# Non-uniformly accelerated rotation

1. Dec 12, 2011

### mattji104

A circular disk having a radius of 4ft. and weighing 40lb. has a small weight of 20 lb attached to it at a distance of 2ft. from its center. If the disk is in the vertical plane and set in rotation about an axis through its center so that its minimum speed is 120 rpm, find its maximum speed.

For this problem I'm pretty sure I need to use equations such as
ω=dθ/dt, α=dω/dt, and most likely α=ω(dω/dθ)

On my first try I was trying to avoid using the derivative equations. I know the minimum will be at the top, and the maximum at the bottom of the wheel. So in the first attempt I solved for the work done by the 20lb weight using -(ΔV), and equating that to the change in kinetic energy of the wheel. My solution was 121.068rpm, but that seems to simple, and frankly wrong.

What I realize is that the 20lb weight is applying a torque to the wheel, so I likely need to integrate over that torque to understand the change in the angular momentum to finally solve for ω when the point has gone ∏rads through a rotation from the minimum value, 120, at the top

2. Dec 12, 2011

### mattji104

Here are scans of my first two attempts, both did not feel right, but maybe I made mistakes within them, and not entirely.

I'm most confident in that last attempt, but the increase seems intuitively high.

#### Attached Files:

File size:
331.3 KB
Views:
64
File size:
335.5 KB
Views:
59
• ###### scan0004.pdf
File size:
316.7 KB
Views:
58
Last edited: Dec 12, 2011
3. Dec 12, 2011

### Staff: Mentor

Your first approach, using conservation of energy, looked most promising.

You need to make sure that you take into account both objects when you determine the moment of inertia, so that the initial rotational energy will be correct. Since the disk is symmetric about the rotation axis, there's no change in gravitational potential for it as it rotates. But the small mass is a different matter. What's its change in height from the minimum speed position to the maximum speed position?

4. Dec 12, 2011

### mattji104

Well, if my assumption is correct in that the maximum is at the bottom of the rotation, and minimum at the top, the change in height is 4 ft. and the work done is:

mg(h1-h2)=20(32.2)(4)=2576 J

Last edited: Dec 12, 2011
5. Dec 12, 2011

### mattji104

Actually, I'm wrong because 20 lb is the weight not the mass, so
(20/32.2)(32.2)(4)=80 J

6. Dec 12, 2011

### Staff: Mentor

You have to be careful with the units. Using the Imperial US system the "20 lb weight" is going to be equivalent to about 9kg of mass. 20lb is taken as a mass. If you do your calculation in these units your energy won't come out in Joules. It'll be something like "foot pound-force" (ft-lbf), or some equally mind bending unit Divide your energy value by g = 32.174 ft/s2 to render it into ft-lbf units. Or just state that the units are lb*ft2/s2 and leave it be. (Or just convert everything to kg's and m's and work in metric!)

But you've got the right idea, and symbolically the math makes sense. So if you do the math correctly the units should take care of themselves.

7. Dec 12, 2011

### mattji104

OK, attached is my solution going through without making my silly unit mistakes, and assuming the 40 lb and 20 lb are mass. Converted everything to SI units for simplicity, which is something I have tried to avoid to build my versatility, but clearly it's not worth it if I can't get the solution!

Also, I made the assumption that the moment of inertia is only given by the disc, because the description of the 20 lb mass seemed to be a point particle, therefore only affecting the rotational motion but not I, is that proper? And does the answer seem reasonable?

So if you can help me with one last conceptual idea that's bothering me. Was it possible to approach this problem using using the torque? What was bugging me was that I saw the weight of 20 lb will introduce 0 torque at ∏/2, but from there to 3∏/2, it would introduce a non-constant acceleration proportional to its component tangent to the circle, which is what I tried to accomplish in "scan0004.pdf" above. Am I correct in thinking that?

#### Attached Files:

• ###### scan0005.pdf
File size:
338.7 KB
Views:
57
8. Dec 13, 2011

### Staff: Mentor

The small mass is definitely a contributor to the moment of inertia! (Even the disk is comprised of point masses in the calculus limit when you derive its moment of inertia).

Your moment of inertia should include both the disk and the small mass.
Yes, you could approach the problem by way of torque and angular acceleration. It's more tricky since yes, the torque will vary. So you'll need to make it a function of the angle. If the angle is measured from the vertical, and if it's zero when the small mass is at its zenith, then it looks like the torque will vary as the sine of the angle. You'll end up either integrating F*dθ to find the work done (so back to work-energy method), or cast it in the form of a differential equation to solve.