Question

A sling-thrower puts a stone (0.250 kg) in the sling’s pouch (0.010 kg) and then begins to make the stone and pouch move in a vertical circle of radius 0.650 m. The cord between the pouch and the person’s hand has negligible mass and will break when the tension in the cord is 33.0 N or more. Suppose the sling-thrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur?

Answer 1

Answer:

The cord will break at the lower point.

The linear velocity:

$V=8,72m/s$

Explanation:

The cord will break at the lower point, as this point is where the maximum tension will be at any given speed. This is because at the lower point to maintain the circular movement the cord must withstand the centripetal force plus the gravity force that goes on the opposite direction. To explain this, to maintain the circular movement the stone must always be subjected to a resultant force equal to the centripetal force:

(1)$F_{T}=F_{c}=m*r*w^{2}$

This is constant for a given speed and radius. The total force is the sum of all the forces on the stone, in this case the gravitational force(W) and the cord tension(T). On the upper point they are on the same direction:

$F_{TU}=F_{c}=T+W => T=F_{c}-W$

On the lower point they have opposite directions:

(2)$F_{TL}=F_{c}=T-W => T=F_{c}+W$

As the centripetal force and weight are constant for a given speed and radius, the tension on the lower point is bigger.

As the cord will break at 33N at the lower point, from (2):

$T=F_{c}+W=33N$

$F_{c}=33N-W=33N-(0,25kg+0,01kg)*9,8m/s^{2}=30,45N$

replacing on (1):

$F_{c}=30,45N=m*r*w^{2}$

$w=\sqrt{\frac{30,45N}{m*r} } =\sqrt{\frac{30,45N}{0,26kg*0,65m} } =13,42/s$

The linear velocity:

$V=w*r=13,42/s*0,65m=8,72m/s$