Question 1

A $2.00 \mathrm{~kg}$ mass is located at $(4.00 \mathrm{~m}, 0.00 \mathrm{~m}, 0.00 \mathrm{~m})$ and a $4.00 \mathrm{~kg}$ mass is located at $(0.00 \mathrm{~m}, 3.00 \mathrm{~m}$, $0.00 \mathrm{~m}$ ). The rotational inertia of this system of masses about the Z-axis, perpendicular to the X-Y plane, is
A) $68 \mathrm{~kg} \mathrm{~m}^{2}$.
B) $55 \mathrm{~kg} \mathrm{~m}^{2}$.
C) $62 \mathrm{~kg} \mathrm{~m}^{2}$.
D) $50 \mathrm{~kg} \mathrm{~m}^{2}$.
E) $58 \mathrm{~kg} \mathrm{~m}^{2}$.

Accepted Answer

The rotational inertia of a system of masses about an axis is the sum of the rotational inertias of the individual masses about that axis. The rotational inertia of a point mass $m$ about an axis a distance $r$ from the mass is $mr^2$. Therefore, the rotational inertia of the $2.00 \mathrm{~kg}$ mass about the Z-axis is $(2.00 \mathrm{~kg})(4.00 \mathrm{~m})^2 = 32.0 \mathrm{~kg} \mathrm{~m}^{2}$, and the rotational inertia of the $4.00 \mathrm{~kg}$ mass about the Z-axis is $(4.00 \mathrm{~kg})(3.00 \mathrm{~m})^2 = 36.0 \mathrm{~kg} \mathrm{~m}^{2}$. The total rotational inertia of the system about the Z-axis is therefore $32.0 \mathrm{~kg} \mathrm{~m}^{2} + 36.0 \mathrm{~kg} \mathrm{~m}^{2} = 68.0 \mathrm{~kg} \mathrm{~m}^{2}$.

Question 2

A $2.00 \mathrm{~kg}$ mass is located at $(4.00 \mathrm{~m}, 0.00 \mathrm{~m}, 0.00 \mathrm{~m})$ and a $4.00 \mathrm{~kg}$ mass is located at $(0.00 \mathrm{~m}, 3.00 \mathrm{~m}$, $0.00 \mathrm{~m}$ ). The rotational inertia of this system of masses about the $\mathrm{X}$-axis, perpendicular to the Z-Y plane, is
A) $28 \mathrm{~kg} \mathrm{~m}^{2}$.
B) $36 \mathrm{~kg} \mathrm{~m}^{2}$.
C) $33 \mathrm{~kg} \mathrm{~m}^{2}$.
D) $23 \mathrm{~kg} \mathrm{~m}^{2}$.
E) $41 \mathrm{~kg} \mathrm{~m}^{2}$.

Accepted Answer

The rotational inertia (moment of inertia) $I$ about an axis for point masses is given by $I = \sum m_i r_i^2$, where $m_i$ is the mass of the $i$th point and $r_i$ is the distance of the $i$th point from the axis of rotation. For rotation about the X-axis, only the Y and Z coordinates contribute to the distance $r$. The first mass is on the X-axis, so its contribution is $0$ because its distance from the X-axis is $0$. The second mass is 3.00 m away from the X-axis (all in the Y direction), so its contribution is $4.00 \mathrm{~kg} \times (3.00 \mathrm{~m})^2 = 36 \mathrm{~kg} \cdot \mathrm{m}^2$.

Question 3

What is the rotational inertia of a solid iron disk of mass $41.0 \mathrm{~kg}$ with a thickness of $5.00 \mathrm{~cm}$ and radius of $30.0 \mathrm{~cm}$, about an axis perpendicular to the disk and passing through its center?
A) $0.980 \mathrm{~kg} \mathrm{~m}^{2}$
B) $0.761 \mathrm{~kg} \mathrm{~m}^{2}$
C) $2.29 \mathrm{~kg} \mathrm{~m}^{2}$
D) $1.85 \mathrm{~kg} \mathrm{~m}^{2}$

Accepted Answer

D

Question 4

A $4.00 \mathrm{~kg}$ mass is located at $(2.00 \mathrm{~m}, 2.00 \mathrm{~m}, 0.00 \mathrm{~m})$ and a $3.00 \mathrm{~kg}$ mass is located at $(-1.0 \mathrm{~m}, 3.00$ $\mathrm{m}, 0.00 \mathrm{~m}$ ). The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is
A) $24 \mathrm{~kg} \mathrm{~m}^{2}$.
B) $56 \mathrm{~kg} \mathrm{~m}^{2}$.
C) $62 \mathrm{~kg} \mathrm{~m}^{2}$.
D) $43 \mathrm{~kg} \mathrm{~m}^{2}$.
E) $36 \mathrm{~kg} \mathrm{~m}^{2}$.

