10.42 . CP A small block on a frictionless, horizontal surface has a mass of 0.0
ID: 2298620 • Letter: 1
Question
10.42 . CP A small block on a
frictionless, horizontal surface
has a mass of 0.0250 kg. It is
attached to a massless cord passing
through a hole in the surface
(Fig. E10.42). The block is originally
revolving at a distance of
0.300 m from the hole with an
angular speed of 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. (a) Is the angular momentum
of the block conserved? Why or why not? (b) What is the new
angular speed? (c) Find the change in kinetic energy of the block.
(d) How much work was done in pulling the cord?
Explanation / Answer
a) Yes, angular momentum is conserved.
Reason: - The axis of rotation is vertical axis passing through the hole.
The force with which the cord is pulled is along the line which coincides with this axis.
Therefore, the arm length of the force = 0
Therefore, torque = 0
Since torque = 0, therefore angular momentum is conserved.
b)Let I1 = initial moment of inertia,
w1 = initial angular velocity,
I2 = final moment of inertia,
w2 = final angular velocity
r1 = initial radius,
r2 = final radius,
m = mass of the block
Then I1 = m*r1^2, I2 = m*r2^2
Initial angular momentum = I1 * w1 = m*r1^2*w1
Final angular momentum = I2 * w2 = m*r2^2*w2
By conservation of momentum,
m * r2^2 * w2 = m * r1^2 * w1
Divide both sides by m:-
r2^2 * w2 = r1^2 * w1
Or w2 = (r1/r2)^2 * w1
Or w2 = (0.3/0.15)^2 * 1.75 rad/s = 2^2 * 1.75 rad/s
= 4 * 1.75 rad/s = 7 rad/s
Ans: 7 rad/s
c)Initial speed v1 = w1 * r1 = 1.75 * 0.3 m/s = 5.25 m/s
Final speed v2 = w2 * r2 = 7 * 1.5 m/s = 10.5 m/s
Initial kinetic energy = 1/2 mv1^2 = 1/2 * 0.0250 * 5.25^2 = 0.3445 J
Final kinetic energy = 1/2 mv2^2 = 1/2 * 0.0250 * 10.5^2 = 1.3781 J
Change in kinetic energy = 1.3781 - 0.3445 = 1.0336 J
d) Work done = change in kinetic energy = 1.0336 J
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