Angular velocity uniform rod problem

[Poutine]

Hi all, recently joined and having abit of trouble with a problem (several actually but I managed to figure out how to get started on one of them).

In any case the problem says:
Four thin, uniform rods each of mass M and length d = .75 m, are rigidly connected to a vertical axle to form a turnstile. The turnstile rotates clockwise about the axle, which is attached to a floor, with initial angular velocity w = -2.0 rad/s. A mud ball of mass m = M/4 and inital speed vi = 15 m/s is thrown and sticks to the end of one rod at an angle 60 degrees. Find the final angular velocity of the ball-turnstile system.

Now I think I have to use the Conservation of Angular momentum. As such I need to find the inertia of the turnstile which I think turns out to be IT=(4/3)*M*d^2

I think the formula I'm going end up with will probably be
It * wf + Ib * wf = IT* wi + angular momentum of the ball before the contact

The thing is I don't know how to get further from here.

 

[quantumdude]

I don't have a table of rotational inertias for different shapes in front of me, so I won't comment on that until tomorrow, after I look it up.

But as for where you go after that, you're basically done. You will have something like:

I_iω_i=I_fω_f

You know everything except ω_f
(Don't forget to treat the mudball as a point mass, so I_mud=mr².)

 

[Poutine]

Okay you answered my main question which is what is Imud? But I don't understand why that is true. Why is it mr^2, I was thinking it might have been 2/5 mr^2 but it seems I was wrong but I don't understand why.

Just to be certain the angular momentum before contact would be (M/4)*v*d*cos 60 right?

 

[quantumdude]

Originally posted by Poutine
Okay you answered my main question which is what is Imud? But I don't understand why that is true. Why is it mr^2, I was thinking it might have been 2/5 mr^2 but it seems I was wrong but I don't understand why.

They don't give you the radius of the ball of mud, so that tells you to treat it as a point (it must be very small compared to the length of the rods). The rotational inertia of a point is I=mr².

Just to be certain the angular momentum before contact would be (M/4)*v*d*cos 60 right?

No, the cos(60) has nothing to do with it, as that only has to do with the configuration the system is in when the mud does make contact. The angular momentum beforehand is L_i=I_iω_i.