Mouse Jumps Onto An Exercise Wheel

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Homework Help Overview

The problem involves a mouse jumping onto a vertical exercise wheel, which is initially at rest and can rotate without friction. The discussion centers around determining the angular velocity of the system immediately after the mouse jumps on and the maximum height the mouse reaches while riding the wheel.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and energy conservation as methods to approach the problem. The original poster attempts to apply these principles but expresses uncertainty about the second part of the problem regarding the maximum height of the mouse.

Discussion Status

Some participants have provided feedback on the original poster's calculations and approach, suggesting that the energy conservation method is valid. However, there is no explicit consensus on the correctness of the original poster's final answer or formula.

Contextual Notes

The original poster mentions concerns about the reliability of an answer packet due to previous incorrect solutions, indicating a lack of trust in external resources. There is also a note about the potential need for torque considerations, although energy conservation is emphasized as a primary method.

burnst14
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Homework Statement


The vertical exercise wheel in a mouse cage is initially at rest, but can turn without friction around a horizontal axis through the center of the wheel. The wheel has a moment of inertia I=0.0004kg m2 and radius R = 0.06m An extremely smart pet mouse of mass m = 0.03 kg runs across her cage with initial speed v = 2 m/s, jumps onto the edge of her exercise wheel, holds on tightly, and rotates together with the wheel.

  1. Determine the angular velocity of the "wheel plus mouse turning together" immediately after she jumps on.
  2. Determine the maximum height of the mouse as she rides the wheel.

Homework Equations

The Attempt at a Solution


I solved the first one and got 7.09 rad/s by:
L = Rmv = Iω
0.06*0.03*2 = (0.0004+0.03*0.062
ω = 7.09 rad/s

However part 2, I can't wrap my head around. I have an answer packet, but it has already had a couple incorrect solutions in it so I'm not quick to trust it.
I tried 1/2Iω2 = mgh

KE = 1/2(5.08E-4)(7.09)2 = 0.03*9.81*h

where h is the maximum height and 0.03 kg is the mass of only the mouse because the wheel is a homogenous cylinder and has equal mass on all "sides".

I feel like I'm missing torque though.
Any ideas?

Thanks guys
 
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The approach via energy conservation is good. Torque will vary along the ride, but you don't have to worry about that as energy is conserved and you can calculate it both before and after.
 
So my answer is correct then? Or is my formula missing a piece?

Thanks!
 
I don't see a final answer, but ω and the approach look good. You should add units, however.
 

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