Question

A sling-thrower puts a stone (0.250 kg). In the slings 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 Nmore. 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?

Short answer

a) Breaking of the cord will occur at the lowest point of the circle.

b) v =8.73 m/s

Step-by-step solution

Given

Total mass of the object

$$\begin{gathered} m=0.250\mathrm{kg}+0.01\mathrm{kg} \\ =0.260\mathrm{kg} \end{gathered}$$

Radius of the circle: $R=0.650\mathrm{m}$

Maximum tension in the cord:T =33.0 N

Understanding the concept

The problem deals with the Newton’s second law of motion which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Also, it involves centripetal force.Use the concept of centripetal force and Newton’s second law.

Draw the free body diagram

Find out if the breaking will occur at the lowest point of the circle or at the highest point

(a)

The tension in the cord will be the greatest at the lowest point of the swing.

From the free body diagram, we can note that,

Tension in the cord at the lowest position is,

$T-mg=\frac{m{v}^2}{R}$ [Applying Newton’s second law]

$T=mg+\frac{m{v}^2}{R}$ (i)

Hence the breaking of the cord will occur at the lowest point of the circle.

Calculate the speed of the stone at which the breaking occurs

(b)

Now to find the speed of the stone (v) at which the breaking will occur:

Given condition is, that at the breaking point T =33.0N

Using this value in equation (i) and solving for speed v

$$\begin{gathered} \Rightarrow 33.0\mathrm{N}-\left(0.26\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\right)=\left(\frac{0.26\mathrm{kg}\times {\mathrm{v}}^2}{0.650\mathrm{m}}\right) \\ \Rightarrow 30.452\mathrm{N}=\left(\frac{0.26\mathrm{kg}\times {\mathrm{v}}^2}{0.650\mathrm{m}}\right) \\ {\mathrm{v}}^2=\frac{30.452\mathrm{N}\times 0.650\mathrm{m}}{0.26\mathrm{kg}} \\ =76.13{\left(\mathrm{m}/\mathrm{s}\right)}^2 \\ \Rightarrow \mathrm{v}=8.73\mathrm{m}/\mathrm{s} \end{gathered}$$