A small block on a frictionless horizontal surface has a mass of 2.15×10 2 kg .

Question

A small block on a frictionless horizontal surface has a mass of 2.15×102 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 0.265 m from the hole with an angular speed of 1.95 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.165 m . You may treat the block as a particle.

A)Is angular momentum conserved

B)What is the new angular speed

C)Find te change in kinetic energy of the block

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

Explanation / Answer

A)

yes the angular momentum is conserved

B)

m = mass of small block = 0.0215 kg

r1 = initial distance from hole = 0.265 m

r2 = final distance from hole = 0.165 m

W1 = initial angular velocity = 1.95 rad/s

W2 = final angular velocity

Using conservation of angular momentum

m r12 W1 = m r22 W2                                     mr2 = moment of inertia

(0.265)2 (1.95) = (0.165)2 W2    

W2 = 2.24 rad/s

c)

initial rotational KE = (0.5) m r12 W21

final rotational KE = (0.5) m r22 W22      

change in KE = (0.5) m r12 W21 - (0.5) m r22 W22      = (0.5) m (r12 W21 - r22 W22  )

change in KE = (0.5) (0.0215) ((0.265)2 (1.95)2 - (0.165)2 (2.24)2 ) = 0.00140

d)

work done = change in kinetic energy = 0.00140

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