Whirling stone in vertical circle.

[Yoonique]

Homework Statement

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.

Homework Equations

ΣF=ma
a_rad = v_t²/r
v_t = rω
x-direction right = positive
y-direction up = positive
assuming it spins at counter clockwise direction = positive

The Attempt at a Solution

At the top of the circle:
ΣF=ma
-T-mg = ma_rad = -mv_top²/r
v_top² = (T+mg)/m

At the bottom of the circle:
ΣF=ma
4T-mg = mv_bottom²/r
v_bottom² = (4T-mg)/m

At an angle θ counter clockwise from the vertical axis:
-mgsinθ = ma_t
-gsinθ = dV_t/dθ x dθ/dt
∫ -rgsinθ dθ = ∫ V_t dV_t
rgcosθ = 0.5V_t² + c

When θ = 0, and r =1.0, V_t² = v_bottom² = (4T-mg)/m
c = g - (4T-mg)/2m
Therefore, V_t² = (4T-mg)/m + 2gcosθ - 2g

When θ = 2π, V_t² = v_top² = (T+mg)/m
(4T-mg)/m - 4g = (T+mg)/m
T= 2mg

v_top² = (T+mg)/m = 3g

v_top = -√(3g)Is my answer correct?

 

[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 [itex]\theta=\pi[/itex] not [itex]\theta=2\pi[/itex]

 

[AlephNumbers]

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.

Is that an exact reproduction of the problem statement?

 

[Yoonique]

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 [itex]\theta=\pi[/itex] not [itex]\theta=2\pi[/itex]

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

 

[Yoonique]

Is that an exact reproduction of the problem statement?

Yes it is the exact question.

 

[AlephNumbers]

Then I think your solution is correct.

 

[AlephNumbers]

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 [itex]\theta=\pi[/itex] not [itex]\theta=2\pi[/itex]

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

 

[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.