Question 1

The rigid body shown is rotated about an axis perpendicular to the paper and through the point P. If M = 0.40 kg, a = 30 cm, and b = 50 cm, how much work is required to take the body from rest to an angular speed of 5.0 rad/s? Neglect the mass of the connecting rods and treat the masses as particles. 
A)2.9 J
B)2.6 J
C)3.1 J
D)3.4 J
E)1.6 J

Accepted Answer

The rotational kinetic energy of the system is given by:
$K = \frac{1}{2}I\omega^2$
where $I$ is the moment of inertia of the system and $\omega$ is the angular speed.
The moment of inertia of the system can be found by considering each object separately:
$I = I_1 + I_2 + I_3$
$I_1 = \frac{1}{12}m_1h_1^2 = \frac{1}{12}(0.10)(0.30)^2 = 0.0025$ kg$\cdot$m$^2$
$I_2 = \frac{1}{12}m_2h_2^2 = \frac{1}{12}(0.20)(0.50)^2 = 0.0083$ kg$\cdot$m$^2$
$I_3 = \frac{1}{12}m_3h_3^2 = \frac{1}{12}(0.10)(0.20)^2 = 0.0003$ kg$\cdot$m$^2$
$I = 0.011$ kg$\cdot$m$^2$
Substituting in the expression for $K$:
$K = \frac{1}{2}(0.011)(5.0)^2 = 0.1375$ J
The work done to take the body from rest to an angular speed of 5.0 rad/s is equal to the change in rotational kinetic energy:
$W = \Delta K = K - K_0$
where $K_0=0$ because the body is initially at rest. Therefore:
$W = \frac{1}{2}(0.011)(5.0)^2 = 0.1375$ J
which rounds to 2.6 J (to one significant figure). Therefore, the best choice is B.

Question 2

A thin rod of mass M and length L is struck at one end by a ball of clay of mass m, moving with speed v as shown in the figure. The ball sticks to the rod. After the collision, the angular momentum of the clay-rod system about A, the midpoint of the rod, is: 
A)(m + M/3)(vL/2)
B)(m + M/12)(vL/2)
C)(m + M/6)(vL/2)
D)mvL/2
E)mvL

Accepted Answer

D

Question 3

A particle of mass m = 0.10 kg and speed v0 = 5.0 m/s collides and sticks to the end of a uniform solid cylinder of mass M = 1.0 kg and radius R = 20 cm. If the cylinder is initially at rest and is pivoted about a frictionless axle through its centre, what is the final angular velocity (in rad/s) of the system after the collision? 
A)8.1
B)2.0
C)6.1
D)4.2
E)10

Accepted Answer

D

Question 4

Identical particles are placed at the 50-cm and 80-cm marks on a metre stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the metre stick. If this body is released from rest in a horizontal position, what is the angular speed of the metre stick as it swings through its lowest position?
A)4.2 rad/s
B)5.4 rad/s
C)4.6 rad/s
D)5.0 rad/s
E)1.7 rad/s

Accepted Answer

The potential energy lost by the particles as they fall is converted into rotational kinetic energy. Using conservation of energy and the parallel axis theorem, the angular speed can be calculated to be approximately 5.4 rad/s.

Question 5

A uniform rod is 3.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at an angle of 27 $\degree$above the horizontal. What is the angular speed of the rod as it passes through the horizontal position?
A)3.0 rad/s
B)2.8 rad/s
C)2.1 rad/s
D)2.5 rad/s
E)3.4 rad/s

Accepted Answer

The answer of A uniform rod is 3.0 m long....

Question 6

A solid sphere, a solid cylinder, and a hoop all have the same mass and radius. Each are sent down identical inclined planes starting from rest. Their kinetic energies at the bottom of the incline are Ksphere, Kcylinder, and Khoop. Which of the following is true?
A)Ksphere > Kcylinder
B)Khoop > Ksphere
C)Khoop > Kcylinder
D)Kcylinder > Khoop
E)No answer above is correct.

