Question

A certain string can withstand a maximum tension of 40 Nwithout breaking. A child ties a 0.37 kgstone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 M, slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? (b) What is the speed of the stone as the string breaks?

Short answer

$$\begin{gathered} \mathrm{a})40\mathrm{N} \\ \mathrm{b})9.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Step-by-step solution

Given

$$\begin{gathered} \mathrm{Max}.\mathrm{Tension}\mathrm{force}\mathrm{T}=40\mathrm{N} \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{stone}\mathrm{M}=0.37\mathrm{kg} \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{R}=0.91\mathrm{m} \end{gathered}$$

Understanding the concept

This problem is based on the Newton’s second law of motion. Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.

Formula:

$\mathbf{F}\mathbf{=}\mathbf{ma}$

Calculate the stone on its path when the string breaks

(a)

The tension is greatest at the bottom of the swing.

By Newton’s second law:

$$\begin{gathered} T-mg=\frac{m{v}^2}{R} \\ T=m\left(g+\frac{v^2}{R}\right) \end{gathered}$$

Increasing the speed eventually leads to the tension at the bottom of the circle reaching that breaking value of 40 N.

Calculate the speed of the stone as the string breaks

(b)

Solving the above equation for the speed, we find

$$\begin{gathered} v=\sqrt{R\left(\frac{T}{m}-g\right)} \\ =\sqrt{\left(0.91\mathrm{m}\right)\left(\frac{40\mathrm{N}}{0.37\mathrm{kg}}-9.8\mathrm{m}/{\mathrm{s}}^2\right)} \\ =9.5\mathrm{m}/\mathrm{s} \end{gathered}$$