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

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Homework Help Overview

The problem involves a turntable with a mouse walking from its edge to the center, focusing on the concepts of rotational motion and work done. The context includes calculating the new rotation speed and the work done by the mouse as it moves.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and kinetic energy in relation to the mouse's movement. Questions arise about the calculations for angular momentum and the kinetic energy of the mouse at different positions.

Discussion Status

Some participants have provided guidance on equating initial and final angular momentum and kinetic energy. There is acknowledgment of a mistake in the inertia calculation for the mouse, and the discussion is exploring the implications of this correction.

Contextual Notes

Participants are working under the constraints of specific rotational inertia values and the requirement that the answer for work done must be in millijoules. There is a focus on ensuring the correct interpretation of kinetic energy contributions from both the turntable and the mouse.

thebest100
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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.
 
Last edited:
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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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