A small block on a frictionless, horizontal surface has a mass of 2.80×10 2 kg .

Question

A small block on a frictionless, horizontal surface has a mass of 2.80×102 kg . It is attached to a massless cord passing through a hole in the surface (Figure 1) .The block is originally revolving at a distance of0.300 m from the hole with an angular speed of2.25 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.

Is angular momentum of the block conserved?

What is the new angular speed?

Find the change in kinetic energy of the block.

Explanation / Answer

a)

Yes, angular momentum is conserved as there is no torque to change the angular speed

b)

Initial angular momentum ,

Li = m*Wi*r^2 = 0.028*2.25*0.3^2 = 5.67*10^-3 kg.m^2/s

Final angular momentum,

Lf = m*Wf*r'^2

By conservation of angular momentum,Lf = Li

So, m*Wf*r'^2 = m*Wi*r^2

So, Wf*0.15^2 = 2.25*0.3^2

So, Wf = 9 rad/s2 <------- new angular speed

c)

Change in Kinetic energy = 0.5*(m*r'^2)*Wf^2 - 0.5*(m*r'^2)*Wi^2

= 0.5*0.028*(0.15^2*9^2 - 0.3^2*2.25^2) = 0.019 J

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