How Does the String's Path Affect Angular Momentum and Final Speed?

[csnsc14320]

Homework Statement

Mass m is attached to a post of radius R by a string. Initially it is a distance r from the center of the post and is moving tangentially with speed v_o. There are two cases: In case (a) the string passes through a hole in the center of the post at the top, while the string is gradually shortened by drawing it through the hole, and case (b) the string wraps around the outside of the post.

What quantities are conserved in each case? Find the final speed of the mass when it hits the post for each case.

Homework Equations

The Attempt at a Solution

So, for case (a):
Angular momentum is conserved: I_o \omega_o = I_f \omega_f

m R^2 \frac{v_o^2}{(R)^2} = m r^2 \frac{v_f^2}{r^2}

Solving for v_f we should get the right answer.

Now, case B is where I'm stumped. My initial thought is that angular momentum is conserved again, but then I end up with the exact same case I have in (a). Does the fact that the mass is rotating about the circumference of the rod, as opposed to the center of the rod, affect the speed of the mass? And, if so, how would I get started on how to find that?

 

[chrisk]

Solve for v_f and see what you get. Does this make sense?

 

[csnsc14320]

Solve for v_f and see what you get. Does this make sense?

Oops.

I_o \omega_o = I_f \omega_f

m R^2 \frac{v_o}{R} = m r^2 \frac{v_f}{r}

R v_o = r v_f

v_f = \frac{R}{r} v_o

Didn't mean to square my angular velocity in my first post.

So, knowing this for case (a), how do we obtain v_f for case (b)?

The r_o = R and r_f = r, and v_o is the same, why wouldn't the answer be the same as in case a?

Unless angular momentum isn't conserved? But I see no torque provided to cause a change in angular momentum in the first place.

 

[chrisk]

How is the length of the string changing when pulled through the center compared to wrapping around a pole with a radius of "a"?