Accepted Answer

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Question 5

A $4.00 \mathrm{~kg}$ mass is located at $(2.00 \mathrm{~m}, 2.00 \mathrm{~m}, 0.00 \mathrm{~m})$ and a $3.00 \mathrm{~kg}$ mass is located at $(-1 \mathrm{~m}, 3.00$ $\mathrm{m}, 0.00 \mathrm{~m}$ ). The rotational inertia of this system of masses about the Y-axis, perpendicular to the X-Z plane, is
A) $32 \mathrm{~kg} \mathrm{~m}^{2}$.
B) $29 \mathrm{~kg} \mathrm{~m}^{2}$.
C) $40 \mathrm{~kg} \mathrm{~m}^{2}$.
D) $24 \mathrm{~kg} \mathrm{~m}^{2}$.
E) $19 \mathrm{~kg} \mathrm{~m}^{2}$.

Accepted Answer

The answer of A $4.00 \mathrm{~kg}$ mass is located at...

Question 6

A $6.00 \mathrm{~kg}$ mass is located at $(2.00 \mathrm{~m}, 2.00 \mathrm{~m}, 2.00 \mathrm{~m})$ and a $5.00 \mathrm{~kg}$ mass is located at $(-1.0 \mathrm{~m}, 3.00 \mathrm{~m}$, $-2.00 \mathrm{~m}$ ). The rotational inertia of this system of masses about the Z-axis, perpendicular to the $\mathrm{X}-\mathrm{Y}$ plane, is
A) $98 \mathrm{~kg} \mathrm{~m}^{2}$.
B) $79 \mathrm{~kg} \mathrm{~m}^{2}$.
C) $60 \mathrm{~kg} \mathrm{~m}^{2}$.
D) $112 \mathrm{~kg} \mathrm{~m}^{2}$.
E) $85 \mathrm{~kg} \mathrm{~m}^{2}$.

Accepted Answer

The rotational inertia of a system of masses about an axis is the sum of the rotational inertias of the individual masses about that axis. The rotational inertia of a point mass $m$ about an axis a distance $r$ from the mass is $mr^2$.Therefore, the rotational inertia of the 6.00 kg mass about the Z-axis is:$$I_1 = (6.00 \mathrm{~kg})(2.00 \mathrm{~m})^2 = 24.0 \mathrm{~kg} \mathrm{~m}^2$$The rotational inertia of the 5.00 kg mass about the Z-axis is:$$I_2 = (5.00 \mathrm{~kg})[(1.0 \mathrm{~m})^2 + (3.00 \mathrm{~m})^2 + (2.00 \mathrm{~m})^2] = 55.0 \mathrm{~kg} \mathrm{~m}^2$$Therefore, the total rotational inertia of the system of masses about the Z-axis is:$$I = I_1 + I_2 = 24.0 \mathrm{~kg} \mathrm{~m}^2 + 55.0 \mathrm{~kg} \mathrm{~m}^2 = 79.0 \mathrm{~kg} \mathrm{~m}^2$$Therefore, the correct answer is B) $79 \mathrm{~kg} \mathrm{~m}^{2}$.

Question 7

A $6.00 \mathrm{~kg}$ mass is located at $(2.00 \mathrm{~m}, 2.00 \mathrm{~m}, 2.00 \mathrm{~m})$ and a $5.00 \mathrm{~kg}$ mass is located at $(-1.0 \mathrm{~m}, 3.00 \mathrm{~m}$, $-2.00 \mathrm{~m}$ ). The rotational inertia of this system of masses about the X-axis, perpendicular to the Z-Y plane, is
A) $167 \mathrm{~kg} \mathrm{~m}^{2}$.
B) $281 \mathrm{~kg} \mathrm{~m}^{2}$.
C) $113 \mathrm{~kg} \mathrm{~m}^{2}$.
D) $69 \mathrm{~kg} \mathrm{~m}^{2}$.
E) $85 \mathrm{~kg} \mathrm{~m}^{2}$.

