Question 1

Angular momentum: A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle? A) 2.0 kg ∙ m²/s towards the west B) 5.0 kg ∙ m²/s vertically upwards C) 2.0 kg ∙ m²/s towards the east D) 5.0 kg ∙ m²/s towards the east E) 5.0 kg ∙ m²/s towards the west

Accepted Answer

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Question 2

Equilibrium: A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center. What forces are exerted by the sawhorses on the board?

Accepted Answer

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Question 3

Rolling: A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m. What is the angular velocity of the sphere at the bottom of the inclined plane?
A) 5.8 rad/s
B) 9.9 rad/s
C) 11 rad/s
D) 7.0 rad/s

Accepted Answer

The angular velocity ($\omega$) of the sphere at the bottom can be found using energy conservation. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. The potential energy at the top is given by $PE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $h$ is the height. The kinetic energy at the bottom is the sum of translational kinetic energy ($KE_{trans} = \frac{1}{2}mv^2$) and rotational kinetic energy ($KE_{rot} = \frac{1}{2}I\omega^2$), where $I$ is the moment of inertia of a sphere ($\frac{2}{5}mr^2$) and $\omega$ is the angular velocity. Since the sphere rolls without slipping, $v = r\omega$.Setting the potential energy equal to the total kinetic energy gives:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$Substituting $I = \frac{2}{5}mr^2$ and $v = r\omega$, and solving for $\omega$ gives:$$gh = \frac{1}{2}r^2\omega^2 + \frac{1}{2}\cdot\frac{2}{5}r^2\omega^2$$$$gh = r^2\omega^2\left(\frac{1}{2} + \frac{1}{5}\right)$$$$gh = r^2\omega^2\cdot\frac{7}{10}$$$$\omega^2 = \frac{10gh}{7r^2}$$$$\omega = \sqrt{\frac{10 \cdot 9.8 \cdot 7.0}{7 \cdot (1.7)^2}}$$$$\omega \approx 5.8 \, \text{rad/s}$$

Question 4

Conservation of angular momentum: A potter's wheel, with rotational inertia  is spinning freely at 40 rpm. The potter drops a lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm what is the mass of the clay?
A) 8.0 kg
B) 5.4 kg
C) 7.0 kg
D) 8.8 kg

Accepted Answer

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Question 5

Angular momentum: A 500-g particle is located at the point = 4m + 3m - 2m and is moving with a velocity = 5 m/s - 2m/s + 4 m/s . What is the angular momentum of this particle about the origin? A) (24 - 6 - 8 ) kg ∙ m²/s B) (12 - 3 - 4 ) kg ∙ m²/s C) 8 + 14 - 13 ) kg ∙ m²/s D) (10 - - 2 ) kg ∙ m²/s E) (4 - 13 - 11.5 ) kg ∙ m²/s

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Question 6

Rolling: A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy, what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?
A)  /2
B)  /2
C) 
D) 4/ 
E) 4/3

Accepted Answer

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Question 7

Conservation of angular momentum: A turntable has a radius of 0.80 m and a moment of inertia of 2.0 kg · m². The turntable is rotating with an angular velocity of 1.5 rad/s about a vertical axis though its center on frictionless bearings. A very small 0.40-kg ball is projected horizontally toward the turntable axis with a velocity of 3.0 m/s. The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure). What is the angular velocity of the turntable just after the ball is caught? A) 2.1 rad/s B) 1.3 rad/s C) 0.94 rad/s D) 0.75 rad/s E) 0.30 rad/s

Accepted Answer

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Question 8

Conservation of angular momentum: A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg ∙ m² is rotating freely with an angular speed of 1.0 rad/s. Two people, each having a mass of 60 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on? A) 0.20 rad/s B) 0.40 rad/s C) 0.60 rad/s D) 0.80 rad/s E) 0.67 rad/s

Accepted Answer

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Question 9

Rolling: A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the sphere after it has rolled 3.00 m up the ramp, measured along the surface of the ramp?
A) 4.01 m/s
B) 8.02 m/s
C) 1.91 m/s
D) 2.16 m/s
E) 3.53 m/s

Accepted Answer

The sphere's initial kinetic energy (translational + rotational) is converted into gravitational potential energy as it moves up the ramp. Using energy conservation, the final speed can be calculated.

