A student sits on a rotating stool holding two 3.2-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.36 m from the rotation axis.

Question

A student sits on a rotating stool holding two 3.2-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.36 m from the rotation axis.

(a) Find the new angular speed of the student.
Answer in rad/s

(b) Find the kinetic energy of the student before and after the objects are pulled in.
before Answer in J
after Answer in J

Answer 1

To solve this problem, we can use the principle of conservation of angular momentum. According to this principle, the initial angular momentum of the system should be equal to the final angular momentum after the objects are pulled in.

(a) Let's start by calculating the initial angular momentum of the system before the objects are pulled in. The formula for angular momentum is:

L = moment of inertia * angular speed

Given:
Moment of inertia (I) = 3.0 kg·m^2
Initial angular speed (w1) = 0.75 rad/s

Initial angular momentum (L1) = I * w1

Calculating, we have:
L1 = 3.0 kg·m^2 * 0.75 rad/s
L1 = 2.25 kg·m^2/s

Now, let's find the final angular momentum of the system after the objects are pulled in. The moment of inertia will change as the objects are brought closer to the rotation axis. The new moment of inertia (I') can be calculated as the sum of the moment of inertia of the student plus stool and the moment of inertia of the objects.

Given:
Mass of each object (m) = 3.2 kg
Distance from rotation axis (r1) = 1.0 m
New distance from rotation axis (r2) = 0.36 m

The moment of inertia (I') of the new system can be calculated using the parallel-axis theorem, which states:

I' = İ + m * d^2

Where:
İ is the moment of inertia of the student plus stool
m is the mass of each object
d is the change in distance from the rotation axis (r1 - r2)

Calculating:
İ = 3.0 kg·m^2
d = 1.0 m - 0.36 m = 0.64 m

I' = 3.0 kg·m^2 + (2 * 3.2 kg) * (0.64 m)^2
I' = 3.0 kg·m^2 + 4.096 kg·m^2
I' = 7.096 kg·m^2

Now, let's find the final angular speed (w2) using the conservation of angular momentum:

L1 = L2
I * w1 = I' * w2

Substituting the values we have:
3.0 kg·m^2 * 0.75 rad/s = 7.096 kg·m^2 * w2

Solving for w2:
w2 = (3.0 kg·m^2 * 0.75 rad/s) / 7.096 kg·m^2
w2 ≈ 0.317 rad/s

Therefore, the new angular speed of the student is approximately 0.317 rad/s.

(b) To find the kinetic energy before and after the objects are pulled in, we can use the formula:

Kinetic energy (KE) = (1/2) * moment of inertia * angular speed^2

Before:
Kinetic energy before (KE1) = (1/2) * 3.0 kg·m^2 * (0.75 rad/s)^2

After:
Kinetic energy after (KE2) = (1/2) * 7.096 kg·m^2 * (0.317 rad/s)^2

Calculating each expression:
KE1 = 1.125 J (approximately)
KE2 = 0.355 J (approximately)

Therefore, the kinetic energy of the student before the objects are pulled in is approximately 1.125 J, and after the objects are pulled in, it becomes approximately 0.355 J.