Solving for Mass in a Spring Scale Experiment

[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
R = radius of wheel
-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.

 

[LowlyPion]

Is that the problem as stated?

 

[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.

 

[LowlyPion]

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

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?

 

[bodensee9]

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

 

[LowlyPion]

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

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.