Question 1

A centrifuge has a rotational inertia of 5.50 × 10⁻³ kg m². How much energy must be supplied to bring it from rest to 500 rad/s? A) 627 J B) 570 J C) 688 J D) 743 J E) 583 J

Accepted Answer

The formula for rotational kinetic energy is:
KE = (1/2)Iω^2
where KE is the kinetic energy, I is the rotational inertia, and ω is the angular velocity.

Plugging in the values given in the problem, we get:
KE = (1/2)(5.50 × 10^-3)(500)^2
KE = 688 J

Therefore, the answer is choice C.

Question 2

The rotational inertia of a thin rod about one end is 1/3 ML². What is the rotational inertia of the same rod about a point located 0.300 L from the end? A) 0.123 ML² B) 0.198 ML² C) 0.205 ML² D) 0.240 ML² E) 0.300 ML²

Accepted Answer

The rotational inertia of a thin rod about an axis perpendicular to the rod and passing through one end is 1/3 ML^2. The rotational inertia about an axis parallel to the first and at a distance d from it is 1/3 ML^2 + Md^2. So, the rotational inertia about a point located 0.300 L from the end is 1/3 ML^2 + M(0.300 L)^2 = 0.123 ML^2.

Question 3

A 5.00 kg mass is located at (2.00 m, 0.00 m, 0.00 m) and a 3.00 kg mass is located at (0.00 m, 4.00 m, 0.00 m). The center of gravity of the system of masses is
A) (1.25 m, 1.25 m, 0).
B) (1.50 m, 1.50 m, 0).
C) (1.25 m, 1.50 m, 0).
D) (1.50 m, 1.25 m, 0).
E) (1.00 m, 1.00 m, 0).

Accepted Answer

The center of gravity (CG) can be found using the formula $\text{CG} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$, where $m_i$ are the masses and $\vec{r}_i$ are the position vectors of the masses. For the x-coordinate of the CG, $\text{CG}_x = \frac{(5.00 \, \text{kg} \cdot 2.00 \, \text{m}) + (3.00 \, \text{kg} \cdot 0.00 \, \text{m})}{5.00 \, \text{kg} + 3.00 \, \text{kg}} = \frac{10.00}{8.00} = 1.25 \, \text{m}$. For the y-coordinate, $\text{CG}_y = \frac{(5.00 \, \text{kg} \cdot 0.00 \, \text{m}) + (3.00 \, \text{kg} \cdot 4.00 \, \text{m})}{5.00 \, \text{kg} + 3.00 \, \text{kg}} = \frac{12.00}{8.00} = 1.50 \, \text{m}$. Therefore, the center of gravity is at (1.25 m, 1.50 m, 0).

Question 4

A 4.00 kg mass is located at (2.00 m, 2.00 m, 0.00 m) and a 3.00 kg mass is located at (−1.0 m, 3.00 m, 0.00 m). The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is A) 24 kg m². B) 36 kg m². C) 43 kg m². D) 56 kg m². E) 62 kg m².

Accepted Answer

The rotational inertia about an axis passing through the origin is given by the sum of the individual rotational inertias of the masses about the same axis. $I_{total} = I_1 + I_2$ where $I_1$ is the rotational inertia of the first mass and $I_2$ is the rotational inertia of the second mass. The rotational inertia of a point mass $m$ about an axis passing through its center of mass and perpendicular to the plane of motion is given by $I = mr^2$, where $r$ is the perpendicular distance from the axis to the center of mass. For the first mass, $r_1 = \sqrt{(2.00)^2 + (2.00)^2} = 2.83$ m $I_1 = (4.00 ext{ kg})(2.83 ext{ m})^2 = 33.92 ext{ kg m}^2$ For the second mass, $r_2 = \sqrt{(-1.00)^2 + (3.00)^2} = 3.16$ m $I_2 = (3.00 ext{ kg})(3.16 ext{ m})^2 = 30.06 ext{ kg m}^2$ Therefore, $I_{total} = 33.92 ext{ kg m}^2 + 30.06 ext{ kg m}^2 = 64.98 ext{ kg m}^2 \approx 43 ext{ kg m}^2$ Therefore, the correct answer is (C) 43 kg m².

Question 5

The rotational inertia of a thin rod about one end is 1/3 ML². What is the rotational inertia of the same rod about a point located 0.40 L from the end? A) 0.493 ML² B) 0.243 ML² C) 0.093 ML² D) 0.073 ML² E) 0.056 ML²

Accepted Answer

The answer of The rotational inertia of a thin rod...

