Re: Followup question: Rotational Dynamics-hamster and wheel

The original question was:

"After getting a drink of water, a hamster jumps onto an exercise wheel for a run. A few seconds later the hamster is running in place with a speed of 1.4 m/s. Find work done by the hamster to get the exercise wheel moving, assuming it is a hoop of radius 0.13 m and mass 6.5 g.

[Difficulty]
the hamster's mass

[Thoughts]
By work-energy theorem, the amount of work done is equivalent to the finial kinetic energy minus the initial kinetic energy. Since the initial kinetic energy is 0, it follows that the work is equivalent to the final kinetic energy. K = (1/2)mv^2, v = 1.4 m/s, m = 6.5 g + hamster's mass, which is unknown. The radius is not used. I know the moment of inertia I of a hoop is mr^2, and m = 6.5 g and r = 0.13 m. Please help me."

The MadSci reply was:

"With this kind of problem I'm assuming you have studied rotational kinetic energy, which is what you need to think about in this problem, for which the moment of inertia is one of the key parts."

The new question is:

Well, I know the momentum of the system is conserved; that's it, the initial momentum mv of the hamster running towards the wheel should equal to the final momentum of the system with both the hamster and the wheel. However, we are not able to use this approach because the mass of the hamster and its speed (assuming constant) when it was running towards the wheel are unknown.

By the way, what if the speed of the hamster is not constant, but with a constant acceleration a? Would the conservation of momentum still work (since the initial momentum of the hamster is not constant, or can this be done using calculus (like using integrals to sum up the total work of a system if both force and position are not constant)? This physics book I am reading does not involve the implement of calculus, but I am studying a separate calculus textbook on my own.

You are making the problem much harder than it needs to be. The only work we need to figure out is the work to make the wheel spin, so we can actually totally ignore the mass of the hamster! And we only need the speed of the hamster in order to know the linear speed of the rim of the wheel. The work to spin the wheel is the difference between the final and initial rotational kinetic energy, and since the initial rotational kinetic energy was zero we only need to compute the final rotational kinetic energy! It's really that simple.

RKE = 0.5 I w²
where "I" is the moment of inertia of the wheel and "w" is the angular speed in radians per second. For the exercise wheel we may assume that all the mass is at the rim, so
I = m r².
The angular speed is found by
w = v / r
where "v" is the linear speed of the rim of the exercise wheel, which has the same magnitude as the running speed of the hamster. (You must realize that the hamster is actually stationary in space, with the exercise wheel spinning around the hamster!) We have
Work = RKE = 0.5 m r² (v/r)²
which works out to
Work = 0.5 m v²

The reason that this looks so much like a regular linear kinetic energy is that all the mass of the wheel is at the same radius. If the spinning shape had a different mass distribution there would be an additional multiplicative factor. At any rate, for this problem we have
Work = 0.5 * 0.0065kg * (1.4m/s)²
Work = 0.00637 J.

John Link, MadSci Physicist