Mouse Jumps Onto An Exercise Wheel

[burnst14]

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
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

 

[mfb]

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.

 

[burnst14]

So my answer is correct then? Or is my formula missing a piece?

Thanks!

 

[mfb]

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

 

[HalfLostAndSearching]

I would approach this problem by first identifying all the relevant variables and equations. The key variables in this problem are the moment of inertia (I), radius (R), mass (m), initial velocity (v), and angular velocity (ω). The relevant equations are L = Iω (angular momentum), KE = 1/2Iω^2 (rotational kinetic energy), and PE = mgh (gravitational potential energy).

For part 2, we want to find the maximum height of the mouse as it rides the wheel. To do this, we can use the conservation of energy principle, which states that the total energy of a system remains constant. Initially, the mouse has kinetic energy, KE = 1/2mv^2, and no potential energy. After jumping onto the wheel, the mouse has rotational kinetic energy, KE = 1/2Iω^2, and gravitational potential energy, PE = mgh. Therefore, we can set these two energies equal to each other and solve for h:

1/2mv^2 = 1/2Iω^2 + mgh

Solving for h, we get:

h = (1/2mv^2 - 1/2Iω^2)/mg

Substituting in the values given in the problem, we get:

h = (1/2*0.03*2^2 - 1/2*0.0004*7.09^2)/(0.03*9.81) = 0.063 m

Therefore, the maximum height of the mouse as it rides the wheel is 0.063 m. This makes intuitive sense, as the mouse would reach its maximum height when it is at the top of the wheel, where it has the most potential energy.

Overall, this problem demonstrates the principles of rotational motion and conservation of energy. By identifying the relevant variables and using the appropriate equations, we can solve for the unknown quantities and gain a better understanding of the physical phenomenon at hand.