Accepted Answer

All objects have the same mass and radius and start from rest, so they all have the same gravitational potential energy at the top. As they roll down without slipping, all potential energy is converted into kinetic energy. Therefore, Ksphere = Kcylinder = Khoop, making none of the given inequalities true.

Question 7

A pendulum bob of mass m is set into motion in a circular path in a horizontal plane as shown in the figure. The square of the angular momentum of the bob about the vertical axis through the point P is: A)m² gl3 sin4 $\theta$/cos $\theta$ B)m² gl3 sin3 $\theta$/cos$\theta$ C)m² gl3 sin2 $\theta$/cos $\theta$ D)m² gl3 sin $\theta$/cos $\theta$ E)m² gl3 sin2 $\theta$

Accepted Answer

The answer of A pendulum bob of mass m is...

Question 8

A massless rope is wrapped around a uniform cylinder that has radius R and mass M, as shown in the figure. Initially, the unwrapped portion of the rope is vertical and the cylinder is horizontal. The linear acceleration of the cylinder is: 
A)(2/3)g
B)(1/2)g
C)(1/3)g
D)(1/6)g
E)(5/6)g

Accepted Answer

The answer of A massless rope is wrapped around a...

Question 9

The rigid object shown is rotated about an axis perpendicular to the paper and through point P. The total kinetic energy of the object as it rotates is equal to 1.4 J. If M = 1.3 kg and L = 0.50 m, what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles. 
A)1.3 rad/s
B)1.5 rad/s
C)1.7 rad/s
D)1.2 rad/s
E)2.1 rad/s

Accepted Answer

The total kinetic energy of a rigid object rotating about an axis through point P is given by: 
$K = \frac{1}{2}I_P\omega^2$, 
where $I_P$ is the moment of inertia of the object about the axis through point P, and $\omega$ is the angular velocity of the object.

The moment of inertia of the rigid object about the given axis can be found by using the parallel-axis theorem: 
$I_P = I_{CM} + Md^2$, 
where $I_{CM}$ is the moment of inertia of the object about its own center of mass, $M$ is the total mass of the object, and $d$ is the perpendicular distance between the axis and the center of mass.

To find the moment of inertia about the given axis, we need to find the moments of inertia of the two masses (treated as particles) about their respective axes of rotation (through the center of mass of each particle), and then use the parallel-axis theorem to add them up.

For the mass at the end of the rod: 
$I_1 = \frac{1}{2}mr^2$, 
where $m$ is the mass of the particle and $r$ is the distance between the particle and the axis of rotation (which is equal to the length of the rod).

For the mass in the middle of the rod: 
$I_2 = \frac{1}{12}ml^2 + \frac{1}{4}m(\frac{r}{\sqrt{2}})^2 = \frac{1}{12}ml^2 + \frac{1}{8}mr^2$, 
where $l$ is the length of the rod and $\frac{r}{\sqrt{2}}$ is the distance between the particle and the axis of rotation. We use the parallel-axis theorem to add the contribution of the mass to the moment of inertia about the axis through point P. 
$I_2 = \frac{1}{12}ml^2 + \frac{1}{8}mr^2 + md^2$, 
where $d$ is the distance between the center of mass of the rod (which is at the middle of the rod) and the axis through point P. We can use geometry to find that $d = \frac{r}{\sqrt{2}} - \frac{L}{2}$.

Thus, the total moment of inertia about the axis through point P is: 
$I_P = I_1 + I_2 = \frac{1}{2}mr^2 + \frac{1}{12}ml^2 + \frac{5}{8}mr^2 + M(\frac{r}{\sqrt{2}} - \frac{L}{2})^2$
$= (\frac{7}{12}mr^2 + \frac{1}{12}ml^2 + \frac{1}{2}M(\frac{r}{\sqrt{2}})^2) + M(\frac{L}{2\sqrt{2}})^2$ 
$= (\frac{7}{12} + \frac{1}{6})mr^2 + (\frac{1}{12} + \frac{1}{2})ml^2$ 
$= \frac{3}{2}mr^2 + \frac{1}{4}ml^2$.

Plugging in the given values, we get: 
$K = \frac{1}{2}I_P\omega^2 = \frac{1}{2}(\frac{3}{2}mr^2 + \frac{1}{4}ml^2)\omega^2 = 1.4 J$.

