Download Loop the Loop with a Twist

skiladæmi 8
Due: 11:59pm on Wednesday, October 28, 2015
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Loop the Loop with a Twist
In this problem you will consider the motion of a cylinder of radius rcyl that is rolled from a certain height h so that it "loops
the loop," that is, rolls around the track with a loop of radius rloop shown in the figure without losing contact with the track. Unless otherwise stated, assume that friction is sufficient that the
cylinder rolls without slipping. The radius rcyl of the cylinder is
much smaller than the radius rloop of the loop.
Part A
Compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the
same,a greater, or a lesser height in order just barely to complete the loop the loop?
Hint 1. How to approach the question
In order just to complete the loop without falling off, each object will need a certain critical speed v crit at the top
of the loop. The value of v crit will be the same for all objects. Each object will need to be dropped from a certain
height h in order to achieve a speed v crit at the top of the loop. The most straightforward way to find h is to use
conservation of mechanical energy.
Hint 2. The difference between a sliding and a rolling object
The object that slides will have only translational kinetic energy. The rolling object, however, will have both
translational kinetic energy and rotational kinetic energy.
ANSWER:
The rolling object should be released from a greater height.
The rolling object should be released from a lesser height.
The rolling object should be released from exactly the same height.
The answer depends on the moment of inertia of the rolling object.
Correct
Part B
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Find the minimum height h that will allow a solid cylinder of mass m and radius rcyl to loop the loop of radius rloop .
Express h in terms of the radius rloop of the loop.
Hint 1. How to approach the problem
First, determine the minimum speed the cylinder needs to have at the top of the loop in order to stay in contact
with the track. Then, compute the total mechanical energy of the cylinder at the top of the loop (potential plus
kinetic energy). Apply conservation of energy to find the height h from which the object must be released.
Hint 2. Find the critical speed at the top of the loop
To stay in contact with the track, the cylinder must exceed some critical speed v crit at the top of the loop. Find
an expression for v crit .
Express your answer in terms of the loop's radius rloop and the acceleration due to gravity g.
Hint 1. How to approach the problem
Use what you know about circular motion to determine the minimum accleration needed to keep the
object in contact with the track. Then identify all of the forces acting on the object. Apply Newton's
second law to relate these two quantities.
Hint 2. Find the acceleration at the top of the loop
In order to stay in contact with the track, the object must have an appropriate centripetal acceleration.
Find ac , the acceleration of a cylinder in the vertical direction at the top of the loop, in terms of its speed v.
Express ac in terms of the speed v of the object at the top of the loop and the radius rloop of the
loop. Your answer should obey the convention that a positive (negative) value for ac indicates an
upward (downward) acceleration.
ANSWER:
ac
= 2
−(v )
rloop
Hint 3. Find the sum of vertical forces
Assume that the object is traveling at the critical velocity, so that it just barely touches the track at the
top of the loop. Find the sum of the forces acting on the object at the top of the loop.
Express your answer in terms of given quantities and g, the magnitude of the accleration due to
gravity. The variable ac should not appear in your answer.
ANSWER:
∑ Fy
= −mg
ANSWER:
v crit
= −
−
−−
−−
√ (grloop )
Hint 3. Find the potential energy at top of loop
Find the potential energy Utop of the cylinder at the top of the loop, taking the gravitational potential to be zero
at the bottom of the loop.
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Express your answer in terms of m, g, and rloop .
ANSWER:
Utop
= 2mgrloop
Hint 4. Find the kinetic energy at the top of the loop
Find the total kinetic energy Ktop of the cylinder at the top of the loop. Assume that it reaches the top of the
loop with speed v .
Express your answer in terms of v and any other given quantities.
Hint 1. Find the rotational kinetic energy
Find the rotational kinetic energy Kr of the cylinder at the top of the loop.
Express your answer in terms of the cylinder's mass m and its translational speed v .
Hint 1. Formula for rotational kinetic energy
If an object has moment of inertia I with respect to an axis about which it is rotating with angular
velocity ω, its rotational kinetic energy is given by 1
2
Iω
2
.
Hint 2. Formula for the moment of inertia of a cylinder
If the cylinder has mass m and radius r, the appropriate moment of inertia is Icm
=
1
2
2
mr
Hint 3. Find the angular velocity of the cylinder
Find an expression for ω, the magnitude of the angular velocity of the cylinder.