Accepted Answer

The answer of A $6.00 \mathrm{~kg}$ mass is located at...

Question 8

A $6.00 \mathrm{~kg}$ mass is located at $(2.00 \mathrm{~m}, 2.00 \mathrm{~m}, 2.00 \mathrm{~m})$ and a $5.00 \mathrm{~kg}$ mass is located at $(-1.0 \mathrm{~m}, 3.00 \mathrm{~m}$, $-2.00 \mathrm{~m}$ ). The rotational inertia of this system of masses about the Y-axis, perpendicular to the Z-X plane, is
A) $55 \mathrm{~kg} \mathrm{~m}^{2}$.
B) $73 \mathrm{~kg} \mathrm{~m}^{2}$.
C) $60 \mathrm{~kg} \mathrm{~m}^{2}$.
D) $66 \mathrm{~kg} \mathrm{~m}^{2}$.
E) $48 \mathrm{~kg} \mathrm{~m}^{2}$.

Accepted Answer

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Question 9

The  rotational inertia of a thin rod about one end is $1 / 3 \mathrm{ML}^{2}$. What is the rotational inertia of the same rod about a point located $0.300 \mathrm{~L}$ from the end?
A) $0.300 \mathrm{ML}^{2}$
B) $0.205 \mathrm{ML}^{2}$
C) $0.240 \mathrm{ML}^{2}$
D) $0.198 \mathrm{ML}^{2}$
E) $0.123 \mathrm{ML}^{2}$

Accepted Answer

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Question 10

The rotational inertia of a thin rod about one end is $1 / 3 \mathrm{ML}^{2}$. What is the rotational inertia of the same rod about a point located $0.40 \mathrm{~L}$ from the end?
A) $0.093 \mathrm{ML}^{2}$
B) $0.493 \mathrm{ML}^{2}$
C) $0.073 \mathrm{ML}^{2}$
D) $0.243 \mathrm{ML}^{2}$
E) $0.056 \mathrm{ML}^{2}$

Accepted Answer

The rotational inertia of a thin rod about an axis perpendicular to the rod and passing through one end is $1 / 3 \mathrm{ML}^{2}$. The rotational inertia about a parallel axis a distance $d$ from the first axis is given by $$I=I_{CM}+Md^2$$where $I_{CM}$ is the rotational inertia about an axis through the center of mass. For a thin rod, $I_{CM}=1 / 12 \mathrm{ML}^{2}$. So, for a point located $0.40 \mathrm{~L}$ from the end, we have$$I=\frac{1}{12}ML^2+M(0.40L)^2=\frac{1}{3}ML^2$$

Question 11

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

Accepted Answer

Torque is calculated as the product of force and the perpendicular distance from the pivot point, which in this case is $Torque = 40 \mathrm{~N} \times 0.30 \mathrm{~m} = 12 \mathrm{~N} \cdot \mathrm{m}$.

Question 12

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

Accepted Answer

The torque ($\tau$) generated by a force applied at an angle to a lever arm is given by the formula $\tau = rF\sin(\theta)$, where $r$ is the length of the lever arm (in meters), $F$ is the force applied (in Newtons), and $\theta$ is the angle between the force and the lever arm (in degrees). Here, $r = 0.30 \, \text{m}$, $F = 50.0 \, \text{N}$, and $\theta = 70.0^\circ$. Plugging these values into the formula gives $\tau = 0.30 \times 50.0 \times \sin(70.0^\circ) \approx 14.1 \, \text{N} \cdot \text{m}$.

Question 13

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

Accepted Answer

The work done in turning the bolt can be calculated using the formula $W = \tau \theta$, where $W$ is the work done, $\tau$ is the torque applied, and $\theta$ is the angle in radians. First, convert the angle from degrees to radians: $180^\circ = \pi$ radians. Then, calculate the work: $W = 20.0 \, \text{N} \cdot \text{m} \times \pi = 62.8 \, \text{J}$.

Question 14

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

Accepted Answer

The answer of A torque of $15.0 \mathrm{~N} \cdot \mathrm{m}$...

