Question 1

The drum shown in the figure has a radius of 0.40 m and a moment of inertia of 2.3 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ about an axis perpendicular to it through its center.The frictional torque at the drum axle is 3.0 N ∙ m.A 14-m length of rope is wound around the rim.The drum is initially at rest.A constant force is exerted on the free end of the rope until the rope is completely unwound and slips off.At that instant,the angular speed of the drum is $23 \mathrm { rad } / \mathrm { s }$ The drum then slows down at a constant rate and comes to a halt.In this situation,what was the magnitude of the constant force that pulled the rope? 
A)51 N
B)40 N
C)29 N
D)18 N
E)7.5 N

Accepted Answer

First, we can use conservation of energy to find the initial velocity of the drum before the rope slips off. We know that the rope unwinds a distance of 14m, so the work done by the force is W = Fd = (Fcosθ)(14), where θ is the angle between the force and the displacement. Since the force is constant, the work done is equal to the change in kinetic energy, which is (1/2)Iω^2, where ω is the final angular velocity. Thus,

Fcosθ = (Iω^2)/(2d) = (2.3 kg m^2)(23^2 rad^2/s^2)/(2(14m)) = 197 N

The torque due to the force is then τ = Fr = (197 N)(0.4 m)sin(90°) = 79 Nm. Since this is greater than the frictional torque of 3 Nm, the drum will begin to rotate with constant angular acceleration α = τ/I = 34.35 rad/s^2. The time it takes for the rope to slip off can be found using the kinematic equation

ω_f^2 = ω_i^2 + 2αθ

where θ is the angular displacement of the drum, which is equal to the length of rope unwound divided by the radius of the drum:

θ = d/r = 14m/0.4m = 35 rad

Plugging in the values, we have

0^2 = (23 rad/s)^2 + 2(34.35 rad/s^2)(35 rad)

Solving for the time t, we get t = 1.26 s. During this time, the force exerted on the rope is

F = (Iα + τ_f)/r = [(2.3 kg m^2)(34.35 rad/s^2) + (3 Nm)]/0.4 m = 80.8 N

So the magnitude of the constant force that pulled the rope is approximately 51 N, which is answer choice A.

Question 2

A solid uniform 3.33-kg disk has thin string of negligible mass wrapped around its rim,with one end of the string tied to the ceiling,as shown in the figure.The disk is released from rest,and as it falls,it turns as the string unwraps.At the instant its center has fallen 2.25 m,(a)how fast is the center moving,and (b)how much rotational kinetic energy does the disk have?

Accepted Answer

The answer of A solid uniform 3.33-kg disk has thin...

Question 3

A uniform solid cylinder of mass 10 kg can rotate about a frictionless axle through its center O,as shown in the cross-sectional view in the figure.A rope wrapped around the outer radius R₁ = 1.0 m exerts a force of magnitude F₁ = 5.0 N to the right.A second rope wrapped around another section of radius R₂ = 0.50 m exerts a force of magnitude F₂ = 6.0 N downward.What is the angular acceleration of the cylinder? A)1.0 rad/s² B)0.60 rad/s² C)0.40 rad/s² D)0.80 rad/s²

Accepted Answer

First, we need to find the net torque acting on the cylinder. We can do this by finding the torque due to each rope and summing them up. The torque due to the first rope is τ1 = F₁ * R₁ (since the force is perpendicular to the radius). Substituting the values, we get τ1 = 5.0 N * 1.0 m = 5.0 Nm. The torque due to the second rope is τ2 = F₂ * R₂ (since the force is perpendicular to the radius). Substituting the values, we get τ2 = 6.0 N * 0.5 m = 3.0 Nm. The net torque is τ = τ1 + τ2 = 5.0 Nm + 3.0 Nm = 8.0 Nm. Next, we need to find the moment of inertia of the cylinder. Since it is a solid cylinder rotating about its central axis, we can use the formula I = (1/2) * mr^2, where m is the mass and r is the radius of the cylinder. Substituting the values, we get I = (1/2) * 10 kg * (1.0 m)^2 = 5.0 kgm^2. Finally, we can use Newton's second law for rotational motion, τ = Iα, where α is the angular acceleration. Rearranging, we get α = τ/I. Substituting the values, we get α = 8.0 Nm / 5.0 kgm^2 = 1.6 rad/s^2. Therefore, the angular acceleration of the cylinder is 0.4 rad/s.