Express your answer in terms of the cylinder's translational speed v and its radius rcyl .
ANSWER:
ω
v
= rcyl
ANSWER:
Kr
1
= 4
2
mv
ANSWER:
Ktop
= 3
4
ANSWER:
h
= 11
4
2
mv
rloop
Correct
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.
Exercise 10.51
The rotor (flywheel) of a toy gyroscope has mass 0.130 kg . Its moment of inertia about its axis is 1.20 × 10 −4 kg ⋅ m2 .
The mass of the frame is 2.50×10−2 kg . The gyroscope is supported on a single pivot with its center of mass a horizontal
distance of 4.00 cm from the pivot. The gyroscope is precessing
in a horizontal plane at the rate of one revolution in 2.10 s .
Part A
Find the upward force exerted by the pivot.
ANSWER:
F
= 1.52 N Correct
Part B
Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min.
ANSWER:
ω
= 1620 rev/min Correct
PSS 10.1 Rotational Dynamics for Rigid Bodies
Learning Goal:
To practice Problem­Solving Strategy 10.1 Rotational Dynamics for Rigid Bodies.
While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a
room and attempts to swing the heavy door shut before the dragon gets to her . The door is initially perpendicular to the
wall, so it must be turned through 90∘ to close it. The door is 3.00 m tall and 1.25 m wide, and it weighs 750 N . You can
ignore the friction at the hinges. If Exena applies a force of 220 N at the edge of the door and perpendicular to it, how much
time does it take her to close the door?
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Problem­Solving Strategy: Rotational dynamics for rigid bodies
IDENTIFY the relevant concepts:
In some cases you may be able to use an energy approach. However, if the target variable is a force, a torque, an
acceleration, an angular acceleration, or an elapsed time, using Στz = I αz is almost always the best approach.
SET UP the problem using the following steps:
1. Sketch the situation and select the body or bodies to analyze.
2. Draw a free­body diagram for each body and label unknown quantities with algebraic symbols. Show the shape
of the body accurately, with all dimensions and angles that you will need for calculations of torque.
3. Choose coordinate axes for each body, and indicate a positive sense of rotation for each rotating body.
EXECUTE the solution as follows:
⃗ Στz = I αz , or both to each body.
1. Write an equation of motion for each body by applying ΣF ⃗ = ma, 2. There may be geometrical relationships between the motions of two or more bodies, as with a string that
unwinds from a pulley while turning it. Express these relationships in algebraic form.
3. Check that the number of equations matches the number of unknown quantities. Then, solve the equations to
find the target variable(s).
EVALUATE your answer:
Check that the algebraic signs of your results make sense. Whenever possible, check the results for special cases or
extreme values of quantities. Ask yourself: “Does this result make sense?”
IDENTIFY the relevant concepts
You are asked to find the time it takes Exena to close the door (a rigid body that rotates about the hinges). Energy
considerations do not easily provide information about time. Instead, applying Στz = I αz to describe the rotational motion
of the door will be the best approach.
SET UP the problem using the following steps
Part A
⃗ ⃗ In the following free­body diagrams, let F be the force exerted by Exena on the door, F g the weight of the door, and n
the magnitude of the normal force exerted by the hinges. The diagrams show a view of the door from above, with the
edge of the door in contact with the hinges at the top of each diagram. Any force directed out of the plane of the figure
is represented by a dot, and any force directed into the plane of the figure is represented by a cross. Which free­body
diagram correctly describes the system under investigation?
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ANSWER:
diagram A
diagram B
diagram C
diagram D
Correct
If we ignore the friction at the hinges, there are only three forces acting on the door: its weight, the forces exerted
by Exena, and the normal force exerted by the hinges.
Part B
Which set of axes shown in the figure represents the best orientation for the coordinate axes? As in Part A, the
diagrams show a view of the door from above, with the edge of the door in contact with the hinges at the top of each
diagram.
ANSWER:
set A
set B
set C
set D
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Correct
It is most convenient to take a point on the axis of rotation as the origin of the coordinate axes. Here, the axis of
rotation is parallel to the door and passes through the hinges, so the z axis is directed out of the plane of the figure
through the edge of the door at the top of the diagram. It also helpful to choose the positive x axis in the direction
of the force that causes the rotation. Keep in mind that counterclockwise torque is positive.
EXECUTE the solution as follows
Part C
If Exena applies a force of 220 N at the edge of the door and perpendicular to it, how much time t does it take her to
close the door?
