# Angular speed wheel problem

1. Jun 17, 2009

### bodensee9

Hello:

If a person rides on a wheel that has the same speed throughout rotation. Does this mean that magnitude of angular velocity is same throughout rotation? I think if the wheel has the same speed, then the centripetal force is equivalent throughout rotation.

Now, the person decides to carry a spring to weigh himself. The maximum of the spring reads X, and the minimum reads Y.

So, wouldn't I have:

w = angular velocity
N = normal
-N - mg = -mw^2*R (at top)
N - mg = mw^2*R (at bottom)

So, N top = mw^2*R - mg, N bottom = mw^2*R + mg? And N top = Y, and N bottom = X. I am supposed to find m, but somehow there's something wrong with this equation. Thanks.

2. Jun 17, 2009

### LowlyPion

Is that the problem as stated?

3. Jun 18, 2009

### bodensee9

Yes it is. I think mv^2/R would be the same both top and bottom because your speed is the same, no? Thanks.

4. Jun 18, 2009

### LowlyPion

If the wheel is rotating at the same speed, then yes, that is a question that answers itself from what you've given.
If your problem is to find an expression for m ...

Then you also need to consider that F = k*x such that

kΔX = ΔF = mg + m*ω2r - (mg - m*ω2r)

Then express as m?

5. Jun 19, 2009

### bodensee9

Hello:
I'm sorry, but what is F= kx? I thought that was the force that causes a displacement of a spring? Thanks.

6. Jun 19, 2009

### LowlyPion

It is. The displacement of the spring at the bottom is X, at the top it is Y. The Δdisplacement of the spring reading they give you then is (X - Y). And this means the ΔForce is the Force at the Bottom minus the Force it reads at the Top.

You can write then

ΔF = kΔX = mg + m*ω2r - (mg - m*ω2r) = 2*m*ω2r = k*(X - Y)

You can rearrange for m.