A small block on a frictionless horizontal surface has a mass of 0.0255 kg . It

Question

A small block on a frictionless horizontal surface has a mass of 0.0255 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 0.250 m from the hole with an angular speed of 1.85 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.190 m . You may treat the block as a particle.

Part A

Is angular momentum conserved?

Is angular momentum conserved?

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Part B

Why or why not?

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Part C

What is the new angular speed?

Express your answer in radians per second to three significant figures.

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Part D

Find the change in kinetic energy of the block.

Express your answer in joules to three significant figures.

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Part E

How much work was done in pulling the cord?

Express your answer in joules to three significant figures.

Explanation / Answer

mass of the block m=0.0255 kg

angular speed w1=1.85 rad/sec at r1=0.25 m ,

angular speed is w2 at r1=0.19 m ,

A)

here,

angular momentum is conserved,

because, there are no exterforec torque acting on the system

B)

hence,

angular momentum is conserved,

I1*w1=I2*w2

(m*r1^2)*w1=(m*r2^2)*w2

(r1^2)*w1=(r2^2)*w2

(0.25^2)*1.85=(0.19^2)*w2

===> w2=3.203 rad/sec

new angular speed w2=3.20 rad/sec

C)

initial K.E is K1=1/2*I1*w1^2

K1=1/2*(m*r1^2)*w1^2

K1=1/2*(0.0255*0.25^2)*1.85^2

K1=2.73*10^-3 J

final K.E is K2=1/2*I2*w2^2

K2=1/2*(m*r2^2)*w2^2

K2=1/2*(0.0255*0.19^2)*3.20^2

K2=4.71*10^-3 J

change in K.E =K2-K1

=(4.71*10^-3)-(2.73*10^-3)

=1.98*10^-3 J

=1.98 mJ

d)

work done =change in K.E

W=1.98 mJ

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