Question 1

A disk with a rotational inertia of 5.0 kg .m² and a radius of 0.25 m rotates on a frictionless fixed axis perpendicular to the disk and through its center.A force of 8.0 N is applied tangentially to the rim.If the disk starts at rest, then after it has turned through half a revolution its angular velocity is: A)0.57 rad/s B)0.64 rad/s C)0.80 rad/s D)1.6 rad/s E)3.2 rad/s

Accepted Answer

The work done by the force is equal to the change in kinetic energy. Work done = Force x distance = 8.0 N x (π x 0.25 m) = 2π J. The change in kinetic energy = 0.5 x I x ω^2, where I = 5.0 kg m^2. Solving for ω gives ω = sqrt((2 x 2π) / 5) = 1.6 rad/s.

Question 2

A uniform disk, a thin hoop, and a uniform solid sphere, all with the same mass and same outer radius, are each free to rotate about a fixed axis through its center.Assume the hoop is connected to the rotation axis by light spokes.With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown.Rank the objects according to their angular velocities after a given time t, least to greatest. 
A)disk, hoop, sphere
B)disk, sphere, hoop
C)hoop, sphere, disk
D)hoop, disk, sphere
E)sphere, disk, hoop

Accepted Answer

The answer of A uniform disk, a thin hoop, and...

Question 3

A disk with a rotational inertia of 5.0 kg.m² and a radius of 0.25 m rotates on a frictionless fixed axis perpendicular to the disk and through its center.A force of 8.0 N is applied along the rotation axis.The angular acceleration of the disk is: A)0 rad/s² B)0.40 rad/s² C)0.60 rad/s² D)1.0 rad/s² E)2.5 rad/s²

Accepted Answer

The force applied does not produce a torque since it is applied along the rotation axis, meaning there is no angular acceleration.

Question 4

A small disk of radius R₁ is fastened coaxially to a larger disk of radius R₂.The combination is free to rotate on a fixed axle, which is perpendicular to a horizontal frictionless table top, as shown in the overhead view below.The rotational inertia of the combination is I.A string is wrapped around the larger disk and attached to a block of mass m, on the table.Another string is wrapped around the smaller disk and is pulled with a force as shown.The tension in the string pulling the block is: A)R₁F/R₂ B)mR₁R₂F/(I - mR₂²) C)mR₁R₂F/(I + mR₂²) D)mR₁R₂F/(I - mR₁R₂) E)mR₁R₂F/(I + mR₁R₂)

Accepted Answer

The system has two components: the small disk and the large disk with the attached block. The tension in the string pulling the block is equal to the force exerted on the small disk since the strings are assumed to be ideal and inextensible. 
The force exerted on the small disk produces a torque given by τ=F(R1), where R1 is the radius of the small disk. The tension force on the larger disk produces a torque equal to τ=F(R2), where R2 is the radius of the larger disk. 
The rotational inertia of the system is given by I=1/2m(R1^2+R2^2), where m is the mass of the larger disk and block. 
Now, applying Newton's second law for rotational motion, we have:
τnet=Iα
where τnet is the net torque acting on the system and α is its angular acceleration. 
The net torque on the system is given by:
τnet=F(R2-R1)
since the two torques due to the tension forces act in opposite directions. 
We can now solve for the tension in the string pulling the block:
F=(Iα)/(R2-R1)
α=(F(R2-R1))/I
The angular acceleration of the system is the same for both disks since they are rigidly connected. Thus, we can use this equation to solve for the tension in the string pulling the block:
Tension force on block = F = (Iα)/(R2-R1) = (mR1R2F)/(I + mR1^2)
Therefore, the tension in the string pulling the block is given by:

Tension force on block = m*R1*R2*F / (I + m*R1^2) 
which is closest to option C.

Question 5

A block is attached to each end of a rope that passes over a pulley suspended from the ceiling.The blocks do not have the same mass.If the rope does not slip on the pulley, then at any instant after the blocks start moving, the rope:
A)pulls on both blocks, but exerts a greater force on the heavier block
B)pulls on both blocks, but exerts a greater force on the lighter block
C)pulls on both blocks and exerts the same magnitude force on both blocks
D)does not pull on either block
E)pulls only on the lighter block

Accepted Answer

The answer of A block is attached to each end...

Question 6

A cylinder is 0.10 m in radius and 0.20 m in length.Its rotational inertia, about the cylinder axis on which it is mounted, is 0.020 kg . m².A string is wound around the cylinder and pulled with a force of 1.0 N.The angular acceleration of the cylinder is: A)2.5 rad/s² B)5.0 rad/s² C)10 rad/s² D)15 rad/s² E)20 rad/s²

Accepted Answer

The torque exerted on the cylinder is given by:
τ = RF = (0.10 m)(1.0 N) = 0.10 Nm
Using the formula for rotational acceleration:
α = τ/I
where I is the rotational inertia, we get:
α = (0.10 Nm)/(0.020 kg m^2) = 5.0 rad/s^2
Therefore, the angular acceleration of the cylinder is 5.0 rad/s^2, which is choice (B).

