A small block on a frictionless, horizontal surface has a mass of 2.10×10 2 kg .

Question

A small block on a frictionless, horizontal surface has a mass of 2.10×102 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 of2.67 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.

Part A

Is angular momentum of the block conserved?

Is angular momentum of the block conserved?

Part B

What is the new angular speed?

Part C

Find the change in kinetic energy of the block.

Express your answer with the appropriate units.

Part D

How much work was done in pulling the cord?

Express your answer with the appropriate units.

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Figure 1 of 1

Explanation / Answer

m = mass = 2.10*10-2kg

Ri = initial radius = 0.300m

wi = initial angular velocity = 2.67 rad/s

final radius, Rf = 0.150 m

A) yes, angular momentum will conserved.

B) angular momentum , L = I*w

I = moment of interia = m*R2

from conservation of angular momentum: Li = Lf

Ii*wi = If*wf

m*( 0.300)2* 2.67 = m( 0.150)2*wf

wf( final angular velocity) = ( 0.300)2* 2.67 / ( 0.150)2

= 10.68 rad/s

C) K.E = (1/2)*I*w2

change in kinetic energy, K.E = K.E_final - K.E_initial

= ( 1/2)*If *wf2 - (1/2)*Ii*wi2

= (1/2)*m*(0.150)2*(10.68)2 - (1/2)* m*( 0.300)2*(2.67)2

= (1/2)*2.10*10-2 [ (0.150)2*(10.68)2 - ( 0.300)2*(2.67)2 ]

= 0.0202 J

D) work done = change in K.E

= 0.0202 J

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