What Is the Work Done by a Mouse Walking on a Spinning Turntable?

AI Thread Summary
A turntable with a radius of 25 cm and a rotational inertia of 0.0154 kg m^2 spins at 22.0 rpm with a 19.5-g mouse on its edge. As the mouse walks to the center, the new rotation speed is calculated to be 23.7 rpm. The work done by the mouse is determined by the change in kinetic energy, which involves equating initial and final angular momentum. The mouse's kinetic energy at the center is zero, meaning only the turntable's kinetic energy contributes to the final calculation. The discussion emphasizes the importance of using the correct inertia formula for the mouse to solve for work done accurately.
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Homework Statement



A turntable of radius 25 cm and rotational inertia 0.0154 kg m^2 is spinning freely at 22.0 rpm about its central axis, with a 19.5-g mouse on its outer edge. The mouse walks from the edge to the center. Find (a) the new rotation speed and (b) the work done by the mouse.

Homework Equations




W = Kf - Ki
Inertia for a disk is MR^2
Intertia for a ring is (1/2)MR^2


The Attempt at a Solution


solution to part A) 23.7 rpm

i have tried the following but they all seem to be wrong (303 mJ, 3.3 mJ, 127 mJ)

the answer to part B must be in mJ

please give me an idea on how to solve part B.
 
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I haven’t checked your calculations for part A, so I’m writing the method anyway.

The initial angular momentum comprises of the ang mom of the disk and ang mom of the mouse. The final ang mom is only of the disk, since the mouse is at the centre. Equate initial and final angular momenta.

Initial KE of disk plus mouse < Final KE of same. The increase is because of the work done by the mouse.

Now plug in the numbers.
 
Shooting star said:
I haven’t checked your calculations for part A, so I’m writing the method anyway.

The initial angular momentum comprises of the ang mom of the disk and ang mom of the mouse. The final ang mom is only of the disk, since the mouse is at the centre. Equate initial and final angular momenta.

Initial KE of disk plus mouse < Final KE of same. The increase is because of the work done by the mouse.

Now plug in the numbers.

by "Final KE of same" you mean KE of the disk and the mouse

would the KE of the mouse at the center of the disk be 0, so only the KE of disk would be involved in the final right
 
thebest100 said:
by "Final KE of same" you mean KE of the disk and the mouse
Right.

would the KE of the mouse at the center of the disk be 0, so only the KE of disk would be involved in the final right
Yes. Just treat the mouse as a point mass (ignore its rotation).
 
Thank You Shooting Star and Doc Al. I found my mistake, I was using 1/2MR^2 for the Inertia of the mouse, instead of MR^2.

Thank you
 
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