Conservation of energy/Angular momentum for elastic collison

[Oliver Legote]

Homework Statement

A uniform thin rod of Length L and mass M can freely rotate about a point 0 and is at rest in at the vertical. A ball of mass m on a light string of length R, which is also attached about the pivot is deflected by a small angle from the vertical and let go of.

If the collision is totally elastic find the length of the string such that after the collision the ball remains at rest in the vertical

Homework Equations

Conservation of Angular momentum; I₁ϖ₁ = I₂ϖ₂
Conservation of Energy; U1+K1= U2+K2
Ek= 1/2 mv^2
GPE= MGH

The Attempt at a Solution

I attempted to find the GPE of the ball on the string at rest when it is deflected, using V=√(2gh) where H was defined as the change in the y-coordinate of the ball's center of mass: R(1-cosα) where I defined α as the small deflection that was not named.

I then tried to used the moment of inertia of a Rod passing through the end as (1/3) M L².

As the collision is elastic Kinetic energy is conserved, however I cannot for the life of me get an equation in which I can isolate R in terms of L. I can't seem to find the angular speed of the rod after it is struck, which I feel is the way to go.

Any advice would be greatly appreciated, cheers!

 

[haruspex]

V=√(2gh)

You do not need to worry about the starting height. Let the ball has some angular velocity ω_i just before the collision.
Invent other variables as necessary.
What equations can you write?

 

[Oliver Legote]

You do not need to worry about the starting height. Let the ball has some angular velocity ω_i just before the collision.
Invent other variables as necessary.
What equations can you write?

If I let the ball have a given angular velocity I can write conservation of angular momentum? I can give the Rod an angular velocity after the collision of ω_f.

Kinetic energy is conserved so ½mv_i²= ½Mv_f²; where Where v=ωr so;
½mω_i^2 R^2 = ½M ω_f^2 L^2

I could use angular momentum of the ball = mvr? Where v=ωr so;

mωr²= 1/3ML²ω_f?

I'm still a bit lost as to how to proceed. I'm still not sure how I can isolate R in terms of L with what I have? Am I missing a variable here?

 

[haruspex]

You have the KE of the rod wrong. You need to consider its rotational energy.
When you have that corrected you will have two equations. What variable in them do you need to eliminate?

 

[Oliver Legote]

So I have the Ek's as equal:

Eqn1: 1/6 ML²ω_F²= 1/2mω_i²R²

Equation 2 is the conservation of angular momentum;
Eqn 2: mω²_iR^2 = 1/3ML²ω_f

Eliminating ω_f I get

mR^2 = 1/3 ML^2EDIT: So final answer as R= √(1/3)M/m *L


Which is still leaving me with the length in terms of the masses, is this okay? Or is my physics still incorrect?

Thanks :)

 

[haruspex]

So I have the Ek's as equal:

Eqn1: 1/6 ML²ω_F²= 1/2mω_i²R²

Equation 2 is the conservation of angular momentum;
Eqn 2: mω²_iR^2 = 1/3ML²ω_f

Eliminating ω_f I get

mR^2 = 1/3 ML^2EDIT: So final answer as R= √(1/3)M/m *L


Which is still leaving me with the length in terms of the masses, is this okay? Or is my physics still incorrect?

Thanks :)

That all looks right.
The question specifies the variables m, M and L, so you should expect all of those to turn up in the answer.