Question

A mass m0.2kg is attached to the end of a massless string. The mass is rotating in a circular orbit of radius r 0.6m on a frictionless table with period T 1.2s. The string is pulled slowly downwards 0.2m through hole in the table. φ What is the new period of circular orbit for mass m? (ii) How much work is done on the mass?

Answer 1

here,

m = 0.2 kg

r1 = 0.6 m

time period , T = 1.2 s

final radius , r2 = 0.6 - 0.2

r2 = 0.4 m

(1)

let the new period be T'

angular velocity , w = 2*pi/T

using conservation of angular momentum

m * r1^2 * w1 = m * r2^2 * w2

m * r1^2 * 2*pi/T = m * r2^2 * 2*pi/T'

0.6^2 /1.2 = 0.4^2 /T'

T' = 0.53 s

the new period pf the circular orbit is 0.53 s

(b)

work done , W = change in kinetic energy

W = 0.5 * I2 * w2 - 0.5 * I1 * w1)

W = 0.5 * m * r2^2 *(2 * pi /T')^2 - 0.5 * m * r1^2 * (2 *pi /T)^2

W = 0.5 * 0.2 * 0.4^2 *(2 * pi /0.53)^2 - 0.5 * 0.2 * 0.6^2 * (2 *pi /1.2)^2

W = 1.26 J

the work done is 1.26 J