Flywheel, rotational mechanics problem

[mrdonkey]

Homework Statement

A flywheel is a homogeneous cylinder of mass M with a hole in the middle where the axle connects with it. The outer radius of the wheel is R. The radius of the axle is r, and the mass of the axle is m. A weight of mass m_w is suspended from a string wound around the axle. The axle is also a homogeneous cylinder.
(a) Calculate the moment of inertia of the flywheel + axle combination in terms of the variables given in the problem. In what follows, call this I.
(b) Express the kinetic energy of the flywheel, and the kinetic energy of the falling weight, in terms of the variables given in the problem, and the instantaneous angular speed ω.
(c) Express ω as a function of h, the height that the falling weight drops, assuming that everything starts from rest. State any other assumptions you make, and why.

http://img367.imageshack.us/img367/5936/untitledvs8.jpg

Homework Equations

I(cylinder) = 0.5MR^2 (in general)
K(linear) = 0.5mv^2
K(rot) = 0.5Iω^2

The Attempt at a Solution

a) I = I(axle) + I(wheel)
I(axle) = 0.5mr^2
I(wheel) = 0.5MR^2
Therefore, I = 0.5(mr^2 + MR^2)

I thought this one was pretty straightforward but somehow it doesn't look right, like I should somehow be compensating for the fact that the wheel is not a complete cylinder, but a cylinder with an inner cylinder removed. But if the total mass of the wheel is M, which is given, this should already be compensated for in the problem, no?

b) The kinetic energy of the flywheel is easy enough to solve for, since K(rot) = 0.5Iω^2, and we don't need to simplify any further as we just solved for I. For the mass, I assumed the velocity of the block to be equal to ωr, as the motion of the weight must be the same as the motion of the rim of the axle, and from circular motion we know that v = ωr. So therefore the kinetic energy of the weight, assuming the string doesn't slip, is 0.5(m_w)(ωr)^2

c) We know that the potential energy of the weight before it is dropped is equal to the kinetic energy at the instant before it finishes its drop, so we can say that U_i = K_f = m_w*g*h, or

0.5m_w(ωr)^2 = m_w*g*h
0.5(ωr)^2 = gh, h = 1/g * 0.5(ωr)^2

This was a bit of a leap though, as it doesn't take into account the kinetic energy of the wheel itself, which it seems I should be doing. I'm just not sure how. Any thoughts on anything I've done? I can sort of grasp at the important concepts this question is trying to get me to think about but I seem to be tripping over some details here or there. Any help would be greatly appreciated.

 

[andrevdh]

(c) Assuming that no friction acts on the axle and not taking drag into consideration the mechanical energy of the system will be conserved.

 

[m2003]

I would suggest checking your calculations again to make sure they are correct. Your approach seems reasonable, but there may be some errors in your equations or assumptions. For example, in part b, the kinetic energy of the weight should be 0.5m_wv^2, not 0.5m_w(ωr)^2. Also, in part c, you are assuming that the weight is released from rest, but in the problem statement it says that the flywheel is initially rotating. This may affect the calculation for the potential energy.

I would also recommend considering the conservation of energy in this problem. The total energy of the system (flywheel + weight) should be equal at all times, so you can set the initial energy equal to the final energy and solve for ω.

Additionally, you may want to consider the relationship between linear and rotational motion in this problem. The weight is falling in a linear motion, but it is also causing the flywheel to rotate. How does this affect the calculation for the kinetic energy of the weight?

Overall, it seems like you are on the right track and just need to carefully check your calculations and assumptions. Don't be afraid to go back and revise your approach if you find any errors or inconsistencies.