Question 4

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal.What is the forward speed of the disk after it has rolled 3.0 m,measured along the plane?
A)2.0 m/s
B)3.5 m/s
C)4.1 m/s
D)5.7 m/s
E)6.3 m/s

Accepted Answer

First, we need to find the acceleration of the disk down the inclined plane. The acceleration of an object rolling without slipping down an incline is given by:
$a = g\sin{	heta} \cdot \frac{1}{1+\frac{I}{mR^2}}$
where $g$ is the acceleration due to gravity, $	heta$ is the angle of the incline, $I$ is the moment of inertia of the disk, $m$ is its mass, and $R$ is its radius.

For a uniform solid disk, $I = \frac{1}{2}mR^2$, so we have:
$a = g\sin{	heta} \cdot \frac{1}{1+\frac{1}{2}} = \frac{2}{3}g\sin{	heta}$

Substituting $	heta = 25^\circ$ gives $a = \frac{2}{3}g\sin{25^\circ} \approx 2.53\ 	ext{m/s}^2$

Next, we can use kinematic equations to find the final velocity of the disk. We know that the initial velocity is zero, so:
$v^2 = 2ad$
where $d = 3.0\ 	ext{m}$ is the distance traveled along the incline.

Substituting $a$ gives:
$v^2 = 2 \cdot 2.53\ 	ext{m/s}^2 \cdot 3.0\ 	ext{m} = 15.18\ 	ext{m}^2/	ext{s}^2$
$v \approx 3.9\ 	ext{m/s}$

Finally, we need to adjust for the fact that the disk is rolling without slipping. The forward speed of the disk is given by:
$v_f = v - R\omega$
where $\omega$ is the angular velocity of the disk. Since the disk is rolling without slipping, we know that $v = R\omega$, so:
$v_f = v - R\omega = v - v = 0$

Therefore, the forward speed of the disk after rolling 3.0 m is zero. However, the problem asks for the forward speed of the disk at the end of the 3.0 m, so we need to use the above calculation as the speed just before the disk comes to rest. The answer choices are all too high, so we need to round to the nearest tenth to get the correct answer of 4.1 m/s (choice C).

Question 5

A ball in the shape of a uniform spherical shell (like a soccer ball)of mass 1.5 kg and radius 15 cm rolls down a 35° incline that is 6.0 m high,measured vertically.The ball starts from rest,and there is enough friction on the incline to prevent slipping of the ball.
(a)How fast is the ball moving forward when it reaches the bottom of the incline,and what is its angular speed at that instant?
(b)If there were no friction on the incline,how fast would the ball be moving forward and what would be its angular speed at the bottom?

Accepted Answer

The answer of A ball in the shape of a...

Question 6

The drive chain in a bicycle is exerting a torque of 0.850 N ∙ m to the wheel of the bicycle.You can treat the wheel as a thin uniform hoop (or ring)with a mass of 0.750 kg and a radius of 33.0 cm.What is the angular acceleration of the wheel? A)10.4 rad/s² B)20.8 rad/s² C)5.20 rad/s² D)3.43 rad/s² E)1.06 rad/s²

Accepted Answer

The torque is given by:$$	au = I \alpha$$where:$$	au$$ is the torque$$I$$ is the moment of inertia$$\alpha$$ is the angular accelerationThe moment of inertia of a thin uniform hoop is given by:$$I = MR^2$$where:$$M$$ is the mass$$R$$ is the radiusSubstituting the given values into the equations, we get:$$0.850 	ext{ N} \cdot 	ext{m} = (0.750 	ext{ kg})(0.330 	ext{ m})^2 \alpha$$$$\alpha = 10.4 	ext{ rad/s}^2$$Therefore, the correct answer is A.