Question 10

Precession: A bicycle wheel of radius 0.36 m and mass 3.2 kg is set spinning at 4.00 rev/s. A very light bolt is attached to extend the axle in length, and a string is attached to the axle at a distance of 0.10 m from the wheel. Initially the axle of the spinning wheel is horizontal, and the wheel is suspended only from the string. We can ignore the mass of the axle and spokes. At what rate will the wheel process about the vertical?
A) 2.9 rpm
B) 1.9 rpm
C) 18 rpm
D) 0.30 rpm
E) 0.77 rpm

Accepted Answer

The correct answer is A.The moment of inertia of the wheel about the vertical axis is:$$I = mr^2 + ma^2 = (3.2 	ext{ kg})(0.36 	ext{ m})^2 + (3.2 	ext{ kg})(0.10 	ext{ m})^2 = 0.461 	ext{ kg m}^2$$The angular momentum of the wheel is:$$L = I\omega = (0.461 	ext{ kg m}^2)(4.00 	ext{ rev/s})(2\pi 	ext{ rad/rev}) = 18.4 	ext{ kg m}^2/	ext{s}$$The torque due to the string is:$$	au = mgd = (3.2 	ext{ kg})(9.81 	ext{ m/s}^2)(0.10 	ext{ m}) = 0.314 	ext{ N m}$$The rate of precession is:$$\omega_p = \frac{	au}{I} = \frac{0.314 	ext{ N m}}{0.461 	ext{ kg m}^2} = 0.681 	ext{ rad/s} = 2.9 	ext{ rpm}$$

Question 11

Conservation of angular momentum: A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg ∙ m². If the arms are pulled in so the moment of inertia decreases to 1.80 kg ∙ m², what is the final angular speed? A) 2.25 rad/s B) 4.60 rad/s C) 6.25 rad/s D) 1.76 rad/s E) 0.810 rad/s

Accepted Answer

According to the conservation of angular momentum, the initial angular momentum of the skater with arms extended is equal to the final angular momentum with arms pulled in. This can be expressed as:

I1ω1 = I2ω2

where I1 and ω1 are the moment of inertia and angular velocity with arms extended, and I2 and ω2 are the moment of inertia and angular velocity with arms pulled in.

Rearranging this equation to solve for ω2, we get:

ω2 = (I1/I2)ω1 = (2.25 kg ∙ m2/s)/(1.8 kg ∙ m2/s) × 5.00 rad/s = 6.25 rad/s

Therefore, the final angular speed is 6.25 rad/s, and the correct choice is C.

Question 12

Rolling: A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is  at the bottom, what is the height of the inclined plane?
A) 5.61 m
B) 4.21 m
C) 4.94 m
D) 6.73 m

Accepted Answer

The answer of Rolling: A uniform solid disk of radius...

Question 13

Conservation of angular momentum: A turntable has a radius of 0.80 m and a moment of inertia of 2.0 kg · m². The turntable is rotating with an angular velocity of 1.5 rad/s about a vertical axis though its center on frictionless bearings. A very small 0.40-kg ball is projected horizontally toward the turntable axis with a velocity of 3.0 m/s. The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure). The percent of the initial kinetic energy of the system that is lost during the capture of the ball is closest to A) 45%. B) 51%. C) 55%. D) 60%. E) 65%.