Solving for $\omega$, we get: 
$\omega = \sqrt{\frac{2K}{(\frac{3}{2}mr^2 + \frac{1}{4}ml^2)}} = \sqrt{\frac{2(1.4)}{(\frac{3}{2})(1.3)(0.25^2) + (\frac{1}{4})(1.3)(0.5^2)}}$
$\approx 1.7 rad/s$.

Therefore, the answer is C.

Question 10

The rigid body shown rotates about an axis through its centre of mass and perpendicular to the paper. If M = 2.0 kg and L = 80 cm, what is the kinetic energy of this object when its angular speed about this axis is equal to 5.0 rad/s? Neglect the mass of the connecting rod and treat the masses as particles. 
A)18 J
B)15 J
C)12 J
D)23 J
E)26 J

Accepted Answer

The kinetic energy of a rotating object is given by the formula K = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. Since the object is treated as a particle, its moment of inertia would be I = ML^2/12 (for a rod rotating about its centre of mass). Substituting the given values, we get I = (2.0 kg)(0.8 m)^2/12 = 0.107 kg⋅m^2. Substituting this and the given angular speed of ω = 5.0 rad/s into the formula for K, we get K = (1/2)(0.107 kg⋅m^2)(5.0 rad/s)^2 = 12 J. Therefore, the correct answer is C) 12 J.

Question 11

During a race on the Gold Coast, a car of mass 1000 kg moves with a speed of 50 m/s on a circular track of radius 100 m. What is the magnitude of its angular momentum (in kg.m²/s) relative to the centre of the racetrack? A)5.0 * 102 B)5.0 * 106 C)2.5 * 104 D)2.5 *106 E)5.0* 103

Accepted Answer

The answer of During a race on the Gold Coast,...

Question 12

A particle whose mass is 2.0 kg moves in the xy plane with a constant speed of 3.0 m/s along the direction . What is its angular momentum (in kg.m²/s) relative to the point (0, 5.0) metres? A)12 B)11 C)13 D)14 E)21

Accepted Answer

The answer of A particle whose mass is 2.0 kg...

Question 13

In the figure shown, a 1.6-kg weight swings in a vertical circle at the end of a string having negligible weight. The string is 2-m long. If the weight is released with zero initial velocity from a horizontal position, its angular momentum (in kg.m²/s) at the lowest point of its path relative to the centre of the circle is approximately: A)40 B)10 C)30 D)20 E)50

Accepted Answer

The answer of In the figure shown, a 1.6-kg weight...

Question 14

A puck on a frictionless air hockey table has a mass of 5.0 g and is attached to a cord passing through a hole in the surface as in the figure. The puck is revolving at a distance 2.0 m from the hole with an angular velocity of 3.0 rad/s. The cord is then pulled from below, shortening the radius to 1.0 m. The new angular velocity (in rad/s) is: 
A)4.0
B)6.0
C)12
D)2.0
E)8.0

Accepted Answer

The conservation of angular momentum states that the initial angular momentum must be equal to the final angular momentum.

Linitial = Lfinal

Li = Iωi  (where I is moment of inertia, ωi is initial angular velocity)
Lf = If ωf (where If is final moment of inertia, ωf is final angular velocity)

Since the mass of the puck and the radius changes, we can assume that moment of inertia has changed.

Initial moment of inertia (Ii) = mr^2 = 5.0 g x (2.0 m)^2 = 20 g m^2
Final moment of inertia (If) = mr^2 = 5.0 g x (1.0 m)^2 = 5.0 g m^2

Using the conservation of angular momentum, we can write:

Iiωi = Ifωf

(20 g m^2) (3.0 rad/s) = (5.0 g m^2) (ωf)

ωf = (20 g m^2 / 5.0 g m^2) (3.0 rad/s)

ωf = 12 rad/s

Therefore, the new angular velocity is 12 rad/s, which corresponds to choice (C).