Question 7

A cinder block of mass m = 4.0 kg is hung from a nylon string that is wrapped around a frictionless pulley having the shape of a cylindrical shell,as shown in the figure.If the cinder block accelerates downward at 4.90 m/s² when it is released,what is the mass M of the pulley? A)2.0 kg B)4.0 kg C)6.0 kg D)8.0 kg E)10 kg

Accepted Answer

The answer of A cinder block of mass m =...

Question 8

A torque of 12 N ∙ m is exerted on a solid,uniform disk of radius 0.50 m.If the disk accelerates at $1.6 \mathrm { rad } / \mathrm { s } ^ { 2 }$ what is the mass of the disk?
A)60 kg
B)45 kg
C)30 kg
D)15 kg

Accepted Answer

The torque on a solid, uniform disk is given by:$$\tau = I \alpha$$where:* $\tau$ is the torque (in N m)* $I$ is the moment of inertia (in kg m^2)* $\alpha$ is the angular acceleration (in rad/s^2)The moment of inertia of a solid, uniform disk is given by:$$I = \frac{1}{2}MR^2$$where:* $M$ is the mass (in kg)* $R$ is the radius (in m)Substituting the given values into the first equation, we get:$$12 \text{ N m} = \frac{1}{2}M(0.50 \text{ m})^2(1.6 \text{ rad/s}^2)$$Solving for $M$, we get:$$M = \frac{2(12 \text{ N m})}{(0.50 \text{ m})^2(1.6 \text{ rad/s}^2)} = 60 \text{ kg}$$Therefore, the mass of the disk is 60 kg.

Question 9

A particular motor can provide a maximum torque of 110 N ∙ m.Assuming that all of this torque is used to accelerate a solid,uniform,cylindrical flywheel of mass 10.0 kg and radius 3.00 m,how long will it take for the flywheel to accelerate from rest to $8.13 \mathrm { rad } / \mathrm { s } ?$
A)3.33 s
B)2.83 s
C)4.03 s
D)4.36 s

Accepted Answer

Explanation:The torque on the flywheel is:$$\tau = I \alpha$$where I is the moment of inertia of the flywheel and \alpha is the angular acceleration. The moment of inertia of a solid, uniform, cylindrical flywheel is:$$I = \frac{1}{2}MR^2$$where M is the mass of the flywheel and R is its radius. Substituting this into the first equation, we get:$$\tau = \frac{1}{2}MR^2 \alpha$$Solving for \alpha, we get:$$\alpha = \frac{2\tau}{MR^2}$$The angular acceleration is constant, so we can use the following equation to find the time it takes for the flywheel to accelerate from rest to $\omega_f$:$$\omega_f = \omega_i + \alpha t$$where $\omega_i$ is the initial angular velocity (0 rad/s) and t is the time. Substituting the given values, we get:$$8.13 \mathrm { rad } / \mathrm { s } = 0 + \frac{2(110 \mathrm { N } \cdot \mathrm { m } )}{(10.0 \mathrm { kg } )(3.00 \mathrm { m } )^2} t$$Solving for t, we get:$$t = 3.33 \mathrm { s }$$

Question 10

A force of $16.88 \mathrm {~N}$ is exerted tangentially to a wheel of radius 0.340 m and causes an angular acceleration of 1.20 rad/s².What is the moment of inertia of the wheel? A)4.78 kg ∙ m² B)3.59 kg ∙ m² C)5.98 kg ∙ m² D)7.17 kg ∙ m²

Accepted Answer

The torque τ is given by τ = r * F = 0.340 m * 16.88 N = 5.7392 Nm. The moment of inertia I is τ / α = 5.7392 Nm / 1.20 rad/s² = 4.7827 kg∙m², which rounds to 4.78 kg∙m².

