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

1.26812 N

Step-by-step 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=(2\pi r)/(v)\\\Rightarrow v=(2\pi r)/(t)\\\Rightarrow v=(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=(mv^2)/(r)\\\Rightarrow T=(0.6* 1.3792^2)/(0.9)\\\Rightarrow T=1.26812\ \text{N}`

The tension in the string is 1.26812 N.