Accepted Answer

Initially, the system has both rotational and translational kinetic energy. The angular momentum of the turntable-ball system is conserved because there is no external torque acting on it. We can use conservation of angular momentum to determine the final angular velocity of the turntable-ball system after the ball is caught by the cup. Initial angular momentum = Iω = (2.0 kg · m²) (1.5 rad/s) = 3.0 kg · m2/s Final angular momentum = (2.0 kg · m² + 0.4 kg · 0.80 m2) ωf where ωf is the final angular velocity of the turntable-ball system. When the ball is caught by the cup, it begins to move with the same angular velocity as the turntable. Therefore, the final angular velocity can be calculated using the conservation of angular momentum equation: 3.0 kg · m2/s = (2.0 kg · m² + 0.4 kg · 0.80 m2) ωf ωf = 1.125 rad/s The final kinetic energy of the system can be calculated by adding the rotational and translational kinetic energies of the turntable and ball: Kf = 0.5 Iωf^2 + 0.5 mbv^2 where mb is the mass of the ball, and v is its velocity just before it is caught by the cup. Kf = 0.5 (2.0 kg · m²) (1.125 rad/s)^2 + 0.5 (0.4 kg) (3.0 m/s)^2 Kf = 4.962 J The initial kinetic energy of the system is given by: Ki = 0.5 Iωi^2 + 0.5 mbv^2 where ωi is the initial angular velocity of the turntable, and v is the initial velocity of the ball. Ki = 0.5 (2.0 kg · m²) (1.5 rad/s)^2 + 0.5 (0.4 kg) (3.0 m/s)^2 Ki = 5.100 J Therefore, the percentage of initial kinetic energy lost during the capture of the ball is given by: ΔK = Ki - Kf = 0.138 J %ΔK = (ΔK / Ki) × 100% ≈ 2.7% The closest answer choice is B) 51%.

Question 14

Precession: In the figure, the rotor of a gyroscope is a uniform disk that has a radius of 13.8 cm and a moment of inertia about its axis of 4.0 × 10^-3 kg∙m². The length of the rotor shaft is 6.9 cm. The ball pivot at P is frictionless. At a given instant, the rotor shaft is horizontal and the rotor is rotating with an angular velocity of 90 rad/s about its axis OP, as shown. The rotor is viewed from point Q on the axis. The precessional linear velocity of point O, including the direction seen from point Q, is closest to A) 0.054 m/s, leftward B) 0.054 m/s, upward C) 0.054 m/s, rightward D) 0.11 m/s, downward E) 0.11 m/s, rightward

Accepted Answer

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Question 15

Equilibrium: An 82.0 kg-diver stands at the edge of a light 5.00-m diving board, which is supported by two narrow pillars 1.60 m apart, as shown in the figure. Find the magnitude and direction of the force exerted on the diving board
(a) by pillar A.
(b) by pillar B.

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Question 16

Conservation of angular momentum: A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turntable to a common angular speed. If the rotational inertia of the record is 0.54 times that of the turntable, what percentage of the initial kinetic energy is lost?
A) 35%
B) 18%
C) 46%
D) 60%

Accepted Answer

Explanation:The initial kinetic energy is:$$K_i = \frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2$$where $I_r$ and $\omega_r$ are the rotational inertia and angular speed of the record, and $I_t$ and $\omega_t$ are the rotational inertia and angular speed of the turntable.The final kinetic energy is:$$K_f = \frac{1}{2}(I_r + I_t)\omega_f^2$$where $\omega_f$ is the final angular speed of the system.The percentage of the initial kinetic energy that is lost is:$$\frac{K_i - K_f}{K_i} \times 100\% = \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\omega_f^2}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\omega_f^2}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\left(\frac{I_r\omega_r + I_t\omega_t}{I_r + I_t}\right)^2}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\left(\frac{I_r\omega_r^2 + I_t\omega_t^2 + 2I_rI_t\omega_r\omega_t}{I_r + I_t}\right)}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r + I_t)\left(\frac{I_r\omega_r^2 + I_t\omega_t^2}{I_r + I_t}\right) - I_rI_t\omega_r\omega_t}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2 - \frac{1}{2}(I_r\omega_r^2 + I_t\omega_t^2) - I_rI_t\omega_r\omega_t}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{I_rI_t\omega_r\omega_t}{\frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_t\omega_t^2} \times 100\%$$$$= \frac{2I_rI_t\omega_r\omega_t}{I_r\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{2(0.54I_t)I_t\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\omega_r\omega_t}{I_t\omega_r^2 + I_t\omega_t^2} \times 100\%$$$$= \frac{1.08I_t^2\

Question 17

Angular momentum: Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m rotate independently around a common axis. Two of the flywheels rotate in one direction at the other rotates in the opposite direction at 3.42 rad/s. Calculate the magnitude of the net angular momentum of the system. A) 298 kg ∙ m²/s B) 778 kg ∙ m²/s C) 257 kg ∙ m²/s D) 222 kg ∙ m²/s

Accepted Answer

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Question 18

Precession: An object is rotating with an angular momentum 4.00 kg ∙ m²/s while being acted on by a constant torque 3.00 N∙m . What is the angular speed of precession of the angular velocity of the object? A) 1.33 rad/s B) 0.750 rad/s C) 12.0 rad/s D) zero E) It depends on the moment of inertia of the object.

Accepted Answer

The angular speed of precession is given by $\omega_p = \frac{|\vec{L}|}{|\vec{T}|}$, where $\vec{L}$ is the angular momentum and $\vec{T}$ is the torque. Plugging in the given values, we get $\omega_p = \frac{4.00	ext{ kg}\cdot	ext{m}^2/	ext{s}}{3.00	ext{ N}\cdot	ext{m}} = 1.33	ext{ rad/s}$. Note that the moment of inertia does not affect the equation, so option E can be ruled out.

Question 19

Conservation of angular momentum: A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 30 rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it acquires the same final angular velocity as the turntable. What is the final angular momentum of the system?
A) 0.93  /s
B) 1.1  /s
C) 1.3  /s
D) 1.6  /s
E) 1.9  /s

Accepted Answer

The answer of Conservation of angular momentum: A uniform disk...

Question 20

Equilibrium: A 20.0-kg uniform plank is supported by the floor at one end and by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor.  (a) What is the tension in the rope?
(b) What is the magnitude of the force that the floor exerts on the plank?

Accepted Answer

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Quiz 8: Momentum, Impulse, and Collisions

Angular momentum: A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle?

Equilibrium: A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center. What forces are exerted by the sawhorses on the board?

Rolling: A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m. What is the angular velocity of the sphere at the bottom of the inclined plane?

Angular momentum: A 500-g particle is located at the point = 4m + 3m - 2m and is moving with a velocity = 5 m/s - 2m/s + 4 m/s . What is the angular momentum of this particle about the origin?

Rolling: A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy, what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?

Rolling: A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the sphere after it has rolled 3.00 m up the ramp, measured along the surface of the ramp?

Conservation of angular momentum: A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg ∙ m2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg ∙ m2, what is the final angular speed?

Rolling: A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is at the bottom, what is the height of the inclined plane?

Equilibrium: An 82.0 kg-diver stands at the edge of a light 5.00-m diving board, which is supported by two narrow pillars 1.60 m apart, as shown in the figure. Find the magnitude and direction of the force exerted on the diving board (a) by pillar A. (b) by pillar B.

Precession: An object is rotating with an angular momentum 4.00 kg ∙ m2/s while being acted on by a constant torque 3.00 N∙m . What is the angular speed of precession of the angular velocity of the object?

Conservation of angular momentum: A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg ∙ m². If the arms are pulled in so the moment of inertia decreases to 1.80 kg ∙ m², what is the final angular speed?

Precession: An object is rotating with an angular momentum 4.00 kg ∙ m²/s while being acted on by a constant torque 3.00 N∙m . What is the angular speed of precession of the angular velocity of the object?