A small trolley of mass 2.0 kg 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 at a speed of 300 rpm, the length of the stretched spring is 40 cm. If the original length of the spring is 35 cm, determine the force constant of the spring.

Answer

15795

Explanation

Mass of trolley, m = 2.0 kg
Frequency of rotation of turn table ,
`v = (300)/(60) = 5 ` rps
Extension produced in the string
` y = 40 - 35 = 5 cm = 5 xx 10^(-2) m`
When the turn - table is set into rotation, the tension (restoring force) in spring is equal to the centripetal force. Thus
Restoring force = Centripetal force
` F = Ky = mr omega ^(2) = mr ( 2 pi v)^(2)`
`K = ( 4 pi ^(2) v^(2) mr)/( y)`
[r = length of stretched spring = 40 cm ]
`= ( 4 xx 9.87 xx 5^(2) xx 2.0 xx 40 xx 10 ^(-2))/(5 xx 10^(-2))`
`= 15759 Nm^(-1)`