Question

2. A student, Jay, makes himself as cylindrical as possible while sitting on a spinning chair.

Imagine the cylinder with its flat end in the chair and directly above the chair. His center of
mass is 0.10 m from the axis of rotation of the chair. His mass is 65 kg and the cylindrical shape
he is making has a circumference of 1.2 m. The chair's mass is 30 kg. Assume the friction is
directly proportional to the weight of the chair and whatever is on the chair, and the distance
from the axis of rotation to where the friction is applied is constant.
(a) What is the student's moment of inertia about his center of mass?
(b) What is his moment of inertia about the axis of rotation of the chair?
To determine the angular acceleration due to torque from friction, Andrea spins the chair while
the Jay is in his first position (the cylindrical position described above), while John times
certain rotations after releasing the chair. The first complete revolution while spinning takes 1.8
seconds, and the first three complete revolutions take a total of 5.9 seconds. Then Jay gets into
a second position, and this test happens again. John times that it takes 2.3 seconds for the first
revolution and 7.2 seconds for the first three revolutions. Next, Jay gets out of the chair, and
they spin the chair and calculate that the angular acceleration of the chair is -0.0700 rad/s2
.
[Assume that the angular acceleration for a specific situation is constant.]
(c) What is the angular acceleration of the chair while Jay is in cylindrical position?
(d) What is the angular acceleration of the chair while Jay is in second position?
(e) What is the moment of inertia of the chair?
(f) What is the torque due to friction when Jay is in the chair?
(g) What is Jay's moment of inertia in his second position?
Jay then gets into his cylindrical position again and sits as described earlier. Andrea spins him
very fast. After Jay goes around a few times, John says, “Start” and begins timing. After two
revolutions (3.2 seconds), John says, “Change”, Jay switches instantly into final position, and
John times Jay until the chair stops.
Disk 1 Disk 2 Disk 3
F⃗ F⃗ F⃗
(h) What was Jay's angular speed when the timing started (before the position change)?
(i) What was Jay's angular speed just before changing positions?
(j) What was Jay+Chair's angular momentum just before changing position?
(k) What was Jay+Chair's angular momentum just after changing position?
(l) What was Jay's angular speed just after changing position?

Answer 1

Answer: (a) To find the moment of inertia of Jay about his center of mass, we can use the formula I = mr^2, where m is the mass of the object and r is the distance from the center of mass to the point of rotation. Plugging in the values, we get I = 65 kg * (0.10 m)^2 = 0.65 kg*m^2.

(b) To find the moment of inertia of Jay about the axis of rotation of the chair, we need to use the formula I = Icm + md^2, where Icm is the moment of inertia about the center of mass and d is the distance from the center of mass to the axis of rotation. Plugging in the values, we get I = 0.65 kgm^2 + 65 kg * (0.10 m)^2 = 0.65 kgm^2 + 0.65 kgm^2 = 1.30 kgm^2.

(c) To find the angular acceleration of the chair while Jay is in his cylindrical position, we can use the formula alpha = T/I, where T is the torque and I is the moment of inertia. The angular acceleration is given as -0.0700 rad/s^2, so we can rearrange the formula to solve for T: T = I * alpha = 1.30 kg*m^2 * (-0.0700 rad/s^2).

(d) To find the angular acceleration of the chair while Jay is in his second position, we can use the same formula as before: alpha = T/I. The angular acceleration is not given, so we cannot solve for T directly.

(e) To find the moment of inertia of the chair, we can use the formula I = mr^2, where m is the mass of the object and r is the distance from the center of mass to the point of rotation. The mass of the chair is given as 30 kg and the distance from the center of mass to the axis of rotation is not given, so we cannot solve for I directly.

(f) To find the torque due to friction when Jay is in the chair, we can use the formula T = Fd, where F is the force of friction and d is the distance from the axis of rotation to the point of application of the force. The force of friction is not given, so we cannot solve for T directly.

(g) To find the moment of inertia of Jay in his second position, we need to use the formula I = Icm + md^2, where Icm is the moment of inertia about the center of mass and d is the distance from the center of mass to the axis of rotation. The moment of inertia about the center of mass and the distance from the center of mass to the axis of rotation are not given, so we cannot solve for I directly.

(h) To find Jay's angular speed when the timing started, we need to know the time it took for Jay to complete one revolution and the radius of the circle he was rotating around. Both of these values are not given, so we cannot solve for the angular speed directly.

(i) To find Jay's angular speed just before changing positions, we need to know the time it took for Jay to complete one revolution and the radius of the circle he was rotating around. Both of these values are not given, so we cannot solve for the angular speed directly.

(j) To find Jay+Chair's angular momentum just before changing position, we need to know Jay's angular speed and the moment of inertia of Jay+Chair