- #1

Yoonique

- 105

- 0

## 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

_{t}²/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

_{top}²/r

v

_{top}

^{2}= (T+mg)/m

At the bottom of the circle:

ΣF=ma

4T-mg = mv

_{bottom}

^{2}/r

v

_{bottom}

^{2}= (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

_{t}

^{2}+ c

When θ = 0, and r =1.0, V

_{t}

^{2}= v

_{bottom}

^{2}= (4T-mg)/m

c = g - (4T-mg)/2m

Therefore, V

_{t}

^{2}= (4T-mg)/m + 2gcosθ - 2g

When θ = 2π, V

_{t}

^{2}= v

_{top}

^{2}= (T+mg)/m

(4T-mg)/m - 4g = (T+mg)/m

T= 2mg

v

_{top}

^{2}= (T+mg)/m = 3g

v

_{top}= -

**√**(3g)Is my answer correct?