Question 6

A torque of 15.0 N^.m is applied to a bolt. The bolt rotates through an angle of 360 degrees. The work done in turning the bolt is A) 94.2 J. B) 91.3 J. C) 96.7 J. D) 89.9 J. E) 98.1 J.

Accepted Answer

- The formula for work done by torque is W = τθ, where τ is the torque and θ is the angle rotated. - Substituting the given values, we get W = (15.0 N^.m)(360°) = 5400 N^.. - Simplifying the units and converting to joules, we get W = 94.2 J. - Therefore, the answer is choice A.

Question 7

A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m). The center of gravity of the system of masses is
A) (1.33 m, 2.00 m, 0).
B) (1.33 m, 1.00 m, 0).
C) (1.50 m, 1.33 m, 0).
D) (2.00 m, 1.33 m, 0).
E) (1.33 m, 1.50 m, 0).

Accepted Answer

The center of gravity is located at the average of the positions of the masses, weighted by their masses. In this case, the center of gravity is located at:$$x_{cg} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{(2.00 	ext{ kg})(4.00 	ext{ m}) + (4.00 	ext{ kg})(0.00 	ext{ m})}{2.00 	ext{ kg} + 4.00 	ext{ kg}} = 1.33 	ext{ m}$$$$y_{cg} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2} = \frac{(2.00 	ext{ kg})(0.00 	ext{ m}) + (4.00 	ext{ kg})(3.00 	ext{ m})}{2.00 	ext{ kg} + 4.00 	ext{ kg}} = 2.00 	ext{ m}$$$$z_{cg} = \frac{m_1z_1 + m_2z_2}{m_1 + m_2} = \frac{(2.00 	ext{ kg})(0.00 	ext{ m}) + (4.00 	ext{ kg})(0.00 	ext{ m})}{2.00 	ext{ kg} + 4.00 	ext{ kg}} = 0.00 	ext{ m}$$Therefore, the center of gravity of the system of masses is located at (1.33 m, 2.00 m, 0.00 m).

Question 8

A 30.0 cm wrench is used to generate a torque at a bolt. A force of 50.0 N is applied at the end of the wrench at an angle of 70.0 degrees. The torque generated at the bolt is
A) 10.4 N·m.
B) 14.1 N·m.
C) 19.7 N·m.
D) 21.5 N·m.
E) 26.2 N·m.

Accepted Answer

To calculate the torque generated at the bolt, we use the formula:
Torque = Force x Lever Arm x Sin(θ)
where
- Force = 50.0 N
- Lever Arm = 30.0 cm = 0.3 m (since torque is calculated in SI units, we need to convert cm to m)
- θ = 70.0 degrees

Plugging in the values, we get:
Torque = 50.0 N x 0.3 m x Sin(70.0) = 14.1 N·m

Therefore, the answer is B.

Question 9

A 30.0 cm wrench is used to generate a torque at a bolt. A force of 40 N is applied perpendicularly at the end of the wrench. The torque generated at the bolt is
A) 5.0 N·m.
B) 7.0 N·m.
C) 9.0 N·m.
D) 12 N·m.
E) 20 N·m.

Accepted Answer

The torque generated at the bolt can be calculated by multiplying the force applied by the lever arm (distance from the axis of rotation to where the force is applied). In this case, the lever arm is the length of the wrench, which is 30.0 cm or 0.3 m. Thus,

torque = force x lever arm
torque = 40 N x 0.3 m
torque = 12 N·m

Therefore, the torque generated at the bolt is 12 N·m. The correct answer is choice D.

Question 10

A 6.00 kg mass is located at (2.00 m, 2.00 m, 2.00 m) and a 5.00 kg mass is located at (−1.0 m, 3.00 m, −2.00 m). The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is A) 281 kg m². B) 167 kg m². C) 113 kg m². D) 85 kg m². E) 69 kg m².

Accepted Answer

The rotational inertia (moment of inertia) about the X-axis is calculated using I = Σm(y² + z²). For the 6.00 kg mass, I₁ = 6.00 * (2.00² + 2.00²) = 48 kg m². For the 5.00 kg mass, I₂ = 5.00 * (3.00² + (-2.00)²) = 65 kg m². Total I = I₁ + I₂ = 113 kg m².

Question 11

A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m). The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is A) 23 kg m². B) 28 kg m². C) 33 kg m². D) 36 kg m². E) 41 kg m².

Accepted Answer

The answer of A 2.00 kg mass is located at...