Question 15

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

Accepted Answer

The center of gravity (CG) can be calculated using the formula $\text{CG} = \frac{\sum m_i x_i}{\sum m_i}$ for each dimension (x, y, z). For the x-coordinate: CG_x = $\frac{(2 \times 4) + (4 \times 0)}{2 + 4} = \frac{8}{6} = 1.33 \, \text{m}$. For the y-coordinate: CG_y = $\frac{(2 \times 0) + (4 \times 3)}{2 + 4} = \frac{12}{6} = 2.00 \, \text{m}$. The z-coordinate is 0 since all masses are in the x-y plane. Therefore, the center of gravity is at $(1.33 \, \text{m}, 2.00 \, \text{m}, 0)$.

Question 16

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

Accepted Answer

The center of gravity is calculated using the formula: $(x_{cg}, y_{cg}, z_{cg}) = \left(\frac{\sum m_ix_i}{\sum m_i}, \frac{\sum m_iy_i}{\sum m_i}, \frac{\sum m_iz_i}{\sum m_i}\right)$. For the given masses, $x_{cg} = \frac{5 \times 2 + 3 \times 0}{5 + 3} = 1.25$ and $y_{cg} = \frac{5 \times 0 + 3 \times 4}{5 + 3} = 1.50$. Thus, the center of gravity is $(1.25 \mathrm{~m}, 1.50 \mathrm{~m}, 0)$.

Question 17

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

Accepted Answer

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Question 18

A $5.00 \mathrm{~kg}$ mass is located at $(1.0 \mathrm{~m}, 0.00 \mathrm{~m}, 3.00 \mathrm{~m})$, a $2.00 \mathrm{~kg}$ mass is located at $(0.00 \mathrm{~m}, 3.00 \mathrm{~m},-2.00$ $\mathrm{m})$, and a $3.00 \mathrm{~kg}$ mass is located at $(-1.0 \mathrm{~m},-2.00 \mathrm{~m}, 0.00 \mathrm{~m})$. The center of gravity of the system of masses is
A) $(10 / 10 \mathrm{~m}, 2 / 10 \mathrm{~m}, 3 / 10 \mathrm{~m})$.
B) $(1 / 10 \mathrm{~m}, 10 / 10 \mathrm{~m}, 1 / 10 \mathrm{~m})$.
C) $(3 / 10 \mathrm{~m}, 2 / 10 \mathrm{~m}, 10 / 10 \mathrm{~m})$.
D) $(2 / 10 \mathrm{~m}, 0 \mathrm{~m}, 11 / 10 \mathrm{~m})$.
E) $(2 / 10 \mathrm{~m}, 10 / 10 \mathrm{~m}, 0 \mathrm{~m})$.

Accepted Answer

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Question 19

A $6.00 \mathrm{~kg}$ mass is located at $(1.0 \mathrm{~m},-2.00 \mathrm{~m}, 3.00 \mathrm{~m})$, a $5.00 \mathrm{~kg}$ mass is located at $(1.0 \mathrm{~m}, 3.00 \mathrm{~m},-2.00$ $\mathrm{m})$, and a $4.00 \mathrm{~kg}$ mass is located at $(-1.0 \mathrm{~m},-2.00 \mathrm{~m}, 2.00 \mathrm{~m})$. The center of gravity of the system of masses is
A) $(5 / 15 \mathrm{~m},-17 / 15 \mathrm{~m}, 5 / 15 \mathrm{~m})$.
B) $(12 / 15 \mathrm{~m},-5 / 15 \mathrm{~m}, 16 / 15 \mathrm{~m})$.
C) $(7 / 15 \mathrm{~m},-5 / 15 \mathrm{~m}, 16 / 15 \mathrm{~m})$.
D) $(16 / 15 \mathrm{~m},-1 / 15 \mathrm{~m}, 17 / 15 \mathrm{~m})$.
E) $(5 / 15 \mathrm{~m},-1 / 15 \mathrm{~m}, 17 / 15 \mathrm{~m})$.