Express your answer in seconds to three significant figures.
Hint 1. How to approach the problem
To find the time it takes Exena to shut the door, you need to first find the angular acceleration of the door, which
can be found using the equation for rotational motion. You can then use the equations associated with rotational
kinematics to solve for time.
Hint 2. Find the net torque acting on the door
Find the net torque τnet around the z axis (i.e., the direction coming out of the plane of the diagram shown in
Part B at the origin) acting on the door.
Express your answer in newton­meters to four significant figures.
Hint 1. The definition of torque
The tendency of a force to cause or change rotational motion around a chosen axis is measured by the
torque, which is the product of the magnitude of the force F and the perpendicular distance l between
the axis of rotation and the line of action of the force, or
.
τ = Fl
The distance l is also called the moment arm of the force F .
Hint 2. Determine which forces exert a torque
As shown in the free­body diagram in Part A, there are three forces acting on the door: the weight, the
normal force, and the force applied by Exena. Which one of these forces exerts a nonzero torque on the
door around the z axis (i.e., the direction coming out of the plane of the diagram at the origin)?
Check all that apply.
ANSWER:
the normal force
the weight
the force applied by Exena
Hint 3. Find the moment arm of the force applied by Exena
What is the moment arm l of the force applied by Exena? Recall that Exena applies a force perpendicular
to the door and the door is 3.00 m high and 1.25 m wide.
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Express your answer in meters to four significant figures.
ANSWER:
l
= 1.250 m ANSWER:
τnet
= 275.0 N ⋅ m Hint 3. Find the moment of inertia
What is the moment of inertia I of the door with respect to an axis along its vertical edge?
Express your answer in kilogram­meters squared to four significant figures.
Hint 1. The moment of inertia of a rectangular plate
The moment of inertia I of a thin rectangular plate of width a and height h with respect to an axis along
the edge of length h is
I =
1
3
Ma
2
,
where M is the mass of the plate.
Hint 2. Find the mass of the door
You are given the weight of the door. What is the mass m of the door?
Express your answer in kilograms to four significant figures.
ANSWER:
m
= 76.45 kg ANSWER:
I
= 39.82 kg ⋅ m2 Hint 4. Find the angular acceleration of the door
What is α, the angular acceleration of the door?
Express your answer in radians per second per second to three significant figures.
Hint 1. The equation for rotational motion
As mentioned in the IDENTIFY step, to study the rotational motion of the door, it is most convenient to
use the equation Στz = I αz , which states that the net torque acting on a body is equal to the product of
its angular acceleration α and its moment of inertia I .
ANSWER:
α
= 6.91 rad/s2 Typesetting math: 100%
Hint 5. Identify the rotational kinematics equation needed to find the time
As established in Part A, the door undergoes only rotational motion. Assume that Exena turns the door an angle
θ in a time interval t . Let l be the width of the door; g the gravitational acceleration; v and a, respectively, the
speed and acceleration of the center of mass of the door; and α its angular acceleration. Then, which of the
following expressions will be most useful in solving this problem?
ANSWER:
θ =
1
θ =
1
l =
a
αt
2
gt
2
1
2
at
2
2
2
= αl
ANSWER:
t
= 0.674 s Correct
EVALUATE your answer
Part D
How long would it take Exena to close the door if she doubled the magnitude of the force applied to the door?
Express your answer in seconds to three significant figures.
ANSWER:
t
= 0.477 s Correct
This makes sense. If she pushes twice as hard, it will take less time to shut the door.
Rotating Spheres
Two massive spheres are mounted on a light rod that can be rotated by a string wrapped around a central cylinder, forming
a winch as shown in the figure. A force of magnitude F is applied to the string to turn the system. With respect to the
variables given in the figure, the equation for the magnitude of the angular acceleration α is
α=
rF
2
3mR
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.
Assume that the spheres are small enough that they may be considered point masses and that the masses of the rod and
cylinder can be neglected.
Part A
If the sphere on the left is moved closer to the central cylinder and placed at a distance R/2 from the axis of rotation,
what is the magnitude of the angular acceleration α of the modified system? Assume that the rest of the system
doesn't change.
Hint 1. How to approach the problem
The rotational analogue of Newton's second law states that the angular acceleration of the system is equal to
the net torque exerted on the system divided by the system's moment of inertia. By moving one of the spheres
closer to the central cylinder, will either of these two quantities change?
Hint 2. Find the net torque
Write an expression for the magnitude of the net torque τnet exerted on the system.
Express your answer in terms of some or all of the variables F , m, r, and R.
Hint 1. Torque
The magnitude of the torque τ of a force of magnitude F with respect to a point O is defined as the
product of F times the perpendicular distance l between the line of action of the force and the point O.
That is,
.
τ = Fl
ANSWER:
τnet
= rF
Hint 3. Find the moment of inertia
When the sphere of mass m on the right is at a distance R from the axis of rotation, and the sphere of mass 2m on the left is at a distance R/2 from the axis of rotation,what is the moment of inertia I of the system with
respect to the axis of rotation?
Express your answer in terms of some or all of the variables F , m, r, and R.
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Hint 1. Moment of inertia
The moment of inertia I of a system of objects with respect to a given axis is defined as the sum of the
product of the mass mi of each object in the system and the square of its perpendicular distance ri from
the axis, or
2
I = m1 r
1
2
+ m2 r
2
2
+ m3 r
3
+⋯
,
where r1 is the distance of the object of mass m1 from the axis of rotation, and so on.
ANSWER:
I
= 3
2
2
mR
ANSWER:
α =
rF
2
3mR
α =
3rF
2
2mR
α =
2rF
2
3mR
α =
rF
2
2mR
Correct
If one of the spheres is moved closer to the axis of rotation, the angular acceleration of the system increases
because now the system has a smaller moment of inertia. Specifically, the moment of inertia of the modified
system is half the moment of inertia of the original system, which means that the angular acceleration of the
modified system is twice that of the original system.
Consider again the original system. Instead of applying the force to the string, a force with the same magnitude F is applied
to the rod at a point R/2 from the sphere of mass 2m and in a direction perpendicular to the rod, as shown in the figure.
Part B
What is the magnitude of the angular acceleration α of the system?
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Hint 1. How to approach the problem
As you did in the previous part, use the rotational analogue of Newton's second law to find the angular
acceleration of the system. Application of the force of magnitude F directly to the rod still causes the rod to
rotate around an axis that passes through its midpoint, as in the original system. However, now the
perpendicular distance between the line of action of F and the axis of rotation is different. Will this affect either
the net torque or the moment of inertia of the system?
Hint 2. Find the net torque
When the force of magnitude F is applied to the rod at R/2 from the sphere of mass 2m and in a direction
perpendicular to the rod, what is the magnitude of the net torque τnet exerted on the system?
Express your answer in terms of some or all of the variables F , m, r, and R.
ANSWER:
τnet
= RF
2
Hint 3. Find the moment of inertia
What is the moment of inertia I of the system with respect to the axis of rotation, when the force F is applied to
the rod at R/2 from the sphere of mass m and in a direction perpendicular to the rod?
Express your answer in terms of some or all of the variables F , m, r, and R.
ANSWER:
I
= 3mR
2
ANSWER:
α =
rF
2
3mR
α =
α =
α =
F
6mR
F
6m
F
2
3mR
Correct
In conclusion, the angular acceleration of a rotating system will change if either the moment of inertia of the
system or the net torque on it changes.
Exercise 10.12
A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shown
in . The pulley is a uniform disk with mass 11.0 kg and radius 39.0 cm and turns on frictionless bearings. You measure
that the stone travels a distance 12.9 m during a time interval of 2.00 s starting from rest.
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Part A
Find the mass of the stone.
Express your answer with the appropriate units.
ANSWER:
m
= 10.6 kg
Correct
Part B
Find the tension in the wire.
Express your answer with the appropriate units.
ANSWER:
T
= 35.5 N
Correct
Alternative Exercise 10.123
A horizontal plywood disk with mass 6.60 kg and diameter 1.06 m pivots on frictionless bearings about a vertical axis
through its center. You attach a circular model­railroad track of negligible mass and average diameter 0.960 m to the disk.
A 1.10 −kg , battery­driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch
on the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 m/s relative to the
tracks.
Part A
Find the magnitude and direction of the angular velocity of the disk relative to the earth.
ANSWER:
ω disk
= 0.268 rad/s Typesetting math: 100%
Correct
Part B
ANSWER:
counterclockwise
clockwise
Correct
Exercise 10.42
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 Correct
Score Summary:
Your score on this assignment is 99.2%.
You received 6.94 out of a possible total of 7 points.
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