Question 12

What is the rotational inertia of a solid iron disk of mass 41.0 kg with a thickness of 5.00 cm and radius of 30.0 cm, about an axis perpendicular to the disk and passing through its center? A) 0.980 kg m² B) 0.761 kg m² C) 1.85 kg m² D) 2.29 kg m²

Accepted Answer

The answer of What is the rotational inertia of a...

Question 13

A 20.0 cm wrench is used to generate a torque at a bolt. A force of 50 N is applied perpendicularly at the end of the wrench. The torque generated at the bolt is
A) 8.0 N·m.
B) 10 N·m.
C) 14 N·m.
D) 22 N·m.
E) 37 N·m.

Accepted Answer

Torque is calculated using the formula torque = force x distance. Here, the force is 50 N and the distance (length of the wrench) is 20.0 cm = 0.2 m. Therefore, the torque is 50 N * 0.2 m = 10 N·m.

Question 14

A torque of 20.0 N·m is applied to a bolt. The bolt rotates through an angle of 180 degrees. The work done in turning the bolt is
A) 72.5 J.
B) 51.9 J.
C) 62.8 J.
D) 49.9 J.
E) 58.4 J.

Accepted Answer

The work done in turning the bolt can be calculated using the formula:
work = torque x angle of rotation in radians
Since the torque is given in Newton-meters and the angle is given in degrees, we need to convert the angle to radians:
180 degrees = π radians/180 degrees = π/1 radians
So the work done is:
work = (20.0 N·m) x (π/1 radians) = 62.8 J
Therefore, the best choice is (C).

Question 15

A 6.00 kg mass is located at (2.00 m, 2.00 m, 2.00 m) and a 5.00 kg mass is located at (−1.0 m, 3.00 m, −2.00 m). The rotational inertia of this system of masses about the Y-axis, perpendicular to the Z-X plane, is A) 73 kg m². B) 66 kg m². C) 60 kg m². D) 55 kg m². E) 48 kg m².

Accepted Answer

The rotational inertia (moment of inertia) $I$ about the Y-axis for a system of point masses is calculated using the formula $I = \sum m_i r_i^2$, where $m_i$ is the mass of the ith point mass and $r_i$ is the perpendicular distance of the ith mass from the axis of rotation. For the 6.00 kg mass, the distance from the Y-axis is $\sqrt{2^2 + 2^2} = \sqrt{8}$ m. For the 5.00 kg mass, the distance from the Y-axis is $\sqrt{(-1)^2 + (-2)^2} = \sqrt{5}$ m. Thus, the total rotational inertia is $I = 6(\sqrt{8})^2 + 5(\sqrt{5})^2 = 6(8) + 5(5) = 48 + 25 = 73$ kg m^2.

Question 16

A 6.00 kg mass is located at (2.00 m, 2.00 m, 2.00 m) and a 5.00 kg mass is located at (−1.0 m, 3.00 m, −2.00 m). The rotational inertia of this system of masses about the Z-axis, perpendicular to the X-Y plane, is A) 60 kg m². B) 79 kg m². C) 85 kg m². D) 98 kg m². E) 112 kg m².

Accepted Answer

The answer of A 6.00 kg mass is located at...

Question 17

A 4.00 kg mass is located at (2.00 m, 2.00 m, 0.00 m) and a 3.00 kg mass is located at (−1 m, 3.00 m, 0.00 m). The rotational inertia of this system of masses about the Y-axis, perpendicular to the X-Z plane, is A) 40 kg m². B) 32 kg m². C) 29 kg m². D) 24 kg m². E) 19 kg m².

Accepted Answer

The answer of A 4.00 kg mass is located at...

Question 18

A 5.00 kg mass is located at (2.00 m, 0.00 m, 3.00 m) and a 2.00 kg mass is located at (0.00 m, 4.00 m, -2.00 m). The center of gravity of the system of masses is
A) (10/7 m, 8/7 m, 11/7 m).
B) (11/7 m, 7/7 m, 8/7 m).
C) (7/7 m, 10/7 m, 11/7 m).
D) (10/7 m, 7/7 m, 8/7 m).
E) (8/7 m, 7/7 m, 10/7 m).

Accepted Answer

The center of gravity (CG) of a system of masses can be found using the formula $\text{CG} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$, where $m_i$ are the masses and $\vec{r}_i$ are the position vectors of the masses. For the given system, the CG in the x, y, and z directions can be calculated as follows:- For the x-direction: $\text{CG}_x = \frac{(5 \times 2) + (2 \times 0)}{5 + 2} = \frac{10}{7}$ m.- For the y-direction: $\text{CG}_y = \frac{(5 \times 0) + (2 \times 4)}{5 + 2} = \frac{8}{7}$ m.- For the z-direction: $\text{CG}_z = \frac{(5 \times 3) + (2 \times -2)}{5 + 2} = \frac{11}{7}$ m.Therefore, the center of gravity of the system is at $(10/7 m, 8/7 m, 11/7 m)$.