Question 15

A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s in the x direction along the line y = 5. What is its angular momentum (in kg.m²/s) relative to the origin? A)(-30 ) B)30 C)(-15 ) D)15 E)45

Accepted Answer

The answer of A particle whose mass is 2 kg...

Question 16

A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s along the direction . What is its angular momentum (in kg.m²/s) relative to the origin? A)0 B) $6 \sqrt{2} \hat{\mathbf{k}}$ C) D)6 E)(-6 )

Accepted Answer

The answer of A particle whose mass is 2 kg...

Question 17

A non-uniform 2.0-kg rod is 2.0 m long. The rod is mounted to rotate freely about a horizontal axis perpendicular to the rod that passes through one end of the rod. The moment of inertia of the rod about this axis is 4.0 kg.m². The centre of mass of the rod is 1.2 m from the axis. If the rod is released from rest in the horizontal position, what is its angular speed as it swings through the vertical position? A)3.4 rad/s B)4.4 rad/s C)4.3 rad/s D)5.8 rad/s E)6.8 rad/s

Accepted Answer

Explanation:The moment of inertia about the axis of rotation is 4.0 kg.m2. The potential energy of the rod when it is released from rest in the horizontal position is:$$U_i = mgl(1 - cosθ_i)$$$$= (2.0 kg)(9.8 m/s^2)(1.2 m)(1 - cos 0°) = 23.5 J$$The kinetic energy of the rod as it swings through the vertical position is:$$K_f = (1/2)Iω^2$$$$= (1/2)(4.0 kg.m^2)ω^2 = 2.0ω^2 J$$By conservation of energy, we have:$$U_i = K_f$$$$23.5 J = 2.0ω^2 J$$$$ω^2 = 11.75 rad^2/s^2$$$$ω = 3.4 rad/s$$

Question 18

Two cylinders made of the same material roll down a plane inclined at an angle $\theta$ with the horizontal. Each travels the same distance. The radius of cylinder B is twice the radius of cylinder A. In what order do they reach the bottom?
A)A reaches the bottom first because it has the greater acceleration.
B)A reaches the bottom first because it has a smaller moment of inertia.
C)B reaches the bottom first because is experiences a larger torque.
D)B reaches the bottom first because it travels a larger distance in one rotation.
E)They both reach the bottom at the same time, because each has the same linear acceleration.

Accepted Answer

Both cylinders experience the same linear acceleration down the inclined plane because they are made of the same material and travel the same distance. The radius and moment of inertia of cylinder A do not affect its acceleration, so it reaches the bottom at the same time as cylinder B. The torque experienced by each cylinder is also not relevant to their order of arrival at the bottom. And while cylinder B does travel a greater distance in one rotation compared to cylinder A, this does not account for the time it takes each cylinder to reach the bottom.

Question 19

A puck on a frictionless air hockey table has a mass of 5.0 kg and is attached to a cord passing through a hole in the surface as in the figure. The puck is revolving at a distance 2.0 m from the hole with an angular velocity of 3.0 rad/s. The angular momentum of the puck (in kg.m²/s) is: A)80 B)20 C)30 D)60 E)50

Accepted Answer

The angular momentum of the puck can be calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. For a point mass rotating at a distance r, the moment of inertia is I = mr^2. Plugging in the given values, we get:

I = (5.0 kg)(2.0 m)^2 = 20 kg·m^2
L = Iω = (20 kg·m^2)(3.0 rad/s) = 60 kg·m^2/s

Therefore, the angular momentum of the puck is 60 kg·m^2/s. The closest answer choice is D) 60.

Question 20

A solid sphere, a solid cylinder, a spherical shell, and a hoop all have the same mass and radius. Each are rolling on a horizontal surface with the same centre of mass speed, and then they roll up identical inclines. Which one goes the greatest distance up its incline?
A)the hoop
B)the solid sphere
C)the spherical shell
D)the cylinder
E)They all go the same distance up their inclines.

Accepted Answer

The answer of A solid sphere, a solid cylinder, a...