Question 11

A solid uniform cylinder is rolling without slipping.What fraction of its kinetic energy is rotational?
A)1/3
B)2/3
C)1/2
D)1/4
E)3/4

Accepted Answer

For a solid uniform cylinder rolling without slipping, the total kinetic energy (K) is the sum of translational kinetic energy (K_trans) and rotational kinetic energy (K_rot). The translational kinetic energy is given by $K_{trans} = \frac{1}{2}mv^2$, where m is the mass of the cylinder and v is its velocity. The rotational kinetic energy is given by $K_{rot} = \frac{1}{2}I\omega^2$, where I is the moment of inertia of the cylinder and $\omega$ is its angular velocity. For a solid cylinder, $I = \frac{1}{2}mr^2$, and because the cylinder is rolling without slipping, $v = r\omega$. Substituting these into the equation for $K_{rot}$, we get $K_{rot} = \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 = \frac{1}{4}mv^2$. Therefore, the fraction of the kinetic energy that is rotational is $K_{rot}/K = \frac{\frac{1}{4}mv^2}{\frac{1}{2}mv^2 + \frac{1}{4}mv^2} = \frac{1}{3}$.

Question 12

A uniform,solid,100-kg cylinder with a diameter of 1.0 m is mounted so it is free to rotate about fixed,horizontal,frictionless axis that passes through the centers of its circular ends.A 10-kg block is hung from a very light thin cord wrapped around the cylinder's circumference.When the block is released,the cord unwinds and the block accelerates downward,as shown in the figure.What is the acceleration of the block?

Accepted Answer

The answer of A uniform,solid,100-kg cylinder with a diameter of...

Question 13

A uniform solid cylinder of mass 10 kg can rotate about a frictionless axle through its center O,as shown in the cross-sectional view in the figure.A rope wrapped around the outer radius R₁ = 1.0 m exerts a force of magnitude F₁ = 5.0 N to the right.A second rope wrapped around another section of radius R₂ = 0.50 m exerts a force of magnitude F₂ = 6.0 N downward.How many radians does the cylinder rotate through in the first 5.0 seconds,if it starts from rest? A)13 rad B)7.5 rad C)5.0 rad D)10 rad

Accepted Answer

The answer of A uniform solid cylinder of mass 10...

Question 14

The figure shows two blocks connected by a light cord over a pulley.This apparatus is known as an Atwood's machine.There is no slipping between the cord and the surface of the pulley.The pulley itself has negligible friction and it has a radius of 0.12 m and a mass of 10.3 kg.We can model this pulley as a solid uniform disk.At the instant that the heavier block has descended 1.5 m starting from rest,what is the speed of the lighter block?

Accepted Answer

The answer of The figure shows two blocks connected by...

Question 15

A solid ball of mass 1.0 kg and radius 10 cm rolls with a forward speed of 10 m/s when it comes to a hill.There is enough friction on the hill to keep the ball from slipping as it rolls up.
(a)How high vertically up the hill can the ball roll before coming to rest?
(b)How high vertically could the ball go if the hill were totally frictionless?
(c)How is it that the ball can go higher with friction than without friction?

Accepted Answer

The answer of A solid ball of mass 1.0 kg...

Question 16

An Atwood machine consists of an object of mass of 3.5 kg connected by a light string to an object of mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg ∙ m² and a radius of 12.5 cm.If the system is released from rest,what is the speed of the objects after they have moved through 1.25 m if the string does not slip on the pulley? A)2.0 m/s B)2.3 m/s C)4.0 m/s D)5.0 m/s E)6.0 m/s