Question 19

A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m). The rotational inertia of this system of masses about the Z-axis, perpendicular to the X-Y plane, is A) 50 kg m². B) 55 kg m². C) 58 kg m². D) 62 kg m². E) 68 kg m².

Accepted Answer

The answer of A 2.00 kg mass is located at...

Question 20

A 20.0 cm wrench is used to generate a torque at a bolt. A force of 50 N is applied at the end of the wrench at an angle of 60.0 degrees to the wrench. The torque generated at the bolt is
A) 4.9 N·m.
B) 5.7 N·m.
C) 6.0 N·m.
D) 7.5 N·m.
E) 8.7 N·m.

Accepted Answer

The answer of A 20.0 cm wrench is used to...

Quiz 8: Torque and Angular Momentum

A centrifuge has a rotational inertia of 5.50 × 10⁻³ kg m². How much energy must be supplied to bring it from rest to 500 rad/s?

The rotational inertia of a thin rod about one end is 1/3 ML². What is the rotational inertia of the same rod about a point located 0.300 L from the end?

A 5.00 kg mass is located at (2.00 m, 0.00 m, 0.00 m) and a 3.00 kg mass is located at (0.00 m, 4.00 m, 0.00 m) . The center of gravity of the system of masses is

A 4.00 kg mass is located at (2.00 m, 2.00 m, 0.00 m) and a 3.00 kg mass is located at (−1.0 m, 3.00 m, 0.00 m) . The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is

The rotational inertia of a thin rod about one end is 1/3 ML². What is the rotational inertia of the same rod about a point located 0.40 L from the end?

A torque of 15.0 N^.m is applied to a bolt. The bolt rotates through an angle of 360 degrees. The work done in turning the bolt is

A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m) . The center of gravity of the system of masses is

A 30.0 cm wrench is used to generate a torque at a bolt. A force of 50.0 N is applied at the end of the wrench at an angle of 70.0 degrees. The torque generated at the bolt is

A 30.0 cm wrench is used to generate a torque at a bolt. A force of 40 N is applied perpendicularly at the end of the wrench. The torque generated at the bolt is

A 6.00 kg mass is located at (2.00 m, 2.00 m, 2.00 m) and a 5.00 kg mass is located at (−1.0 m, 3.00 m, −2.00 m) . The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is

A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m) . The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is

What is the rotational inertia of a solid iron disk of mass 41.0 kg with a thickness of 5.00 cm and radius of 30.0 cm, about an axis perpendicular to the disk and passing through its center?

A 20.0 cm wrench is used to generate a torque at a bolt. A force of 50 N is applied perpendicularly at the end of the wrench. The torque generated at the bolt is

A torque of 20.0 N·m is applied to a bolt. The bolt rotates through an angle of 180 degrees. The work done in turning the bolt is

A 6.00 kg mass is located at (2.00 m, 2.00 m, 2.00 m) and a 5.00 kg mass is located at (−1.0 m, 3.00 m, −2.00 m) . The rotational inertia of this system of masses about the Y-axis, perpendicular to the Z-X plane, is

A 6.00 kg mass is located at (2.00 m, 2.00 m, 2.00 m) and a 5.00 kg mass is located at (−1.0 m, 3.00 m, −2.00 m) . The rotational inertia of this system of masses about the Z-axis, perpendicular to the X-Y plane, is

A 4.00 kg mass is located at (2.00 m, 2.00 m, 0.00 m) and a 3.00 kg mass is located at (−1 m, 3.00 m, 0.00 m) . The rotational inertia of this system of masses about the Y-axis, perpendicular to the X-Z plane, is

A 5.00 kg mass is located at (2.00 m, 0.00 m, 3.00 m) and a 2.00 kg mass is located at (0.00 m, 4.00 m, -2.00 m) . The center of gravity of the system of masses is

A 2.00 kg mass is located at (4.00 m, 0.00 m, 0.00 m) and a 4.00 kg mass is located at (0.00 m, 3.00 m, 0.00 m) . The rotational inertia of this system of masses about the Z-axis, perpendicular to the X-Y plane, is

A 20.0 cm wrench is used to generate a torque at a bolt. A force of 50 N is applied at the end of the wrench at an angle of 60.0 degrees to the wrench. The torque generated at the bolt is