Chapter 10, Problem 23A

A string is used to whirl a 2-kg toy in a horizontal circle of radius 2.5 m. Show that when the toy moves at 3 m/s the tension in the string (the centripetal force) is 7.2 N.

To determine

To show: Tension in the string is 7.2 N.

Explanation of Solution

Given:

Mass of toy, $m=2\text{kg}$

Tangential velocity, 13y

  $v=3\text{m/s}$

Length of string, $r=2.5\text{m}$

Formula Used:

The equation for centripetal force is,

  $F_c=\frac{m{v}^2}{r}$

Where, $m$ is mass, $v$ is tangential velocity and $r$ is radius of circular path.

Calculation:

When the string with ball is swung into horizontal circular path, the tension in the string is equal to the centripetal force. So,

  $\begin{array}{l}T=F_c=\frac{m{v}^2}{r}\\ =\frac{2\times 3^2}{2.5}=7.2\text{N}\\ \Rightarrow T=7.2\text{N}\end{array}$

Conclusion:

Hence, it is proved that the tension in the string is $T=7.2\text{N}$