Chapter 10, Problem 25A

A 2-kg iron ball is swung in a horizontal circular path at the end of a 1.6-m length of rope. Assume the rope is very nearly horizontal and the ball’s speed is 10 m/s. Calculate the tension in the rope.

To determine

Tension in the rope.

Answer to Problem 25A

  $T=125\text{N}$

Explanation of Solution

Given:

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

Velocity, $v=10\text{m/s}$

Length of rope, $r=1.6\text{m}$

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 iron ball is swung into horizontal circular path, the tension in the rope is equal to the centripetal force. So,

  $\begin{array}{c}T=F_c\\ =\frac{m{v}^2}{r}\\ =\frac{2\times {10}^2}{1.6}\\ =125\text{N}\end{array}$

Conclusion:

Therefore, the tension in the rope is $T=125\text{N}$