Mouse Jumps Onto An Exercise Wheel

Click For Summary
The discussion focuses on a physics problem involving a mouse jumping onto an exercise wheel. The mouse has a mass of 0.03 kg and runs at an initial speed of 2 m/s, resulting in an angular velocity of 7.09 rad/s for the wheel and mouse system after the jump. The user is attempting to calculate the maximum height the mouse reaches while riding the wheel using energy conservation principles, specifically the equation 1/2Iω² = mgh. There is some uncertainty regarding the inclusion of torque in the calculations, but the consensus is that energy conservation can be applied without needing to account for torque variations during the ride. The user is encouraged to ensure that units are included in their final answers.
burnst14
Messages
53
Reaction score
2

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
 
Physics news on Phys.org
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
5
Views
1K
Replies
67
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
19
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K