# Whirling stone in vertical circle.

1. Apr 4, 2015

### Yoonique

1. The problem statement, all variables and given/known data
A stone, attached to one end of an inelastic string whirls around in a vertical circle of 1.0 m radius. The tension in the string when the stone is at the bottom of the circle is found to be four times the tension in the string when the stone is at the top. Find the velocity of the stone when it is at the top of the circle.

2. Relevant equations
ΣF=ma
vt = rω
x-direction right = positive
y-direction up = positive
assuming it spins at counter clockwise direction = positive

3. The attempt at a solution
At the top of the circle:
ΣF=ma
vtop2 = (T+mg)/m

At the bottom of the circle:
ΣF=ma
4T-mg = mvbottom2/r
vbottom2 = (4T-mg)/m

At an angle θ counter clockwise from the vertical axis:
-mgsinθ = mat
-gsinθ = dVt/dθ x dθ/dt
∫ -rgsinθ dθ = ∫ Vt dVt
rgcosθ = 0.5Vt2 + c

When θ = 0, and r =1.0, Vt2 = vbottom2 = (4T-mg)/m
c = g - (4T-mg)/2m
Therefore, Vt2 = (4T-mg)/m + 2gcosθ - 2g

When θ = 2π, Vt2 = vtop2 = (T+mg)/m
(4T-mg)/m - 4g = (T+mg)/m
T= 2mg

vtop2 = (T+mg)/m = 3g

vtop = -(3g)

2. Apr 4, 2015

### Delta²

seems correct to me, though it would ve been a bit more simple if you had use conservation of energy for the top and bottom position.

Small corection (doesnt affect the final result since it seems you used cos(2pi)=-1), for the top position it is $\theta=\pi$ not $\theta=2\pi$

Last edited: Apr 4, 2015
3. Apr 4, 2015

### AlephNumbers

Is that an exact reproduction of the problem statement?

4. Apr 4, 2015

### Yoonique

Thanks, noted my mistake. I was required to do it by forces though.

5. Apr 4, 2015

### Yoonique

Yes it is the exact question.

6. Apr 4, 2015

### AlephNumbers

Then I think your solution is correct.

7. Apr 4, 2015

### AlephNumbers

How could you use conservation of energy if the circular motion is not uniform?

8. Apr 4, 2015

### Delta²

Well the work of the tension is zero ( string is inelastic, tension is always perpendicular to velocity), we have only work of the weight involved which is a conservative force so conservation of energy applies.