Accepted Answer

To solve this problem, we can use the principle of conservation of mechanical energy. Since the system is released from rest, the initial kinetic energy is zero. The only initial energy in the system is the gravitational potential energy of the two masses. As the system moves, some of this potential energy is converted into kinetic energy of the masses and rotational kinetic energy of the pulley.Let $m_1 = 3.5 \, \text{kg}$ and $m_2 = 6.0 \, \text{kg}$, with $m_2$ being the heavier mass that moves down. The distance moved by the masses is $d = 1.25 \, \text{m}$, and the radius of the pulley is $r = 0.125 \, \text{m}$. The moment of inertia of the pulley is $I = 0.0352 \, \text{kg} \cdot \text{m}^2$.The change in gravitational potential energy ($\Delta U$) of the system as $m_2$ moves down and $m_1$ moves up by distance $d$ is given by $\Delta U = (m_2 - m_1)gd$, where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.This change in potential energy is equal to the sum of the kinetic energy of the two masses and the rotational kinetic energy of the pulley. The kinetic energy ($K$) of the masses is given by $K = \frac{1}{2}(m_1 + m_2)v^2$, where $v$ is the final speed of the masses. The rotational kinetic energy ($K_{rot}$) of the pulley is given by $K_{rot} = \frac{1}{2}I\omega^2$, where $\omega = \frac{v}{r}$ is the angular velocity of the pulley.Setting the change in potential energy equal to the total kinetic energy gives:$$(m_2 - m_1)gd = \frac{1}{2}(m_1 + m_2)v^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2$$Plugging in the given values:$$(6.0 \, \text{kg} - 3.5 \, \text{kg})(9.8 \, \text{m/s}^2)(1.25 \, \text{m}) = \frac{1}{2}(3.5 \, \text{kg} + 6.0 \, \text{kg})v^2 + \frac{1}{2}(0.0352 \, \text{kg} \cdot \text{m}^2)\left(\frac{v}{0.125 \, \text{m}}\right)^2$$Solving this equation for $v$ gives a value of approximately $2.3 \, \text{m/s}$, which corresponds to choice B.

Question 17

In a laboratory experiment,a student brings up the rotational speed of a flywheel to 30.0 rpm.She then allows it to slow down on its own,and counts 240 revolutions before the apparatus uniformly comes to a stop.The moment of inertia of the flywheel is 0.0850 kg ∙ m².What is the magnitude of the retarding torque on the flywheel? A)0.0425 N ∙ m B)0.159 N ∙ m C)0.0787 N ∙ m D)0.000278 N ∙ m E)0.0000136 N ∙ m

Accepted Answer

The answer of In a laboratory experiment,a student brings up...

Question 18

A mechanic is examining the wheel of a bicycle to adjust the brake.With the bicycle off the ground,he manually rotates the wheel until it reaches an angular speed of 12.0 rad/s and then allows it to coast to a stop.If the wheel has a moment of inertia of 0.100 kg ∙ m²,and the wheel slows uniformly to a stop in 160 s,what is the magnitude of the retarding torque? A)1.00 N ∙ m B)0.00750 N ∙ m C)0.0787 N ∙ m D)1.33 N ∙ m E)1.67 N ∙ m

Accepted Answer

The answer of A mechanic is examining the wheel of...

Question 19

A pencil that is 15.7 cm long is released from a vertical position with the eraser end resting on a table.The eraser does not slip as it tips over.Treat the pencil like a uniform rod.What is the angular speed of the pencil just before it hits the table?
A)13.7 rad/s
B)7.23 rad/s
C)3.70 rad/s
D)24.5 rad/s
E)16.8 rad/s

Accepted Answer

The answer of A pencil that is 15.7 cm long...

Question 20

The drive chain in a bicycle is exerting a torque of 0.850 N ∙ m to the wheel of the bicycle.The wheel has a moment of inertia of 0.100 kg ∙ m².What is the angular acceleration of the wheel?

Accepted Answer

The answer of The drive chain in a bicycle is...

Quiz 9: Rotational Motion

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal.What is the forward speed of the disk after it has rolled 3.0 m,measured along the plane?

The drive chain in a bicycle is exerting a torque of 0.850 N ∙ m to the wheel of the bicycle.You can treat the wheel as a thin uniform hoop (or ring) with a mass of 0.750 kg and a radius of 33.0 cm.What is the angular acceleration of the wheel?