Question 7

A torque of 170 N∙m does 4700 J of work on a rotating flywheel.If the flywheel's initial kinetic energy is 1500 J, what is its final kinetic energy?
A)1500 J
B)3200 J
C)4700 J
D)6200 J
E)cannot be calculated without knowing the rotational inertia of the flywheel

Accepted Answer

Work done by torque = change in kinetic energy 
4700 J = final kinetic energy - 1500 J 
Final kinetic energy = 6200 J

Question 8

A certain wheel has a rotational inertia of 12 kg∙m².As it turns through 5.0 rev its angular velocity increases from 5.0 rad/s to 6.0 rad/s.If the net torque is constant its value is: A)0.015 N∙m B)0.18 N∙m C)0.57 N∙m D)2.1 N∙m E)13 N∙m

Accepted Answer

We can use the equation for rotational kinetic energy:

$K_{rot}=\frac{1}{2}I\omega^2$

The initial kinetic energy is:

$K_{i}=\frac{1}{2}(12	ext{ kg}\cdot	ext{m}^2)\cdot (5.0	ext{ rad/s})^2=750	ext{ J}$

The final kinetic energy is:

$K_{f}=\frac{1}{2}(12	ext{ kg}\cdot	ext{m}^2)\cdot (6.0	ext{ rad/s})^2=1080	ext{ J}$

The change in kinetic energy is:

$\Delta K=K_{f}-K_{i}=330	ext{ J}$

We know that the net torque is constant, so we can use the equation for work done by torque:

$W_{net}=\Delta K=	au\Delta	heta$

We want the torque, so we can solve for it:

$	au=\frac{\Delta K}{\Delta	heta}=\frac{330	ext{ J}}{(5.0	ext{ rev})\cdot(2\pi	ext{ rad/rev})}=\boxed{2.1	ext{ N}\cdot	ext{m}}$

Question 9

A small disk of radius R₁ is mounted coaxially with a larger disk of radius R₂.The disks are securely fastened to each other and the combination is free to rotate on a fixed axle that is perpendicular to a horizontal frictionless table top, as shown in the overhead view below.The rotational inertia of the combination is I.A string is wrapped around the larger disk and attached to a block of mass m, on the table.Another string is wrapped around the smaller disk and is pulled with a force as shown.The acceleration of the block is: A)R₁F/mR₂ B)R₁R₂F/(I - mR₂²) C)R₁R₂F/(I + mR₂²) D)R₁R₂F/(I - mR₁R₂) E)R₁R₂F/(I + mR₁R₂)

Accepted Answer

The answer of A small disk of radius R₁ is...

Question 10

A disk with a rotational inertia of 5.0 kg∙m² and a radius of 0.25 m rotates on a fixed axis perpendicular to the disk and through its center.A force of 2.0 N is applied tangentially to the rim.As the disk turns through half a revolution the work done by the force is: A)1.6 J B)2.5 J C)6.3 J D)10 J E)40 J

Accepted Answer

Work done = Force × distance = 2 N × (π × 0.25 m) = 1.6 J

Question 11

A constant torque of 260 N∙m acts on a flywheel.If the flywheel makes 25 complete revolutions, how much work has been done by the torque on the flywheel? A)1.7 J B)41 J C)600 J D)6.5 x 10³ J E)4.1 x 10⁴ J

Accepted Answer

The answer of A constant torque of 260 N∙m acts...

Question 12

A 0.70-kg disk with a rotational inertia given by MR²/2 is free to rotate on a fixed horizontal axis suspended from the ceiling.A string is wrapped around the disk and a 2.0-kg mass hangs from the free end.If the string does not slip then as the mass falls and the cylinder rotates the suspension holding the cylinder pulls up on the mass with a force of: A)6.9 N B)9.8 N C)16 N D)26 N E)29 N

Accepted Answer

The answer of A 0.70-kg disk with a rotational inertia...

Question 13

A disk starts from rest and rotates about a fixed axis, subject to a constant net torque.The work done by the torque during the second revolution is ______ as the work done during the first revolution.
A)the same
B)twice as much
C)half as much
D)four times as much
E)one fourth as much

Accepted Answer

The work done by torque is given by the equation W = τΔθ, where τ is the torque applied and Δθ is the change in angle. Since the disk starts from rest, it completes one revolution in 2π radians. During the second revolution, the angle changes by another 2π radians. However, the torque applied is constant and does not change. Therefore, the work done during the second revolution is the same as that during the first revolution, as the angle and torque are the same in both cases.

Question 14

An 8.0-cm radius disk with a rotational inertia of 0.12 kg.m² is free to rotate on a horizontal axis.A string is fastened to the surface of the disk and a 10-kg mass hangs from the other end.The mass is raised by using a crank to apply a 9.0-N.m torque to the disk.The acceleration of the mass is: A)0.50 m/s² B)3.9 m/s² C)6.0 m/s² D)12 m/s² E)20 m/s²

Accepted Answer

The angular acceleration of the disk can be calculated using the torque equation:
τ = Iα
where τ is the torque applied, I is the rotational inertia, and α is the angular acceleration.
Rearranging the equation, we have:
α = τ/I
Substituting the given values, we get:
α = (9.0 N∙m)/(0.12 kg∙m^2) = 75 rad/s^2
The angular acceleration of the disk is also equal to the linear acceleration of the mass divided by the radius of the disk.
a = αr
Substituting the given values, we get:
a = (75 rad/s^2)(0.08 m) = 6.0 m/s^2
Therefore, the acceleration of the mass is 6.0 m/s^2 or choice B.

Question 15

A disk has a rotational inertia of 6.0 kg∙m² and a constant angular acceleration of 2.0 rad/s².If it starts from rest the work done during the first 5.0 s by the net torque acting on it is: A)0 J B)30 J C)60 J D)300 J E)600 J

Accepted Answer

The formula to calculate work done by torque is:

Work = (1/2) * I * ω^2

Where I is the rotational inertia, ω is the angular velocity, and * represents multiplication.

As the disk starts from rest, the initial angular velocity is 0 rad/s. We are given that the disk has a constant angular acceleration of 2.0 rad/s^2. Using the formula,

ω = ω0 + αt

where ω0 is the initial angular velocity, α is the angular acceleration, and t is the time elapsed.

Plugging in the given values, we get:

ω = 0 + (2.0 rad/s^2) * (5.0 s) = 10.0 rad/s

Now, we can calculate the work done as:

Work = (1/2) * I * ω^2

Plugging in the given values, we get:

Work = (1/2) * (6.0 kg∙m^2) * (10.0 rad/s)^2 = 300 J

Therefore, the correct answer is D) 300 J.

Question 16

A disk starts from rest and rotates around a fixed axis, subject to a constant net torque.The work done by the torque during the second 5 s is ______ as the work done during the first 5 s.
A)the same
B)half as much
C)twice as much
D)three times as much
E)four times as much

Accepted Answer

The work done by the torque is related to the change in kinetic energy, which depends on the square of the angular velocity. Since the disk starts from rest and the torque is constant, the angular velocity increases linearly with time. Therefore, the kinetic energy, and hence the work done, increases with the square of the time. Thus, the work done in the second 5 seconds is three times the work done in the first 5 seconds.

Question 17

A disk with a rotational inertia of 2.0 kg.m² and a radius of 0.40 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center.A force of 5.0 N is applied tangentially to the rim.The angular acceleration of the disk is: A)0.40 rad/s² B)0.60 rad/s² C)1.0 rad/s² D)2.5 rad/s² E)10 rad/s²

Accepted Answer

We can use the formula:
τ = Iα
where τ is the torque applied, I is the rotational inertia, and α is the angular acceleration.

We are given τ = 5.0 N, and I = 2.0 kg·m². Solving for α, we get:

α = τ / I = 5.0 N / (2.0 kg·m²) = 2.5 rad/s²

Therefore, the angular acceleration of the disk is 2.5 rad/s², which is closest to choice C.

Question 18

A circular saw is powered by a motor.When the saw is used to cut wood, the wood exerts a torque of 0.80 N∙m on the saw blade.If the blade rotates with a constant angular velocity of 20 rad/s the work done on the blade by the motor in 1.0 min is:
A)0 J
B)480 J
C)960 J
D)1500 J
E)1800 J

Accepted Answer

The answer of A circular saw is powered by a...

Question 19

A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter 0.40 m and hangs vertically, as shown.The rotational inertia of the flywheel is 0.50 kg∙m².When the block is released and the cord unwinds, the acceleration of the block is: A)0.15 g B)0.56 g C)0.84 g D)1.0 g E)1.3 g

Accepted Answer

The answer of A 16 kg block is attached to...

Question 20

A thin circular hoop of mass 1.0 kg and radius 2.0 m is rotating about an axis through its center and perpendicular to its plane.It is slowing down at the rate of 7.0 rad/s².The net torque acting on it is: A)7.0 N∙m B)14 N∙m C)28 N∙m D)44 N∙m E)none of these

Accepted Answer

The answer of A thin circular hoop of mass 1.0...

Quiz 10: Rotation

A disk with a rotational inertia of 5.0 kg.m2 and a radius of 0.25 m rotates on a frictionless fixed axis perpendicular to the disk and through its center.A force of 8.0 N is applied along the rotation axis.The angular acceleration of the disk is:

A block is attached to each end of a rope that passes over a pulley suspended from the ceiling.The blocks do not have the same mass.If the rope does not slip on the pulley, then at any instant after the blocks start moving, the rope:

A cylinder is 0.10 m in radius and 0.20 m in length.Its rotational inertia, about the cylinder axis on which it is mounted, is 0.020 kg . m2.A string is wound around the cylinder and pulled with a force of 1.0 N.The angular acceleration of the cylinder is:

A torque of 170 N∙m does 4700 J of work on a rotating flywheel.If the flywheel's initial kinetic energy is 1500 J, what is its final kinetic energy?

A certain wheel has a rotational inertia of 12 kg∙m2.As it turns through 5.0 rev its angular velocity increases from 5.0 rad/s to 6.0 rad/s.If the net torque is constant its value is:

A disk with a rotational inertia of 5.0 kg∙m2 and a radius of 0.25 m rotates on a fixed axis perpendicular to the disk and through its center.A force of 2.0 N is applied tangentially to the rim.As the disk turns through half a revolution the work done by the force is:

A constant torque of 260 N∙m acts on a flywheel.If the flywheel makes 25 complete revolutions, how much work has been done by the torque on the flywheel?

A disk starts from rest and rotates about a fixed axis, subject to a constant net torque.The work done by the torque during the second revolution is ______ as the work done during the first revolution.

An 8.0-cm radius disk with a rotational inertia of 0.12 kg.m2 is free to rotate on a horizontal axis.A string is fastened to the surface of the disk and a 10-kg mass hangs from the other end.The mass is raised by using a crank to apply a 9.0-N.m torque to the disk.The acceleration of the mass is:

A disk has a rotational inertia of 6.0 kg∙m2 and a constant angular acceleration of 2.0 rad/s2.If it starts from rest the work done during the first 5.0 s by the net torque acting on it is:

A disk starts from rest and rotates around a fixed axis, subject to a constant net torque.The work done by the torque during the second 5 s is ______ as the work done during the first 5 s.

A disk with a rotational inertia of 2.0 kg.m2 and a radius of 0.40 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center.A force of 5.0 N is applied tangentially to the rim.The angular acceleration of the disk is:

A circular saw is powered by a motor.When the saw is used to cut wood, the wood exerts a torque of 0.80 N∙m on the saw blade.If the blade rotates with a constant angular velocity of 20 rad/s the work done on the blade by the motor in 1.0 min is:

A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter 0.40 m and hangs vertically, as shown.The rotational inertia of the flywheel is 0.50 kg∙m2.When the block is released and the cord unwinds, the acceleration of the block is:

A thin circular hoop of mass 1.0 kg and radius 2.0 m is rotating about an axis through its center and perpendicular to its plane.It is slowing down at the rate of 7.0 rad/s2.The net torque acting on it is:

A disk with a rotational inertia of 5.0 kg.m² and a radius of 0.25 m rotates on a frictionless fixed axis perpendicular to the disk and through its center.A force of 8.0 N is applied along the rotation axis.The angular acceleration of the disk is:

A cylinder is 0.10 m in radius and 0.20 m in length.Its rotational inertia, about the cylinder axis on which it is mounted, is 0.020 kg . m².A string is wound around the cylinder and pulled with a force of 1.0 N.The angular acceleration of the cylinder is:

A certain wheel has a rotational inertia of 12 kg∙m².As it turns through 5.0 rev its angular velocity increases from 5.0 rad/s to 6.0 rad/s.If the net torque is constant its value is:

A disk with a rotational inertia of 5.0 kg∙m² and a radius of 0.25 m rotates on a fixed axis perpendicular to the disk and through its center.A force of 2.0 N is applied tangentially to the rim.As the disk turns through half a revolution the work done by the force is:

An 8.0-cm radius disk with a rotational inertia of 0.12 kg.m² is free to rotate on a horizontal axis.A string is fastened to the surface of the disk and a 10-kg mass hangs from the other end.The mass is raised by using a crank to apply a 9.0-N.m torque to the disk.The acceleration of the mass is:

A disk has a rotational inertia of 6.0 kg∙m² and a constant angular acceleration of 2.0 rad/s².If it starts from rest the work done during the first 5.0 s by the net torque acting on it is:

A disk with a rotational inertia of 2.0 kg.m² and a radius of 0.40 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center.A force of 5.0 N is applied tangentially to the rim.The angular acceleration of the disk is:

A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter 0.40 m and hangs vertically, as shown.The rotational inertia of the flywheel is 0.50 kg∙m².When the block is released and the cord unwinds, the acceleration of the block is:

A thin circular hoop of mass 1.0 kg and radius 2.0 m is rotating about an axis through its center and perpendicular to its plane.It is slowing down at the rate of 7.0 rad/s².The net torque acting on it is: