# Mouse onto the edge of phonograph turntable

1. Mar 4, 2012

### tonit

1. The problem statement, all variables and given/known data
$m$mouse $= 0.06kg$

$r = 0.3m$

$\omega = 0.05 rad / s$

Suppose angular speed does not change.
What is the work needed for the mouse to go to the center.

2. Relevant equations

$I$$o$$\omega$$o$ = $I$$f$$\omega$$f$

3. The attempt at a solution

If $\omega$ doesn't change ,then how can I use the above equation.

Since $I$ of mouse would be equal to $mr^2$ then $\omega$ would change. I am stuck here. Oh, and it's not a homework, it's a problem that I was stuck at some time ago. Thanks for any help

2. Mar 4, 2012

### Staff: Mentor

You can't. If ω is fixed then angular momentum is not conserved.

3. Mar 4, 2012

### tonit

So in this case would the work be equal to: $mr^2$ ?

4. Mar 4, 2012

### Staff: Mentor

That's an expression for rotational inertia, not work.

Can you please state the full problem as it was given?

5. Mar 5, 2012

### tonit

A 60 gm mouse falls onto the outer edge of a phonograph turntable of radius 30 cm rotating at 33rev/min. How much work must it do to walk into the center post? Assume that the angular velocity of the turntable doesn't change.

6. Mar 5, 2012

### SHISHKABOB

Seems like you just need to find out the centripetal force that the mouse feels and then you'll have the work. Though I could be wrong.

7. Mar 5, 2012

### tiny-tim

hi tonit!
yes, work done is the integral of force "dot" displacement

so, first, what is the force needed, as a function of r ?

8. Mar 5, 2012

### schliere

There is a centripetal acceleration associated with rotation at constant angular velocity, $a_c = \omega^2 r$. So if my suspicions are correct, you can interpret that as a force that must be overcome, $F_c = m \omega^2 r$

9. Mar 5, 2012

### tonit

thanks to all of you. now it is all clear :D