Conservation of Angular Momentum / Kinematics.

[holdenthefuries]

Homework Statement

Consider the tetherball shown:

At t=0, the ball, of mass m, is moving in a horizontal circle with velocity Vo and has length Lo. The cord is now slowly drawn in until it has length Lo/3.

Take

Vo = 20 m/sec.
Lo = 1.5 m
m = 2 kg.

What is the final velocity of the ball, and what angle does the cord make with the verticle?

Homework Equations

This is a conservation of angular momentum problem... so IiWi = IfWf...

I am treating the tetherball as a single particle so therefore its moment of inertia = mr^2.

(=)

2((LoSin(Ø)^2))Wi = 2((LfSin(Ø+∆Ø)^2))Wf

R is given at all times by LxSin(Ø+∆Ø)... for our purposes where Lx is either Lo or Lo/3. Of course at Lo, ∆Ø=0.

Also, Tangential velocity = rw...

or:

V = r dØ/dT

The Attempt at a Solution

So, I have been working this problem for so long that I am starting to go a bit crazy, and seem to find myself sputtering around in a rut!

It is clear to me that I have been looking at this problem a bit poorly...

I am trying to set Vdt/r = dØ and integrating both sides... but I'm not sure that is a valid operation due to the sinØ inside the radius. If it is a valid operation, then theta = 20 do after completing the IVP... But from there, how do I find the final time?

I am completely lost on this one... I think.

ANY help would be vastly appreciated.

You don't want to see the pages of failed attempts... this is about where I am right now.

Thanks BUNDLES!
Sean

 

[holdenthefuries]

And as a side... I have calculated Vf in symbolic form to be:

60sinØ/(sin(Ø+∆Ø)

If I could just figure out what theta and time were, I'd be gold! Any hints to this would be perfect. Thanks!

 

[chaoseverlasting]

I think in this case, it would be easier if you assumed the final angle to be theta, and find the final speed (vf) in terms of theta, and using conservation of energy calculate vf (the loss in T=gain in Ug). You'll have two relations in theta and vf to solve for as everything else is a constant.