# Homework Help: Work required to accelerate a horizontal wheel

1. Nov 10, 2009

### Mark617

1. The problem statement, all variables and given/known data
How much work is required to accelerate a horizontal wheel of mass 1150 kg from rest to a rotational rate of 1 revolution in 5 seconds

2. Relevant equations
I = 1/2(m)(r^2)
Rotational KE (ie work) = KE2 - KE1
Rotational KE = (1/2)(I)(Omega Squared)

3. The attempt at a solution

Without a radius or diameter, I find it impossible to get a final answer. Furthermore, it is not stated whether or not the acceleration is uniform, therefore I don't know that I can use the kinematic equations since they rely on a constant acceleration. If acceleration was uniform the angular acceleration would be 0.251327 rad/s^2 but I am not sure how this helps me since it seems everything is dependent on a radius.

We do not use I = m(r^2) because our teacher made it clear the wheel is not a rigid object instead it is a solid cylinder, so we use the above equation for I.

We can derive a frequency of 0.2 rev/s, a period of 5 seconds for one revolution, and an Omega (angular velocity) of 0 rad/s initial, 1.25664 rad/s final.

I come to a final answer of work = 454 x (radius ^2).

Am I missing something? Can I go any further without more information?

This is a basic Physics I class. I am sure there is an answer that can be derived but if it requires some advanced formulas I am not sure I can give that answer as we may not have covered that material (yet).

Thank you for any help that can point me in the right direction / help me understand this question!

Last edited: Nov 10, 2009
2. Nov 10, 2009

### semc

Since v=r $$\omega$$ and you can find the angular velocity, sub the values into the energy equation which cancels out the radius term. Of course this is all based on constant acceleration.

3. Nov 10, 2009

### Mark617

I see what you did there, but I would be left with a linear velocity value in the energy equation which I still cannot find without a radius. (This assumes I understand your response haha).

Substitution would lead to a KE equation as follows : W = (.5)(.5)(M)(r squared)[(v / r) squared].

This would indeed cancel out the r values but alas I am left with a linear velocity. Perhaps I am not understanding, or more likely, my brain is fried due to the time I have put into this one problem while everything else went so smoothly :)

4. Nov 10, 2009

### semc

Ops din't scrutinize your workings sorry about that. From my way of interpreting the question, since they mention horizontal wheel and normally wheels are orthogonal to the floor unless you are driving those cars that can fly...... Anyway since horizontal wheel i assume it ain't moving hence no translational movement .

5. Nov 10, 2009

### Mark617

Exactly how I read it. The center of mass is undergoing no translational motion, so I dropped the (.5)(M)(velocity squared) from the KE equation. So if I was to substitute in the value of zero, the whole thing would be zero and if that happens my head will explode haha.

I guess I just imagine a merry go round - seems to me it would require more work to move the kid on the outside of the merry go round versus the child sitting a few feet from the axis :( I guess my mental block is stemming from always being given the radius or diameter until this point.

6. Nov 10, 2009

### Andrew Mason

You do need the radius. The energy of the spinning wheel is proportional to I for a given angular speed.

$$\Delta KE = I\Delta\omega^2/2$$

which reduces to:

$$\Delta KE = \frac{1}{4}mr^2\Delta\omega^2$$

AM