A student holds two lead weights, each of mass

Question

12 kg. When the students’ arms are extended
horizontally, the lead weights are 0.82 m from
the axis of rotation and the student rotates
with an angular speed of 1.5 rad/sec. The
moment of inertia of student plus stool is
5.5 kg m2
and is assumed to be constant; i.e.,
the student’s arms are massless! Then the
student pulls the lead weights horizontally to
a radius 0.16 m from the axis of rotation.
Calculate the final angular speed of the
system.
Answer in units of rad/s

Answer 1

We can solve this problem using the conservation of angular momentum. The initial angular momentum (L1) is equal to the final angular momentum (L2).

L1 = L2

Angular momentum is the product of moment of inertia (I) and angular velocity (ω).

L1 = I1 * ω1
L2 = I2 * ω2

where I1 and I2 are the moments of inertia before and after the student pulls the weights in, and ω1 and ω2 are the angular velocities before and after, respectively.

Initially, the student has a moment of inertia of 5.5 kg m^2, and each lead weight has a moment of inertia equal to its mass (12 kg) times the square of the distance to the axis of rotation (0.82 m).

I_lead = 2*(12 kg * (0.82 m)²) = 16.2336 kg m^2.

So the total moment of inertia initially is:

I1 = 5.5 kg m^2 + 16.2336 kg m^2 = 21.7336 kg m^2.

After the student pulls the weights, the new moment of inertia of the lead weights becomes:

I_lead_new = 2*(12 kg * (0.16 m)²) = 0.6144 kg m^2.

So the total moment of inertia after the student pulls the weights is:

I2 = 5.5 kg m^2 + 0.6144 kg m^2 = 6.1144 kg m^2.

Now we can apply the conservation of angular momentum.

I1 * ω1 = I2 * ω2

21.7336 kg m^2 * 1.5 rad/s = 6.1144 kg m^2 * ω2

Solving for ω2:

ω2 = (21.7336 kg m^2 * 1.5 rad/s) / 6.1144 kg m^2 = 5.3189 rad/s.

The final angular speed of the system is 5.3189 rad/s.