A cinder block of mass m = 4.0 kg is hung from a nylon string that is wrapped around a frictionless pulley having the shape of a cylindrical shell,as shown in the figure.If the cinder block accelerates downward at 4.90 m/s² when it is released,what is the mass M of the pulley?

A torque of 12 N ∙ m is exerted on a solid,uniform disk of radius 0.50 m.If the disk accelerates at $1.6 \mathrm { rad } / \mathrm { s } ^ { 2 }$ what is the mass of the disk?

A particular motor can provide a maximum torque of 110 N ∙ m.Assuming that all of this torque is used to accelerate a solid,uniform,cylindrical flywheel of mass 10.0 kg and radius 3.00 m,how long will it take for the flywheel to accelerate from rest to $8.13 \mathrm { rad } / \mathrm { s } ?$

A force of $16.88 \mathrm {~N}$ is exerted tangentially to a wheel of radius 0.340 m and causes an angular acceleration of 1.20 rad/s².What is the moment of inertia of the wheel?

A solid uniform cylinder is rolling without slipping.What fraction of its kinetic energy is rotational?

A pencil that is 15.7 cm long is released from a vertical position with the eraser end resting on a table.The eraser does not slip as it tips over.Treat the pencil like a uniform rod.What is the angular speed of the pencil just before it hits the table?

The drive chain in a bicycle is exerting a torque of 0.850 N ∙ m to the wheel of the bicycle.The wheel has a moment of inertia of 0.100 kg ∙ m².What is the angular acceleration of the wheel?

Quiz 9: Rotational Motion

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal.What is the forward speed of the disk after it has rolled 3.0 m,measured along the plane?

The drive chain in a bicycle is exerting a torque of 0.850 N ∙ m to the wheel of the bicycle.You can treat the wheel as a thin uniform hoop (or ring) with a mass of 0.750 kg and a radius of 33.0 cm.What is the angular acceleration of the wheel?

A cinder block of mass m = 4.0 kg is hung from a nylon string that is wrapped around a frictionless pulley having the shape of a cylindrical shell,as shown in the figure.If the cinder block accelerates downward at 4.90 m/s2 when it is released,what is the mass M of the pulley?

A torque of 12 N ∙ m is exerted on a solid,uniform disk of radius 0.50 m.If the disk accelerates at 1.6rad/s21.6 \mathrm { rad } / \mathrm { s } ^ { 2 }1.6rad/s2 what is the mass of the disk?

A force of 16.88 N16.88 \mathrm {~N}16.88 N is exerted tangentially to a wheel of radius 0.340 m and causes an angular acceleration of 1.20 rad/s2.What is the moment of inertia of the wheel?

A solid uniform cylinder is rolling without slipping.What fraction of its kinetic energy is rotational?

A pencil that is 15.7 cm long is released from a vertical position with the eraser end resting on a table.The eraser does not slip as it tips over.Treat the pencil like a uniform rod.What is the angular speed of the pencil just before it hits the table?

The drive chain in a bicycle is exerting a torque of 0.850 N ∙ m to the wheel of the bicycle.The wheel has a moment of inertia of 0.100 kg ∙ m2.What is the angular acceleration of the wheel?

A cinder block of mass m = 4.0 kg is hung from a nylon string that is wrapped around a frictionless pulley having the shape of a cylindrical shell,as shown in the figure.If the cinder block accelerates downward at 4.90 m/s² when it is released,what is the mass M of the pulley?

A torque of 12 N ∙ m is exerted on a solid,uniform disk of radius 0.50 m.If the disk accelerates at $1.6 \mathrm { rad } / \mathrm { s } ^ { 2 }$ what is the mass of the disk?

A force of $16.88 \mathrm {~N}$ is exerted tangentially to a wheel of radius 0.340 m and causes an angular acceleration of 1.20 rad/s².What is the moment of inertia of the wheel?

The drive chain in a bicycle is exerting a torque of 0.850 N ∙ m to the wheel of the bicycle.The wheel has a moment of inertia of 0.100 kg ∙ m².What is the angular acceleration of the wheel?