Accepted Answer

The center of gravity (CG) of a system of masses is calculated using the formula $\vec{r}_{CG} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$, where $m_i$ and $\vec{r}_i$ are the mass and position vector of the $i^{th}$ mass, respectively. For the given system:- The total mass is $6 + 5 + 4 = 15 \, \mathrm{kg}$.- The x-coordinate of the CG is $\frac{(6 \cdot 1) + (5 \cdot 1) + (4 \cdot -1)}{15} = \frac{6 + 5 - 4}{15} = \frac{7}{15} \, \mathrm{m}$.- The y-coordinate of the CG is $\frac{(6 \cdot -2) + (5 \cdot 3) + (4 \cdot -2)}{15} = \frac{-12 + 15 - 8}{15} = \frac{-5}{15} \, \mathrm{m}$.- The z-coordinate of the CG is $\frac{(6 \cdot 3) + (5 \cdot -2) + (4 \cdot 2)}{15} = \frac{18 - 10 + 8}{15} = \frac{16}{15} \, \mathrm{m}$.Thus, the center of gravity is at $(7 / 15 \, \mathrm{m}, -5 / 15 \, \mathrm{m}, 16 / 15 \, \mathrm{m})$.

Question 20

A $1.500 \mathrm{~m}$ long uniform beam of mass $30.00 \mathrm{~kg}$ is supported by a wire as shown in the figure. The beam makes an angle of 10.00 degrees with the horizontal and the wire makes an angle of 30.00 degrees with the beam. A $50.00 \mathrm{~kg}$ mass, $\mathrm{m}$, is attached to the end of the beam. What is the tension in the wire?

A) $1,855 \mathrm{~N}$
B) $1,255 \mathrm{~N}$
C) $1,435 \mathrm{~N}$
D) $2,034 \mathrm{~N}$
E) $1,035 \mathrm{~N}$

Accepted Answer

The answer of A $1.500 \mathrm{~m}$ long uniform beam of...

Quiz 8: Torque and Angular Momentum

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

The rotational inertia of a thin rod about one end is $1 / 3 \mathrm{ML}^{2}$ . What is the rotational inertia of the same rod about a point located $0.300 \mathrm{~L}$ from the end?

The rotational inertia of a thin rod about one end is $1 / 3 \mathrm{ML}^{2}$ . What is the rotational inertia of the same rod about a point located $0.40 \mathrm{~L}$ from the end?

A $30.0 \mathrm{~cm}$ wrench is used to generate a torque at a bolt. A force of $40 \mathrm{~N}$ is applied perpendicularly at the end of the wrench. The torque generated at the bolt is

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

A torque of $20.0 \mathrm{~N} \cdot \mathrm{m}$ is applied to a bolt. The bolt rotates through an angle of 180 degrees. The work done in turning the bolt is

A torque of $15.0 \mathrm{~N} \cdot \mathrm{m}$ is applied to a bolt. The bolt rotates through an angle of 360 degrees. The work done in turning the bolt is

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

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

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

Quiz 8: Torque and Angular Momentum

What is the rotational inertia of a solid iron disk of mass 41.0 kg41.0 \mathrm{~kg}41.0 kg with a thickness of 5.00 cm5.00 \mathrm{~cm}5.00 cm and radius of 30.0 cm30.0 \mathrm{~cm}30.0 cm , about an axis perpendicular to the disk and passing through its center?

The rotational inertia of a thin rod about one end is 1/3ML21 / 3 \mathrm{ML}^{2}1/3ML2 . What is the rotational inertia of the same rod about a point located 0.300 L0.300 \mathrm{~L}0.300 L from the end?

The rotational inertia of a thin rod about one end is 1/3ML21 / 3 \mathrm{ML}^{2}1/3ML2 . What is the rotational inertia of the same rod about a point located 0.40 L0.40 \mathrm{~L}0.40 L from the end?

A 30.0 cm30.0 \mathrm{~cm}30.0 cm wrench is used to generate a torque at a bolt. A force of 40 N40 \mathrm{~N}40 N is applied perpendicularly at the end of the wrench. The torque generated at the bolt is

A 30.0 cm30.0 \mathrm{~cm}30.0 cm wrench is used to generate a torque at a bolt. A force of 50.0 N50.0 \mathrm{~N}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 torque of 20.0 N⋅m20.0 \mathrm{~N} \cdot \mathrm{m}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 torque of 15.0 N⋅m15.0 \mathrm{~N} \cdot \mathrm{m}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