A small trolley of mass 2.0kg resting on a horizontal turn table is connected by a light spring to the centre of the table. When the turn table is set into rotation a speed of 600rpm, the length of stretched spring is 50cm. If the original length of the spring is 42cm, find the force constant of the spring.

Explanation

Here, mass of trolley, `m=2.0kg,`
Frequency of rotation, `v=(600)/(60)=10rps`
Extension produced in string,
`x=50-42=8cm =8xx10^(-2)m`
When the turn table is set into rotation, the tension (restoring force), in the spring is equal to the centrifugal force. Then length of stretched spring, `r=50cm=0.50m.`
`:. F=kx=mr(2piv)^(2)=4pi^(2)mrv^(2)`
or `k=(4pi^(2)mrv^(2))/(x)`
`=(4xx(22//7)^(2)xx2.0xx0.50xx(10)^(2))/(8xx10^(-2))`
`=49387.7Nm^(-1)`