonous!

(have a theta: θ and an omega: ω )

a. Calculate the moment of Inertia, I_z, of the beam and clay ball together around the pivot point, after the clay sticks to the beam.
…
For a, I was thinking of finding the moment of Inertia at the new center of mass after the ball sticks, then using the parallel axis theorem to get I_z, but that seems too messy..

That's far too complicated …

just calculate the moment of inertia of the whole (about the pivot) as the sum of the moments of inertia of its parts.

For b, I think there must be some way to do it with energy conservation, but I really don't know where to start.

How can energy be conserved? It's a perfectly inelastic collision (the two final velocities are the same).

You can use conservation of energy after the collision, but for the collision itself, you need … ?

 

[Poisonous]

Hi Poisonous!

(have a theta: θ and an omega: ω )


That's far too complicated …

just calculate the moment of inertia of the whole (about the pivot) as the sum of the moments of inertia of its parts.


How can energy be conserved? It's a perfectly inelastic collision (the two final velocities are the same).

You can use conservation of energy after the collision, but for the collision itself, you need … ?

For a: I_z = 1/12ML² + (M/2)(L/4)²

right?

For b: For the collision itself you would need conservation of angular momentum? So,

Iclay * ωclay + 0 = Iz * ωz

so, ωz = MVL/8Iz

So you know the rotational kinetic energy at the beginning, and that it should be all potential energy at the point of $$\theta$$, but how do you relate the energies with $$\theta$$?

Thanks for the help.

 

[tiny-tim]

For a: I_z = 1/12ML² + (M/2)(L/4)²

No, 1/12 would be at the end, you want L/4 from the end.

So you know the rotational kinetic energy at the beginning, and that it should be all potential energy at the point of $$\theta$$, but how do you relate the energies with $$\theta$$?

(what happened to that θ i gave you? )

You relate θ to h.