Moment of Inertia of a turntable

[Raziel2701]

Homework Statement

A uniform circular turntable of mass 2m and radius R is at rest in space. Koko throws a lump of putty of mass m and speed v toward the edge of the turntable so that it sticks at the extreme edge of the turntable at R.

Using conservation of L, show that the angular velocity ω of the putty/turntable system after the collision is \omega=\frac{2v}{5R}

Homework Equations

Parallel-axis theorem. Conservation of angular momentum

The Attempt at a Solution

I found the center of mass of the putty/turntable system to be R/3. I set my origin to be there. For the initial angular momentum, that is before the putty sticks to the turntable, I get mv2R/3.

For the final angular momentum I get the moment of inertia times the angular velocity. My problem is in determining what the right moment of inertia is.

Why can't I just use the parallel axis theorem applied to the putty and the turntable to find the moment of inertia? I've done mR^2 for the putty, plus 4/9mR^2 by the parallel axis theorem, plus another mr^2 from the disk(since it's mass is 2, the half cancels) plus 1/9 mR^2 by the parallel axis theorem?

So why is this wrong though?

 

[vela]

Why can't I just use the parallel axis theorem applied to the putty and the turntable to find the moment of inertia? I've done mR^2 for the putty, plus 4/9mR^2 by the parallel axis theorem, plus another mr^2 from the disk(since it's mass is 2, the half cancels) plus 1/9 mR^2 by the parallel axis theorem?

So why is this wrong though?

You have the right idea but simply made a few mistakes. For the turntable, you have
$$I_1 = I_{1,\rm cm} + Md_1^2 = \frac 12 MR^2 + M(R/3)^2 = \frac{11}{18} MR^2$$ where $M$ is the mass of the turntable. Setting $M=2m$, you get $I_1 = (11/9)mR^2$. You used the wrong mass in the $md^2$ term of the parallel-axis theorem.

For the putty, as you're modeling it as a point particle, you have $I_{2,\rm cm} = 0$, not $mR^2$ as you assumed, so its contribution to the moment of inertia of the system is just $I_2 = md_2^2 = m(2R/3)^2 = (4/9) mR^2$.

The moment of inertia of the system about its center of mass is therefore $I=I_1+I_2 = (5/3)mR^2$. Using your value of $L$, you'll get the expected result for the angular velocity.