Question

A small block on a frictionless horizontal surface has a mass of 2.50

Answer 1

m = 2.5*10^-2 kg

Initial radius of circualar motion, R = 0.27 m

Initial angular speed, Wi = 2 rad/s

final radius of circle, r = 0.2 m

Let Angular speed of mass after radius is reduced to 0.2, = Wf

Initial angular momentum, Mi = (m*R^2)*Wi

Final engular momentum, Mf = (m*r^2)*Wf

Now, by conservation of angular momentum,

Mi = Mf

a) Yes angular momentum is conserved

So, (m*R^2)*Wi = (m*r^2)*Wf

So, Wf = (R^2/r^2)*Wi = (0.27^2/0.2^2)*2 = 3.65 rad/s <------------answer(new angular speed)

c)

Change in Kinetic energy = 0.5*(mr^2)*Wf^2 - 0.5*(mR^2)*Wi^2

= 0.5*2.5*10^-2*((3.65*0.2)^2 - (2*0.27)^2) = 3.02*10^-3 J <-------answer

d)

Work done in pulling the cord = 3.02*10^-3 J <------- same as change in Kineitc energy