A student holds two lead weights, each of mass 6.4 kg. When the students’ arms are extended horizontally, the lead weights are 0.84 m from the axis of rotation and the student rotates with an angular speed of 2.4 rad/sec. The moment of inertia of student plus stool is 9.4 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 0.23 m from the axis of rotation. 1). Find the new angular speed of the student? 2). Find the kinetic energy of the rotating system (comprised of student, stool, and weights) before and after he pulls the weights inward?

1) To find the new angular speed of the student, we can use the conservation of angular momentum. Angular momentum (L) is conserved in an isolated system with no external torques, hence:

L_initial = L_final

Angular momentum L is the product of the moment of inertia (I) and the angular velocity (ω). So the equation becomes:

I_initial * ω_initial = I_final * ω_final

Initially, the moment of inertia is the sum of the moment of inertia of the student and the stool (which is given as constant) plus the moment of inertia of the lead weights. Since the arms are considered massless, we only take the weights into account. The moment of inertia of a point mass is given by I = m * r^2, where m is the mass and r is the distance from the axis of rotation. With two weights, the initial moment of inertia due to the weights will be:

I_weights_initial = 2 * m * r_initial^2

Where: m = 6.4 kg (mass of one weight) r_initial = 0.84 m (initial distance from the axis)

Plugging in the numbers:

I_weights_initial = 2 * 6.4 kg * (0.84 m)^2 I_weights_initial = 2 * 6.4 kg * 0.7056 m^2 I_weights_initial = 9.0432 kg*m^2

Adding the moment of inertia of the student and stool (9.4 kg*m^2), we get:

I_initial = I_student+stool + I_weights_initial I_initial = 9.4 kg*m^2 + 9.0432 kg*m^2 I_initial = 18.4432 kg*m^2

When the student pulls the weights inward, the moment of inertia due to the weights is now:

I_weights_final = 2 * m * r_final^2 r_final = 0.23 m

I_weights_final = 2 * 6.4 kg * (0.23 m)^2 I_weights_final = 2 * 6.4 kg * 0.0529 m^2 I_weights_final = 0.6784 kg*m^2

The total final moment of inertia, with the I of the student and the stool remaining the same, is:

I_final = I_student+stool + I_weights_final I_final = 9.4 kg*m^2 + 0.6784 kg*m^2 I_final = 10.0784 kg*m^2

Using angular momentum conservation:

I_initial * ω_initial = I_final * ω_final

We know the initial angular speed (ω_initial) is 2.4 rad/s. Solving for ω_final gives us:

ω_final = (I_initial * ω_initial) / I_final ω_final = (18.4432 kg*m^2 * 2.4 rad/s) / 10.0784 kg*m^2 ω_final ≈ 4.389 rad/s

The new angular speed of the student after pulling the weights in is approximately 4.389 rad/sec.

2) To find the kinetic energy before and after pulling the weights in, we use the formula for rotational kinetic energy:

KE = 0.5 * I * ω^2

Before pulling the weights in, we have:

KE_initial = 0.5 * I_initial * ω_initial^2 KE_initial = 0.5 * 18.4432 kg*m^2 * (2.4 rad/s)^2 KE_initial = 0.5 * 18.4432 kg*m^2 * 5.76 rad^2/s^2 KE_initial ≈ 53.146 kg*m^2/s^2

KE_initial ≈ 53.146 Joules

After pulling the weights in, we have:

KE_final = 0.5 * I_final * ω_final^2 KE_final = 0.5 * 10.0784 kg*m^2 * (4.389 rad/s)^2 KE_final = 0.5 * 10.0784 kg*m^2 * 19.2725 rad^2/s^2 KE_final ≈ 97.044 Joules

After pulling the weights in, we have:

KE_final = 0.5 * I_final * ω_final^2 KE_final = 0.5 * 10.0784 kg*m^2 * (4.389 rad/s)^2 KE_final = 0.5 * 10.0784 kg*m^2 * 19.2725 rad^2/s^2 KE_final ≈ 97.044 Joules

So, the kinetic energy of the system before the student pulls the weights inward is approximately 53.146 Joules, and after pulling the weights inward, it is approximately 97.044 Joules.

Extra: Angular momentum is a very important principle in physics, particularly in systems where rotational motion is involved. It shows that if no external torque acts on the system, the product of the moment of inertia and angular velocity remains constant.

The result may seem counterintuitive because the rotational kinetic energy increases when the student pulls the weights inward, even though no external work is done. This increase in kinetic energy comes from the work the student does against the centrifugal force (perceived effect due to rotation), moving the weights in towards the center, effectively converting some of the student's internal energy into kinetic energy of the system. This is an illustration of the conservation of energy principle, where energy can be transformed from one type to another but is never lost.