A small block on a frictionless surface has a mass of 90 g. It is attached to a
ID: 1440348 • Letter: A
Question
A small block on a frictionless surface has a mass of 90 g.
It is attached to a massless string passing through a hole in a
horizontal surface (see diagram). The block is originally
rotating in a circle of radius 65 cm with angular speed 0.70
rad/s. The string is then pulled from below until the radius of
the circle is 45 cm. You may treat the block as a point particle.
(a) Is the angular momentum of the block conserved? Why or
why not? (b) What is the final angular speed?
(c) What are the initial and final tensions in the string? (d)
What was the change in kinetic energy of the block?
(e) How much work was done in pulling the string?
Explanation / Answer
(a) Yes, angular momentum is conserved.
Explanation:
There is no external torque being applied to change it.
Here, 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 = (0.65/0.45)^2 * 0.70 rad/s = 1.44^2 * 0.70 rad/s
= 2.09 * 0.70 rad/s = 1.46 rad/s
So, the final angular speed = 1.46 rad/s.
(d)Initial speed v1 = w1 * r1 = 0.70 * 0.65 m/s = 0.455 m/s
Final speed v2 = w2 * r2 = 1.46 * 0.45 m/s = 0.657 m/s
Initial kinetic energy = 1/2 mv1^2 = 1/2 * 0.09 * 0.455^2 = 0.00932 J
Final kinetic energy = 1/2 mv2^2 = 1/2 * 0.09 * 0.657^2 = 0.01942 J
Change in kinetic energy = 0.01942 - 0.00932 = 0.0101 J
(c) The tension in the string shall be equal to the centrifugal force, i.e., mv^2/r.
So, initial tension, T1 = mv1^2/r1 = 0.09 * 0.455^2/0.65 = 0.0287 N
Final tension in the string, T2 = mv2^2/r2 = 0.09 * 0.657^2/0.45 = 0.0863 N
(e) Work done = change in kinetic energy = 0.0101 J