Question

An object of mass 0.5 kg is whirled at the end of a string 0.8 m long. If the string makes three revolutions in 1.2 sec, find the tension on the string.

Answer 1

Step 1: Given data

Mass of the object, m = 0.5 kg .

Length of the string, l = 0.8 m.

time of revolution, t = 1.2 sec.

Let, T, be the tension on the string.

Step 2: Finding the velocity

Now velocity,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{Total}\mathrm{path}\mathrm{traveled}\mathrm{in}\mathrm{revolutions}}{\mathrm{Total}\mathrm{time}} \\ v=\frac{2\pi r\times 3}{t}=\frac{2\pi (0.8)\times 3}{1.2}=\frac{2\times \left(3.14\times 0.8\right)\times 3}{1.2} \\ v=12.56m/s. \end{gathered}$$

Step 3: Finding the tension on the string

We know from the concept of circular motion.

$T=\frac{m{v}^2}{R}.$……(1)

where T = tension, m = mass of the object, and R = radius of revolution (length of the string).

Now putting values in equation (1);

$T=\frac{\left(0.5\right)\times {\left(12.56\right)}^2}{1.2}=65.73N.$

Therefore, the tension on the string will be 65.73 N.