Mouse onto the edge of phonograph turntable

[tonit]

Homework Statement

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.

Homework Equations

I_o\omega_o = I_f\omega_f

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

 

[Doc Al]

Homework Equations

I_o\omega_o = I_f\omega_f

The Attempt at a Solution

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

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

 

[tonit]

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

 

[Doc Al]

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

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

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

 

[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.

 

[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.

 

[tiny-tim]

hi 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.

yes, work done is the integral of force "dot" displacement …

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

 

[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

 

[tonit]

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