A small block on a frictionless horizontal surface has a mass of 0.0245 kg . It

Question

A small block on a frictionless horizontal surface has a mass of 0.0245 kg . It is attached to a massless cord passing through a hole in the surface. (See the figure below (Figure 1) .) The block is originally revolving at a distance of 0.270 m from the hole with an angular speed of 2.00 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.115 m . You may treat the block as a particle.

1) Is angular momentum conserved?

2) Why or why not?  The tension in the cord is directed_____ the axis of rotation and produces no torque. There is_______ external torque (toward, away from, no net, a net)

3) What is the new angular speed?

4) Find the change in kinetic energy of the block.

5) How much work was done in pulling the cord?

Explanation / Answer

1)Yes the angular momentum is conserved.

2)The angular momentum is consered as there is no net torque acting on the system.

The tension in the cord is directed towards the axis of rotation and produces no torque. There is no net external torque

3)From conservation of angular momentum

Li = Lf

I1 w1 = I2 w2

w2 = (I1/I2) w1

I = m r^2

I1 = m r1^2 ; I2 = m r2^2

w2 = (m 0.27^2/m 0.115^2) x 2 = 11.025 rad/s

Hence, w2 = 11.025 rad/s

4)Change in KE will be:

delta KE = 1/2 m (r2^2 w2^2 - r1^2 w1^2)

delta KE = 0.5 x 0.0245 (0.115^2 x 11.025^2 - 0.27^2 x 2^2) = 0.0161 J

Hence, delta KE = 0.016 J

5)Work done is Change in KE so W will be:

W = 0.016 J

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