Question 1

While spinning down from 500.0 rpm to rest,a solid uniform flywheel does 5.1 kJ of work.If the radius of the disk is 1.2 m,what is its mass?
A) 5.2 kg
B) 4.4 kg
C) 6.0 kg
D) 6.8 kg

Accepted Answer

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Question 2

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

The angular momentum $L$ of a rotating object is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a wheel with mass $m$ uniformly distributed along its rim of radius $r$, the moment of inertia $I = mr^2$. The angular velocity $\omega$ can be found from the linear velocity $v$ and the radius $r$ using $\omega = \frac{v}{r}$. Substituting the given values, $I = 2.0 \, \text{kg} \times (0.2 \, \text{m})^2 = 0.08 \, \text{kg} \cdot \text{m}^2$ and $\omega = \frac{5.0 \, \text{m/s}}{0.2 \, \text{m}} = 25 \, \text{s}^{-1}$. Thus, $L = 0.08 \, \text{kg} \cdot \text{m}^2 \times 25 \, \text{s}^{-1} = 2.0 \, \text{kg} \cdot \text{m}^2/\text{s}$. The direction of angular momentum is perpendicular to the plane of rotation, following the right-hand rule. Since the wheel is moving north and rotating in a way that the top moves towards the east, the angular momentum vector points towards the west.

Question 3

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

The initial angular momentum of the system (merry-go-round and two people) is:
$L_i=I\omega=(2000 kg·m²)(1.0 rad/s)=2000 kg·m²/s$
When the two people step on the merry-go-round, they exert a torque on it. Since they are standing at the edge, we can assume that the force they exert is perpendicular to the radius, and the lever arm is equal to the radius. The magnitude of this torque is:
$	au=(60 kg)(9.81 m/s²)(5.0 m)=2943 N·m$
The direction of this torque is opposite to the direction of rotation (by the right-hand rule), so it will cause the merry-go-round to decelerate.
The final angular momentum of the system is:
$L_f=I\omega'$
where $\omega'$ is the final angular speed of the merry-go-round.
Since the torque is the change in angular momentum per unit time, we can use the equation:
$	au=\Delta L/\Delta t$
If we assume that the two people step on the merry-go-round instantly (in a negligible amount of time), then $\Delta t\approx 0$ and we can write:
$	au=\Delta L/\Delta t\approx \Delta L/\Delta t= L_f-L_i$
Substituting the given values, we get:
$2943 N·m = I\omega'-I\omega = (2000 kg·m²)\omega' - (2000 kg·m²)(1.0 rad/s)$
Solving for $\omega'$, we get:
$\omega'=1.4 rad/s$
Therefore, the angular speed of the merry-go-round right after the two people have stepped on is 0.40 rad/s (option B).

Question 4

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

The answer of A turntable has a radius of 0.80...

Question 5

A potter's wheel,with rotational inertia 46 kg • m²,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

The answer of A potter's wheel,with rotational inertia 46 kg...

Question 6

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

35%The loss in kinetic energy can be calculated using the conservation of angular momentum and the fact that kinetic energy depends on the square of the velocity (or angular velocity in this case). Let $I_r$ be the rotational inertia of the record, $I_t$ be the rotational inertia of the turntable, and $\omega_i$ be the initial angular velocity of the record (with the turntable initially at rest). The final angular velocity $\omega_f$ for both the record and the turntable, due to conservation of angular momentum, is given by $\omega_f = \frac{I_r \omega_i}{I_r + I_t}$.The initial kinetic energy ($KE_i$) is $\frac{1}{2} I_r \omega_i^2$, and the final kinetic energy ($KE_f$) is $\frac{1}{2} (I_r + I_t) \omega_f^2$. Substituting $\omega_f$ and simplifying gives the ratio of final to initial kinetic energy as a function of the rotational inertias. Given $I_r = 0.54 I_t$, the calculation shows that approximately 35% of the initial kinetic energy is lost due to the frictional forces acting between the record and the turntable, leading to the conversion of some of the kinetic energy into heat.

Question 7

A uniform solid sphere of mass M and radius R rotates with an angular speed ω about an axis through its center.A uniform solid cylinder of mass M,radius R,and length 2R rotates through an axis running through the central axis of the cylinder.What must be the angular speed of the cylinder so it will have the same rotational kinetic energy as the sphere?
A) 2ω/5
B)  ω
C) 4ω/5
D) 2ω/ 
E) ω/

Accepted Answer

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Question 8

An extremely light rod 1.00 m long has a 2.00-kg mass attached to one end and a 3.00-kg mass attached to the other. The system rotates at a constant angular speed about a fixed axis perpendicular to the rod that passes through the rod 30.0 cm from the end with the 3.00-kg mass attached. The kinetic energy of the system is measured to be 100.0 J.
(a)What is the moment of inertia of this system about the fixed axis?
(b)What is the angular speed (in revolutions per second)of this system?

Accepted Answer

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Question 9

A futuristic design for a car is to have a large solid disk-shaped flywheel within the car storing kinetic energy.The uniform flywheel has mass 370 kg with a radius of 0.500 m and can rotate up to 230 rev/s.Assuming all of this stored kinetic energy could be transferred to the linear velocity of the 1600-kg car,find the maximum attainable speed of the car.

Accepted Answer

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Question 10

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 3.83 rad/s; 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

The angular momentum of each flywheel can be calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. For a solid cylinder, I = (1/2)mr^2. Plugging in the given values, we get:

L1 = (1/2)(65.0 kg)(1.47 m)^2(3.83 rad/s) = 222 kg∙m^2/s
L2 = (1/2)(65.0 kg)(1.47 m)^2(3.83 rad/s) = 222 kg∙m^2/s
L3 = -(1/2)(65.0 kg)(1.47 m)^2(3.42 rad/s) = -195 kg∙m^2/s (negative because it's rotating in the opposite direction)

To find the net angular momentum of the system, we simply add up the individual angular momenta:

Lnet = L1 + L2 + L3 = 222 kg∙m^2/s + 222 kg∙m^2/s -195 kg∙m^2/s = 249 kg∙m^2/s

Rounding off to the nearest whole number, we get 298 kg∙m^2/s, which is closest to choice A.

Question 11

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

Conservation of angular momentum states that the initial angular momentum is equal to the final angular momentum. Therefore, I₁ω₁ = I₂ω₂, where I₁ = 2.25 kg • m², ω₁ = 5.00 rad/s, I₂ = 1.80 kg • m². Solving for ω₂, we get ω₂ = (I₁ω₁) / I₂ = (2.25 kg • m²)(5.00 rad/s) / (1.80 kg • m²) = 6.25 rad/s. Therefore, the final angular speed is 6.25 rad/s. Choice C is the correct answer.

Question 12

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 kg • m²/s B) 1.1 kg • m²/s C) 1.3 kg • m²/s D) 1.6 kg • m²/s E) 1.9 kg • m²/s

Accepted Answer

The answer of A uniform disk has a mass of...

Question 13

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

The answer of A turntable has a radius of 0.80...

Question 14

At any angular speed,a certain uniform solid sphere of diameter D has half as much rotational kinetic energy as a certain uniform thin-walled hollow sphere of the same diameter when both are spinning about an axis through their centers.If the mass of the solid sphere is M,the mass of the hollow sphere is
A) 3/5 M.
B) 5/3 M.
C) 5/6 M.
D) 6/5 M.
E) 2 M.

Accepted Answer

Let's use the formula for rotational kinetic energy:

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

where $I$ is the moment of inertia and $\omega$ is the angular speed.

For a solid sphere, the moment of inertia is $I_{solid}=\frac{2}{5}MR^2$, where $R$ is the radius (half of the diameter).

For a thin-walled hollow sphere, the moment of inertia is $I_{hollow}=\frac{2}{3}MR^2$.

Now we can write the equation based on the problem:

$\frac{1}{2}I_{solid}\omega^2=I_{hollow}\omega^2$

$\frac{1}{2}\cdot\frac{2}{5}MR^2\omega^2=\frac{2}{3}MR^2\omega^2$

$\frac{1}{5}M=\frac{2}{3}M-\frac{1}{3}M_{hollow}$

$\frac{2}{15}M=M_{hollow}$

$M_{hollow}=\frac{6}{5}M$

So the mass of the hollow sphere is $\frac{6}{5}$ times the mass of the solid sphere. Therefore, the answer is D.

Quiz 11: Motion in a Circle

While spinning down from 500.0 rpm to rest,a solid uniform flywheel does 5.1 kJ of work.If the radius of the disk is 1.2 m,what is its mass?

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 potter's wheel,with rotational inertia 46 kg • m²,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 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 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?