A small block on a frictionless horizontal surface has a mass of 0.0225 kg . It

Question

A small block on a frictionless horizontal surface has a mass of 0.0225 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 0.255 m from the hole with an angular speed of 1.75 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.145 m . You may treat the block as a particle.

a) Is angular momentum conserved?

b) Why or why not?

c) What is the new angular speed?

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

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

Explanation / Answer

a)

the angular momentum will be conserved

b)

beacause first the force will be zero
next the torque will be zero that is why angular momentum will be conserved.
c)
m r12 = m r22

now the Initial angular momentum = m r12 w1
and the final angular momentum = m r22 w2
at equilibrium or conservation of inertia
m r12 w1 = m r22 w2

r12 w1/ r22 = w2

w2 = r12 w1/ r22

w2 = (0.225 / 0.145)2 X 1.75

new angular speed is w2 = 2.40 X 1.75 = 4.21 rad/s
d)

now the initial speed V1 = w1 X r1 = 1.75 X 0.225 = 3.9 m/s
and the final speed V2 = w2 X r2 = 4.21 X 0.145 = 6.1 m/s
finding initial kinetic energy = 1/2 mV12 = 1/2 X 0.0225 X 3.92 = 0.17 J
and final kinetic energy = 1/2 mV22 = 1/2 X 0.0250 X6.12 = 0.41 J
change in KE = 0.41 - 0.17 = 0.24 J
e)

the work was done in pulling the cord is = change in KE

= 0.41 - 0.17

= 0.24 J

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