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.