Question

A child attaches a 0.6 kg toy to a 0.9 meter length of string and spins it around in uniform circular motion. Calculate the tension in the string if the period of rotation is 4.1 seconds.

Answer 1

Answer:

1.26812 N

Explanation:

$m$ = Mass of toy = $0.6\ \text{kg}$

$r$ = Length of string = $0.9\ \text{m}$

$t$ = Period of rotation = $4.1\ \text{s}$

Time period is given by

$t=\dfrac{2\pi r}{v}\\\Rightarrow v=\dfrac{2\pi r}{t}\\\Rightarrow v=\dfrac{2\pi 0.9}{4.1}\\\Rightarrow v=1.3792\ \text{m/s}$

The rotational velocity is 1.3792 m/s

The tension in the rope will be the centripetal force acting on the toy

$T=\dfrac{mv^2}{r}\\\Rightarrow T=\dfrac{0.6\times 1.3792^2}{0.9}\\\Rightarrow T=1.26812\ \text{N}$

The tension in the string is 1.26812 N.