Quiz 10: Energy and Momentum in Rotating Systems

A thin rod of mass M and length L is struck at one end by a ball of clay of mass m, moving with speed v as shown in the figure. The ball sticks to the rod. After the collision, the angular momentum of the clay-rod system about A, the midpoint of the rod, is:

A uniform rod is 3.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at an angle of 27 $\degree$ above the horizontal. What is the angular speed of the rod as it passes through the horizontal position?

A solid sphere, a solid cylinder, and a hoop all have the same mass and radius. Each are sent down identical inclined planes starting from rest. Their kinetic energies at the bottom of the incline are Ksphere, Kcylinder, and Khoop. Which of the following is true?

A massless rope is wrapped around a uniform cylinder that has radius R and mass M, as shown in the figure. Initially, the unwrapped portion of the rope is vertical and the cylinder is horizontal. The linear acceleration of the cylinder is:

During a race on the Gold Coast, a car of mass 1000 kg moves with a speed of 50 m/s on a circular track of radius 100 m. What is the magnitude of its angular momentum (in kg.m²/s) relative to the centre of the racetrack?

A particle whose mass is 2.0 kg moves in the xy plane with a constant speed of 3.0 m/s along the direction . What is its angular momentum (in kg.m²/s) relative to the point (0, 5.0) metres?

A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s in the x direction along the line y = 5. What is its angular momentum (in kg.m²/s) relative to the origin?

Two cylinders made of the same material roll down a plane inclined at an angle $\theta$ with the horizontal. Each travels the same distance. The radius of cylinder B is twice the radius of cylinder A. In what order do they reach the bottom?

A puck on a frictionless air hockey table has a mass of 5.0 kg and is attached to a cord passing through a hole in the surface as in the figure. The puck is revolving at a distance 2.0 m from the hole with an angular velocity of 3.0 rad/s. The angular momentum of the puck (in kg.m²/s) is:

A solid sphere, a solid cylinder, a spherical shell, and a hoop all have the same mass and radius. Each are rolling on a horizontal surface with the same centre of mass speed, and then they roll up identical inclines. Which one goes the greatest distance up its incline?

Quiz 10: Energy and Momentum in Rotating Systems

A thin rod of mass M and length L is struck at one end by a ball of clay of mass m, moving with speed v as shown in the figure. The ball sticks to the rod. After the collision, the angular momentum of the clay-rod system about A, the midpoint of the rod, is:

A uniform rod is 3.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at an angle of 27 °\degree° above the horizontal. What is the angular speed of the rod as it passes through the horizontal position?

A solid sphere, a solid cylinder, and a hoop all have the same mass and radius. Each are sent down identical inclined planes starting from rest. Their kinetic energies at the bottom of the incline are Ksphere, Kcylinder, and Khoop. Which of the following is true?

A pendulum bob of mass m is set into motion in a circular path in a horizontal plane as shown in the figure. The square of the angular momentum of the bob about the vertical axis through the point P is:

A massless rope is wrapped around a uniform cylinder that has radius R and mass M, as shown in the figure. Initially, the unwrapped portion of the rope is vertical and the cylinder is horizontal. The linear acceleration of the cylinder is:

During a race on the Gold Coast, a car of mass 1000 kg moves with a speed of 50 m/s on a circular track of radius 100 m. What is the magnitude of its angular momentum (in kg.m2/s) relative to the centre of the racetrack?

A particle whose mass is 2.0 kg moves in the xy plane with a constant speed of 3.0 m/s along the direction . What is its angular momentum (in kg.m2/s) relative to the point (0, 5.0) metres?

A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s in the x direction along the line y = 5. What is its angular momentum (in kg.m2/s) relative to the origin?

A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s along the direction . What is its angular momentum (in kg.m2/s) relative to the origin?

Two cylinders made of the same material roll down a plane inclined at an angle θ\thetaθ with the horizontal. Each travels the same distance. The radius of cylinder B is twice the radius of cylinder A. In what order do they reach the bottom?

A puck on a frictionless air hockey table has a mass of 5.0 kg and is attached to a cord passing through a hole in the surface as in the figure. The puck is revolving at a distance 2.0 m from the hole with an angular velocity of 3.0 rad/s. The angular momentum of the puck (in kg.m2/s) is: