Download Exercise 10.2 Torque Magnitude Ranking Task

Exercise 10.2
Part A
Calculate the net torque about point O for the two forces applied as in the figure . The rod and both forces are
in the plane of the page. Take positive torques to be
counterclockwise.
ANSWER:
τ
= ­28.0 N ⋅ m Torque Magnitude Ranking Task
The wrench in the figure has six forces of equal magnitude acting on it.
Typesetting math: 100%
1/22
Chapter 10_assignment 9
Part A
Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench,
measured about an axis centered on the bolt.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Definition of torque
Torque is a measure of the "twist" that an applied force exerts on an object. Mathematically, torque is
defined as
τ = rF sin θ ,
where r is the magnitude of the displacement vector from the rotation axis to the point of application of
the force of magnitude F , and θ is the angle between this displacement and the applied force, as shown
in the figure.
The direction of a torque can be either counterclockwise (as above) or clockwise. This is determined by
the direction the object will rotate under the action of the force.
Hint 2. Maximum torque
Based on the mathematical definition of torque, torque is maximized when the force is large in
magnitude, located a large distance from the axis of interest, and oriented perpendicular to the
displacement r ⃗, which is often referred to as the lever arm of the force.
ANSWER:
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Chapter 10_assignment 9
Exercise 10.8
A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long, and he exerts a 17.0­N force at the end
of the handle at 37 ∘ with the handle (the figure ).
Part A
What is the magnitude of the torque does the machinist exert about the center of the nut?
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Chapter 10_assignment 9
|τ |
= 2.56 N ⋅ m Part B
What is the direction of the torque in part (A).
ANSWER:
counterclockwise
clockwise
Part C
What is the maximum torque he could exert with this force?
ANSWER:
|τ |
= 4.25 N ⋅ m Part D
How should the force mentioned in part (C) be oriented ?
ANSWER:
The force is directed into the page.
The force is directed out of the page.
The force is perpendicular to the wrench.
The force is parallel to the wrench.
Exercise 10.9
The flywheel of an engine has a moment of inertia 2.70 kg ⋅ m2 about its rotation axis.
Part A
rev/min
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Chapter 10_assignment 9
What constant torque is required to bring it up to an angular speed of 350 rev/min in a time of 8.20 s , starting
from rest?
ANSWER:
12.1 N ⋅ m Part B
What is its final kinetic energy?
ANSWER:
1810 J Exercise 10.15
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential
force equal to 81.0 N is applied to the rim of the wheel. The wheel has radius 0.150 m . Starting from rest, the
wheel has an angular speed of 13.5 rev/s after 2.23 s .
Part A
What is the moment of inertia of the wheel?
Express your answer with the appropriate units.
ANSWER:
I
= 0.319 kg⋅m2
Exercise 10.18
We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 67.0 kg and diameter 1.20 m .
Part A
If this tumbler rolls forward at 0.450 rev/s , how much total kinetic energy does he have?
ANSWER:
K
= 145 J 5/22
Chapter 10_assignment 9
Part B
What percent of his total kinetic energy is rotational?
ANSWER:
K rot
K
= 33.3 % Exercise 10.20
A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg . The free
end of the string is held in place and the hoop is released from rest (the figure ). After the hoop has descended 90.0 cm , calculate
Part A
the angular speed of the rotating hoop and
ANSWER:
ω
= 37.1 rad/s Part B
the speed of its center. ANSWER:
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Chapter 10_assignment 9
v
= 2.97 m/s Exercise 10.28
A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is
75.0 m above the foot of the hill. We can model the wheel as a thin­walled cylinder 85.0 cm in diameter and neglect
the small mass of the spokes.
Part A
How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down?
ANSWER:
vf
= 29.3 m/s Part B
How much total kinetic energy does the wheel have when it reaches the bottom of the hill?
ANSWER:
Kf
= 1930 J Exercise 10.29
A size­5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height
of 7.50 m above the base of the hill. We can model this ball as a thin­walled hollow sphere.
Part A
At what rate was it rotating at the base of the hill?
ANSWER:
ω
= 83.1 rad/s 7/22
Chapter 10_assignment 9
Part B
How much rotational kinetic energy did it then have?
ANSWER:
Krot
= 12.5 J Exercise 10.30
The engine of an aircraft propeller delivers an amount of power 177 hp to the propeller at a rotational velocity of
2330 rev/min .
Part A
How much torque does the aircraft engine provide?
ANSWER:
541 N ⋅ m Part B
How much work does the engine do in one revolution of the propeller?
ANSWER:
3400 J Exercise 10.32
An electric motor consumes 8.80 kJ of electrical energy in 1.00 min .
Part A
If one­third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to
the motor output, how much torque will this engine develop if you run it at 2400 rpm ?
ANSWER:
τ
= 0.389 N ⋅ m 8/22
Chapter 10_assignment 9
Exercise 10.35
Part A
Compute the torque developed by an industrial motor whose output is 160 kW at an angular speed of 5000 rev/min .
ANSWER:
τ
= 306 N ⋅ m Part B
A drum with negligible mass, 0.400 m in diameter, is attached to the motor shaft, and the power output of the
motor is used to raise a weight hanging from a rope wrapped around the drum. How heavy a weight can the
motor lift at constant speed?
ANSWER:
w
= 1530 N Part C
At what constant speed will the weight rise?
ANSWER:
v
= 105 m/s ± A Simple Gear System
Gears are important components in many mechanical devices, from mechanical clocks to bicycles. In fact, they are
present whenever a motor produces rotational motion.
An example of a simple gear system is shown in the figure. The bigger wheel (wheel 1) has radius r 1 , while the
smaller one (wheel 2) has radius r 2 . The two wheels have small teeth and are connected through a metal chain so
that when wheel 1 rotates, the chain moves with it and causes wheel 2 to rotate as well.
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Chapter 10_assignment 9
Part A
ω1
Let wheel 1 rotate at a constant angular speed ω1 . Find the ratio ω of the angular speed of wheel 1 to the
2
angular speed of wheel 2.
Express your answer in terms of any or all of the variables r 1 and r 2 .
Hint 1. How to approach the problem
As wheel 1 rotates, the chain is set into motion. Since the chain moves without slipping, its speed must
match the speed of the teeth in contact with it. This movement of the chain is then transferred to the
teeth of wheel 2, which then move at the same speed as the chain.
Write an expression for the speed of the links of the chain in terms of ω1 , and use it to find the angular
speed of wheel 2.
Hint 2. Find the speed of the chain
What is the speed v of the chain when wheel 1 rotates with an angular speed ω1 ?
Express your answer in terms of any or all of the variables ω1 , r 1 , and r 2 .
Hint 1. Relation between linear and rotational motion
The linear speed of an object that rotates with angular speed ω at a distance r from the axis of
rotation is given by
v = ωr.
ANSWER:
v
= ω1 r 1
Hint 3. Find the angular speed of wheel 2
What is the angular speed ω2 of wheel 2 if the chain moves with speed v ?
Express your answer in terms of the variables v and r 2 .
ANSWER:
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Chapter 10_assignment 9
ω2
= v
r2
ANSWER:
ω1
ω2
=
r2
r1
Part B
τ1
The rotation of wheel 1 is caused by a torque τ 1 . Find the ratio τ of the torque acting on wheel 1 to the torque
2
acting on wheel 2.
Express your answer in terms of any or all of the variables r 1 and r 2 .
Hint 1. How to approach the problem
The force that the teeth on wheel 1 exert on the chain is equal to the force that causes the rotation of
wheel 2. Moreover, the chain will exert the same force on the teeth of wheel 2.
Write an expression for the force exerted on the chain in terms of τ 1 , and use it to find the torque on
wheel 2.
Hint 2. Find the force exerted on the chain
If a torque τ 1 causes the rotation of wheel 1, what is the force Fc exerted on the chain by the teeth on
wheel 1?
Express your answer in terms of any or all of the variables τ 1 , r 1 , and r 2 .
Hint 1. Force and torque
When a force F acting on a body causes a rotation of the body, a torque τ is exerted on the
body. The distance d between the point of application of the force and the pivot is called the
lever arm of the force. If the force F acts in a direction perpendicular to the lever arm, the torque
exerted on the body is given by
τ = Fd
.
ANSWER:
Fc
= τ1
r1
Hint 3. Find the torque on wheel 2
What is the torque τ 2 on wheel 2 if the chain exerts a force Fc on the teeth of wheel 2?
Express your answer in terms of the variables Fc and r 2 .
ANSWER:
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Chapter 10_assignment 9
τ2
= Fc r 2
ANSWER:
τ1
τ2
=
r1
r2
Part C
If the power needed to rotate wheel 1 is P1 , what is the ratio 2?
P1
P2
of the power of wheel 1 to the power of wheel
Express your answer in terms of any or all of the variables r 1 and r 2 .
Hint 1. How to approach the problem
Apply the definition of power in rotational motion and use the information found in the previous parts.
Hint 2. Power in rotational motion
When a torque τ acts on a body that rotates with angular speed ω, its power is given by P
= τ ω.
ANSWER:
P1
P2
= 1
While the power transferred to wheel 2 is the same as the power delivered to wheel 1, the angular speed
and torque of wheel 2 depend on the ratio of the radii of the two wheels. This is typical of a gear system.
For this reason, gears are normally described by a quantity called the gear ratio, defined as g =
r1
r2
.
Now suppose that the gear system that you studied in Parts A, B, and C represents the gear wheels on a bicycle.
Wheel 1 is connected to the pedals and wheel 2 to the rear bicycle wheel, as shown in the figure. The gear ratio for
r
this particular system is defined as g = r
, that is, the ratio of the radius r f ront of the front gear wheel to the
f ront
radius r rear of the rear gear wheel.
rear
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Chapter 10_assignment 9
Part D
A cyclist uses 27,000 joules per minute (this is about 385 food Calories per hour) to cycle at a constant
pedaling rate of 450 radians per minute along a flat road using a gear ratio of 4. Find the torque τ rear exerted on
the rear gear wheel.
Express your answer in newton meters.
Hint 1. How to approach the problem
In Part B you found a relationship between the torque on wheel 2 and the torque on wheel 1 in terms of
the gear ratio. Apply the same relationship to the front and rear gear wheels of the bicycle. You can find
the torque on the front gear wheel from the definition of power in rotational motion.
Hint 2. Find an expression for the torque acting on the rear gear wheel
Use the ratio found in Part B to determine which of the following expressions represents the torque τ rear
exerted on the rear gear wheel. Let τ f ront be the torque on the front gear wheel and recall that the gear
ratio is 4.
ANSWER:
τ f ront
τ rear
= 4 ∗ τ f ront
0.25 ∗ τ f ront
8 ∗ τ f ront
Hint 3. Find the torque acting on the front gear wheel
Given that the cyclist uses 27,000 joules per minute to pedal at a rate of 450 radians per minute, find the
torque τ f ront exerted on the front gear wheel.
Express your answer in newton meters.
Hint 1. Power in rotational motion
When a torque τ acts on a body that rotates with angular speed ω, its power is given by P = τ ω.
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Chapter 10_assignment 9
τ f ront
= 60 N ⋅ m ANSWER:
τ rear
= 15.0 N ⋅ m Part E
When the cyclist encounters a steep hill, in order to maintain the same energy consumption and pedaling rate
she changes her gear ratio to 0.7. Which of the following statements describes the change caused by the new
gear ratio?
Hint 1. How to approach the question
You can use the information found in the previous parts to find the angular speed, torque, and power of
the rear gear wheel when the gear ratio changes to 0.7.
ANSWER:
The angular speed of the rear gear increases.
The torque exerted on the rear gear wheel increases.
The torque exerted on the front gear wheel increases.
The power delivered to the rear gear wheel increases.
Without changing the pedaling rate and the energy consumption, a lower gear ratio makes more torque
available to cycle up the hill. On the other hand, the angular speed of the wheels, and correspondingly the
linear speed of the bicycle, decreases.
Exercise 10.39
Part A
Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center
of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a
slender rod rotating with constant angular velocity about one end.
ANSWER:
L
= 4.71×10−6 kg ⋅ m2 /s 14/22
Chapter 10_assignment 9
Exercise 10.44
A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her
rotation axis of 18kg ⋅ m2 . She then tucks into a small ball, decreasing this moment of inertia to 3.6kg ⋅ m2 . While
tucked, she makes two complete revolutions in 1.1 s .
Part A
If she hadn't tucked at all, how many revolutions would she have made in the 1.6 s from board to water?
Express your answer using two significant figures.
ANSWER:
0.58 rev Exercise 10.50
A thin uniform rod has a length of 0.520 m and is rotating in a circle on a frictionless table. The axis of rotation is
perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.4 rad/s and
a moment of inertia about the axis of 2.50×10−3 kg ⋅ m2 . A bug initially standing on the rod at the axis of rotation
decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its
tangential speed is 0.102 m/s . The bug can be treated as a point mass.
Part A
What is the mass of the rod?
Express your answer with the appropriate units.
ANSWER:
m rod
= 2.77×10−2 kg
Part B
What is the mass of the bug?
Express your answer with the appropriate units.
ANSWER:
m bug
= 9.61×10−3 kg
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Chapter 10_assignment 9
Twirling a Baton
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod
of mass 0.120 kg and length 80.0 cm .
Part A
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s . What is its angular
momentum?
Express your answer in kilogram meters squared
per second.
Hint 1. Angular momentum for a rigid body rotating about an axis of symmetry
The angular momentum L⃗ of a rigid body that rotates about an axis of symmetry at angular velocity ω⃗ is
⃗
L = I ω⃗
,
where I is the moment of inertia of the object about the rotational axis.
Hint 2. Moment of inertia
For a uniform rod of length l with mass M , the moment of intertia about an axis passing through its
center perpendicular to the rod is
1
12
Ml
2
.
ANSWER:
2
1.92×10−2 kg ⋅ m /s Part B
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Chapter 10_assignment 9
With a skillful move, the majorette changes the rotation of her baton so that now it is spinning about an axis
passing through its end at the same angular velocity 3.00 rad/s as before. What is the new angular momentum
of the rod?
Express your answer in kilogram meters squared
per second.
Hint 1. How to approach the problem
If you know the moment of inertia of a rod about one of its ends, you can solve this part in the same
way that you solved Part A.
Hint 2. Moment of inertia
For a rod of length l with mass M , the moment of intertia about an axis passing through either end is
1
3
Ml
2
.
ANSWER:
7.68×10−2 kg ⋅ m2 /s Here is another way to solve this problem. There is a theorem that relates the angular momentum L of an
object about an arbitrary axis to the angular momentum of the object about the axis passing through its
center of mass Lcm :
,
where M is the mass of the object, r cm is the length of the position vector of the center of mass with
respect to the point chosen, andv cm is the velocity of the center of mass with respect to the point chosen.
Substituting for the values on the right­hand side would yield the same angular momentum that you
calculated.
L = M r cm v cm + L cm
Record and Turntable
Learning Goal:
To understand how to use conservation of angular momentum to solve problems involving collisions of rotating
bodies.
Consider a turntable to be a circular disk of moment of inertia I
t
rotating at a constant angular velocity ω i around
17/22
Chapter 10_assignment 9
axis of the disk is vertical and the disk is supported by
frictionless bearings. The motor of the turntable is off, so
there is no external torque being applied to the axis.
Another disk (a record) is dropped onto the first such that
it lands coaxially (the axes coincide). The moment of
inertia of the record is Ir . The initial angular velocity of
the second disk is zero.
There is friction between the two disks.
After this "rotational collision," the disks will eventually
rotate with the same angular velocity.
Part A
What is the final angular velocity, ωf , of the two disks?
Express ωf in terms of It , Ir , and ωi .
Hint 1. How to approach the problem
Because there is friction between the record and turntable, you can't use energy conservation. However,
since there are no net external torques acting on the system, angular momentum is conserved.
Hint 2. Initial angular momentum
Find the magnitude, Li , of the (combined) initial angular momenta of the two disks.
Express Li in terms of ωi , It , and/or Ir .
ANSWER:
Li
= It ω i
Hint 3. Final angular momentum
Find the magnitude, Lf , of the (combined) final angular momenta of the two disks.
Express Lf in terms of ωf , It , and/or Ir .
ANSWER:
Lf
= (It + Ir )ω f
ANSWER:
ωf
= It ω i
I t +I r
18/22
Chapter 10_assignment 9
Part B
Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other.
What is the final rotational kinetic energy, Kf , of the two spinning disks?
Express the final kinetic energy in terms of It , Ir , and the initial kinetic energy Ki of the two­disk
system. No angular velocities should appear in your answer.
Hint 1. Initial rotational kinetic energy
What is the initial rotational kinetic energy of the two­disk system, Ki .
Express your answer in terms of It and ωi .
Hint 1. Formula for rotational kinetic energy
The formula for the rotational kinetic energy K of a rigid body with moment of inertia I , spinning
with an angular velocity ω is
K=
1
2
Iω
2
ANSWER:
Ki
= 1
2
I t ωi
2
Hint 2. Final rotational kinetic energy
What is the final rotational kinetic energy of the two­disk system.
Express your answer in terms of It , Ir , and ωf .
ANSWER:
Kf
= (I t +I r )ω
f
2
2
Hint 3. Putting it all together
Use the relationship between ωi and ωf (from the answer to the first part of this problem) to express Kf in terms of ω i . Then use the equation relating Ki and ω i to express Kf in terms of Ki .
ANSWER:
Kf
= Ki I t
I t +I r
19/22
Chapter 10_assignment 9
Some of the energy was converted into heat and sound as the frictional force, torque acted, stopping
relative motion.
Part C
Assume that the turntable deccelerated during time Δt before reaching the final angular velocity ( Δt is the
time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the
same angular velocity). What was the average torque, ⟨τ ⟩ , acting on the bottom disk due to friction with the
record?
Express the torque in terms of It , ωi , ωf , and Δt.
Hint 1. Average angular acceleration
What is the average angular acceleration, ⟨α⟩ , of the bottom disk?
Express ⟨α⟩ in terms of ωi , ωf , and Δt.
Hint 1. Definition of average angular acceleration
The angular acceleration is the rate of change of angular velocity. The average angular
acceleration is the net change in angular velocity
(final angular velocity minus initial angular velocity) divided by the elapsed time.
ANSWER:
⟨α⟩ = ω f −ω i
Δt
Hint 2. Formula for torque
The torque is given by
τ =I
dω
dt
.
ANSWER:
⟨τ ⟩ = I t (ω f −ω i )
Δt
Problem 10.15
Part A
20/22
Chapter 10_assignment 9
A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that
is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends
1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N.
What is the moment of inertia of the wheel?
ANSWER:
3.9 kg · m2
4.1 kg · m2
3.5 kg · m2
3.7 kg · m2
4.3 kg · m2
Problem 10.58
An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque
of 7.00 N ⋅ m is applied to the tire for 2.00 s , the angular speed of the tire increases from 0 to 100 rev/min . The
external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s . Assume the
direction the wheel is rotating is the positive direction.
Part A
Compute the moment of inertia of the wheel about the rotation axis.
ANSWER:
I
= 1.34 kg ⋅ m2 Part B
Compute the friction torque.
ANSWER:
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Chapter 10_assignment 9
τf
= ­0.112 N ⋅ m Part C
Compute the total number of revolutions made by the wheel in the 125­s time interval.
ANSWER:
104 rev Problem 10.98
A runner of mass 56.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis
through its center. The runner's velocity relative to the earth has magnitude 2.90 m/s . The turntable is rotating in
the opposite direction with an angular velocity of magnitude 0.210 rad/s relative to the earth. The radius of the
turntable is 2.70 m , and its moment of inertia about the axis of rotation is 84.0 kg ⋅ m2 .
Part A
Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat
the runner as a particle.)
ANSWER:
0.855 rad/s 22/22