Question 1

The unit kg.m²/s can be used for: A) angular momentum B) rotational kinetic energy C) rotational inertia D) torque E) power

Accepted Answer

The answer of The unit kg.m²/s can be used for: A)...

Question 2

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of its angular momentum around the origin is: A) 6.0 kg∙m²/s B) 9.0 kg∙m²/s C) 11 kg∙m²/s D) 14 kg∙m²/s E) 20 kg∙m²/s

Accepted Answer

The angular momentum is defined as L = Iω, where I is the moment of inertia and ω is the angular velocity. In this problem, we need to first find the moment of inertia of the block about the z-axis.

The moment of inertia of a point mass rotating about an axis at a distance r is I = mr^2. For a block of mass M rotating about a circle of radius R, we can write:

I = MR^2

In this problem, R = 0.50 m and M = 2.0 kg, so:

I = 2.0 kg x (0.50 m)^2 = 0.50 kg·m^2

Next, we can calculate the angular momentum:

L = Iω = (0.50 kg·m^2)(12 rad/s) = 6.0 kg·m^2/s

But this is only the magnitude of the angular momentum. We need to also consider the direction. In this case, the angular momentum vector is perpendicular to both the radius vector from the origin to the block and the velocity vector of the block. Since the block is moving counterclockwise, the angular momentum vector points up along the positive z-axis.

Therefore, the magnitude of the angular momentum around the origin is 6.0 kg·m^2/s.

Question 3

A 6.0-kg particle moves to the right at 4.0 m/s as shown. The magnitude of its angular momentum about the point O is: A) 0 kg∙m²/s B) 288 kg∙m²/s C) 144 kg∙m²/s D) 24 kg∙m²/s E) 249 kg∙m²/s

Accepted Answer

The angular momentum L of a particle about point O is given by L = r x p, where r is the position vector from O to the particle, and p is the momentum vector of the particle. Since the particle is moving to the right, its momentum vector is p = mv to the right. We need to find the position vector r and take its cross product with p to find L.

The position vector r is perpendicular to the direction of motion of the particle and passes through O. Therefore, we can draw a line segment from the particle to O that is perpendicular to the direction of motion, and the magnitude of r is the length of this line segment.

We can use the Pythagorean theorem to find the length of this line segment: r = sqrt((1.0 m)^2 + (2.0 m)^2) = sqrt(5) m. (Note that this is the hypotenuse of a right triangle with legs of length 1.0 m and 2.0 m.)

The direction of r is perpendicular to the plane of the diagram and into the page, so we need to use the right-hand rule to determine its direction for the cross product. If we curl the fingers of our right hand from r to p, our thumb points into the page. Therefore, the direction of L is into the page.

The magnitude of L is given by L = rp sin(theta), where theta is the angle between r and p. Since r and p are perpendicular, sin(theta) = 1, so L = rp = (sqrt(5) m)(6.0 kg)(4.0 m/s) = 144 kg m^2/s. Therefore, the answer is choice C.

Question 4

A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:
A) in the positive y direction
B) in the negative y direction
C) in the positive x direction
D) in the negative x direction
E) in the positive z direction

Accepted Answer

The torque about the origin is given by the cross product of the position vector and the force: 
$\vec{	au} = \vec{r} 	imes \vec{F}$ 
Since the torque is in the negative z direction, this means that the force and position vector must be perpendicular to the z-axis. The only force that satisfies this condition is a force in the negative y direction, since the cross product of any vector in the positive x direction with a vector in the negative y direction will be in the negative z direction. Therefore, the force must be in the negative y direction.

Question 5

A force  = 4.2 N + 3.7 N + 1.2 N acts on a particle located at x = 3.3 m. What is the torque on the particle around the origin?
A) 14 N∙m
B) -4.0 N∙m + 12 N∙m
C) 12 N∙m
D) 14 N∙m - 4.0 N∙m + 12 N∙m
E) cannot be calculated without knowing the mass of the particle

Accepted Answer

The answer of A force  = 4.2 N +...

Question 6

The fundamental dimensions of angular momentum are: A) mass·length·time^-1 B) mass·length^-2·time^-2 C) mass²·time^-1 D) mass·length²·time^-2 E) none of these

Accepted Answer

The answer of The fundamental dimensions of angular momentum are:
A)...

Question 7

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s². At the end of 2.0 s its angular momentum about the origin is: A) 0 kg∙m²/s B) (-36 kg∙m²/s) C) (+48 kg∙m²/s) D) (-96 kg∙m²/s) E) (+96 kg∙m²/s)

Accepted Answer

The answer of A 2.0-kg block starts from rest on...

Question 8

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The component in the xy plane of the angular momentum around the origin has magnitude: A) 0 kg∙m²/s B) 6.0 kg∙m²/s C) 9.0 kg∙m²/s D) 11 kg∙m²/s E) 14 kg∙m²/s

Accepted Answer

The answer of A 2.0-kg block travels around a 0.50-m...

Question 9

A particle moves along the x axis. In order to calculate the torque on the particle, you need to know:
A) the velocity of the particle
B) the rotational inertia of the particle
C) the point about which the torque is to be calculated
D) the kinetic energy of the particle
E) the mass of the particle

Accepted Answer

Torque is defined as the cross product of the lever arm (distance from the pivot point to the point of force application) and the force applied. Therefore, knowing the point about which the torque is to be calculated is essential to determine the lever arm distance, which is crucial for calculating torque. The other options, such as velocity, rotational inertia, kinetic energy, and mass, do not directly influence the calculation of torque without considering the force applied and the lever arm distance from a specific point.

Question 10

The newton.second is a unit of:
A) work
B) angular momentum
C) power
D) linear momentum
E) none of these

Accepted Answer

The answer of The newton.second is a unit of:
A) work
B)...

Question 11

Which of the following is NOT a vector?
A) linear momentum
B) angular momentum
C) rotational inertia
D) torque
E) angular velocity

Accepted Answer

Linear momentum (A), angular momentum (B), torque (D), and angular velocity (E) are all vectors because they have both magnitude and direction. Rotational inertia (C) only has magnitude, so it is not a vector.

Question 12

A particle is located on the x axis at x = 2.0 m from the origin. A force of 25 N, directed 30° above the x axis in the x-y plane, acts on the particle. What is the torque about the origin on the particle?
A) 50 N∙m, in the positive z direction
B) 25 N∙m, in the positive z direction
C) 50 N∙m, in the negative z direction
D) 25 N∙m, in the negative z direction
E) There is no torque about the origin.

Accepted Answer

The answer of A particle is located on the x...

Question 13

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, a distance of 0.75 m from the origin. The z component of the angular momentum around the origin is: A) 6.0 kg∙m²/s B) 9.0 kg∙m²/s C) 11 kg∙m²/s D) 14 kg∙m²/s E) 20 kg∙m²/s

Accepted Answer

The answer of A 2.0-kg block travels around a 0.50-m...

Question 14

Two objects are moving in the x, y plane as shown. The magnitude of their total angular momentum (about the origin O) is: A) 0 kg∙m²/s B) 6 kg∙m²/s C) 12 kg∙m²/s D) 30 kg∙m²/s E) 78 kg∙m²/s

Accepted Answer

The answer of Two objects are moving in the x,...

Question 15

Possible units of angular momentum are: A) kg.m/s B) kg.m²/s² C) kg.m/s² D) kg.m²/s E) none of these

Accepted Answer

Angular momentum has units of kgm²/s. 
Therefore, we need a combination of units that includes kg, m, and s, raised to some powers that will give us kgm²/s. 
The only choice that satisfies this criteria is D, which has kg, m, and s units, each raised to the correct powers to give us kgm²/s.

Question 16

A 2.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s. The magnitude of its angular momentum about the center of the circle is: A) 6.0 kg∙m²/s B) 12 kg∙m²/s C) 48 kg∙m²/s D) 72 kg∙m²/s E) 576 kg∙m²/s

Accepted Answer

The answer of A 2.0-kg block travels around a 0.50-m...

Question 17

The angular momentum vector of Earth, due to its daily rotation, is directed:
A) tangent to the equator toward the east
B) tangent to the equator toward the west
C) north
D) south
E) toward the sun

Accepted Answer

The angular momentum vector of Earth due to its daily rotation is directed north because it points perpendicular to the plane of rotation, which, for Earth's rotation around its axis, is along the axis from the south pole to the north pole.

Question 18

Which of the following is a vector quantity?
A) angular speed
B) rotational inertia
C) rotational kinetic energy
D) mass
E) torque

Accepted Answer

Torque is a vector quantity as it has both magnitude and direction. It is a measure of the force that is causing an object to rotate around an axis. The other options listed are scalar quantities as they do not have direction associated with them.

Question 19

A 15-g paper clip is attached to the rim of a phonograph record with a diameter of 30 cm, spinning at 3.5 rad/s. The magnitude of its angular momentum is: A) 1.2 *10^-3 kg∙m²/s B) 4.7 * 10^-3 kg∙m²/s C) 7.9 * 10^-³ kg∙m²/s D) 1.6 * 10^-² kg∙m²/s E) 1.2 kg∙m²/s

Accepted Answer

The answer of A 15-g paper clip is attached to...

Question 20

A particle moves along the x axis. In order to calculate the angular momentum of the particle, you need to know:
A) the size of the particle
B) the rotational inertia of the particle
C) the point about which the angular momentum is to be calculated
D) the kinetic energy of the particle
E) the acceleration of the particle

Accepted Answer

The angular momentum of a particle is defined as the cross product of the particle's position vector and its linear momentum vector with respect to a point. Therefore, in order to calculate the angular momentum, you need to know the point about which the angular momentum is being calculated. The size of the particle, rotational inertia, kinetic energy, and acceleration are not directly related to the calculation of angular momentum.

Quiz 11: Rolling, Torque, and Angular Momentum

The unit kg.m2/s can be used for:

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of its angular momentum around the origin is:

A 6.0-kg particle moves to the right at 4.0 m/s as shown. The magnitude of its angular momentum about the point O is:

A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:

A force = 4.2 N + 3.7 N + 1.2 N acts on a particle located at x = 3.3 m. What is the torque on the particle around the origin?

The fundamental dimensions of angular momentum are:

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s2. At the end of 2.0 s its angular momentum about the origin is:

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The component in the xy plane of the angular momentum around the origin has magnitude:

A particle moves along the x axis. In order to calculate the torque on the particle, you need to know:

The newton.second is a unit of:

Which of the following is NOT a vector?

A particle is located on the x axis at x = 2.0 m from the origin. A force of 25 N, directed 30° above the x axis in the x-y plane, acts on the particle. What is the torque about the origin on the particle?

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, a distance of 0.75 m from the origin. The z component of the angular momentum around the origin is:

Two objects are moving in the x, y plane as shown. The magnitude of their total angular momentum (about the origin O) is:

Possible units of angular momentum are:

A 2.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s. The magnitude of its angular momentum about the center of the circle is:

The angular momentum vector of Earth, due to its daily rotation, is directed:

Which of the following is a vector quantity?

A 15-g paper clip is attached to the rim of a phonograph record with a diameter of 30 cm, spinning at 3.5 rad/s. The magnitude of its angular momentum is:

A particle moves along the x axis. In order to calculate the angular momentum of the particle, you need to know:

The unit kg.m²/s can be used for:

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s². At the end of 2.0 s its angular momentum about the origin is: