Chapter 10, Problem 53A

From the equation $F=\frac{m{v}^2}{r}$, calculate the tension in a 2-m length of string that whirls a 1-kg mass at 2 m/s in a horizontal circle.

To determine

Tension in the string.

Answer to Problem 53A

  $T=2\text{N}$

Explanation of Solution

Given:

Mass, $m=1\text{kg}$

Tangential velocity, $v=2\text{m/s}$

Length of string, $r=2\text{m}$

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

  $\begin{array}{c}T=F_c\\ =\frac{m{v}^2}{r}\\ =\frac{1\times 2^2}{2}\\ =2\text{N}\end{array}$

Conclusion:

Therefore, the tension in the string is $T=2\text{N}$