OpenStax-CNX module: m14393
1
Role of friction in rolling (check
∗
your understanding)
Sunil Kumar Singh
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 2.0
Abstract
Objective questions, contained in this module with hidden solutions, help improve understanding of
the topics covered under the module "Uniform circular motion".
The questions have been selected to enhance understanding of the topics covered in the module titled "
Role of friction in rolling motion
1
". All questions are multiple choice questions with one or more correct
answers. The understanding level questions seek to unravel the fundamental concepts involved, whereas
application level are relatively dicult, which may interlink concepts from other topics.
Each of the questions is provided with solution. However, it is recommended that solutions may be seen
only when your answers do not match with the ones given at the end of this module.
1 Understanding level (Role of friction in rolling motion)
Exercise 1
(Solution on p. 5.)
In the front wheel driven car,
(a)
(b)
(c)
(d)
friction at the rear wheel is in the direction of motion
friction at the rear wheel is in the opposite direction of motion
friction at the front wheel is in the direction of motion
friction at the front wheel is in the opposite direction of motion
Exercise 2
(Solution on p. 6.)
A sphere can roll on
∗ Version
(a)
a smooth horizontal plane
(c)
a rough horizontal plane
(b)
(d)
a smooth incline plane
a rough incline plane
1.3: Apr 14, 2007 10:33 am -0500
† http://creativecommons.org/licenses/by/2.0/
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2 Application level (Role of friction in rolling)
Exercise 3
(Solution on p. 6.)
A disk of mass M and radius R is at rest on a horizontal surface. An external force F acts
tangentially at the highest point as shown in the gure. If the disk rolls along a straight line as a
result, then nd the acceleration of the disk.
A disk in rolling motion
Figure 1:
external force F acts tangentially at the highest point.
(a)
Exercise 4
2F
3M
(b)
F
2M
(c)
4F
3M
(d)
2F
5M
(Solution on p. 7.)
A horizontal force,F, acts through the center of mass of a disk. If coecient of friction between
the surfaces be µ and mass of the disk be M, then what should be the minimum value of F so
that disk starts sliding.
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A disk in rolling motion
Figure 2:
The external force F acts through center of mass of the disk.
(a) µM g (b) 3µM g (c)
3µM g
2
(d)
µM g
3
Exercise 5
(Solution on p. 9.)
A hollow metal cylinder is placed on a thin rectangular plastic sheet, kept on a horizontal plane
table. The plastic sheet is pulled to right rapidly from the bottom of the cylinder. As a consequence,
the cylinder starts rotating and is carried along with the plastic sheet to the left as shown in the
gure. Here,
Motion of cylinder on a thin plastic sheet
Figure 3:
The plastic sheet is pulled out from beneath rapidly.
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(a)
rotation of cylinder is anticlockwise
(b)
motion of cylinder is pure rolling
(c)
the cylinder has tendency to slide to the right
(d)
the friction acts from right to left
3 Answers
1. (b) and (c)
2. (a), (c) and (d)
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3. (c)
4. (b) 5. (c) and (d)
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Solutions to Exercises in this Module
Solution to Exercise (p. 1)
The engine imparts torque to the front wheel to rotate. Friction converts angular acceleration into linear
acceleration by providing a translational force to accelerate and a torque to counteract angular acceleration.
The two functions as outlined here dictate that friction is in the direction of motion.
The wheel rolls in forward direction
Figure 4:
The friction acts in the direction of motion at the front wheel.
The translation of car, now, pulls the rear wheel to translate as well.
Friction, here, converts linear
acceleration into angular acceleration - by providing a torque to impart angular acceleration and a translation
force to decelerate the wheel in translation. The two functions as outlined here dictate that friction is in the
opposite direction of motion.
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The wheel rolls in forward direction
Figure 5:
The friction acts in the opposite direction of motion at the rear wheel.
Hence, options (b) and (c) are correct.
Solution to Exercise (p. 1)
Rolling at constant velocity does not require external force and friction. Hence, a circular body can roll
on a smooth horizontal plane with constant velocity. In the case of an incline, gravity works on the body
through the center of mass to impart acceleration. This forms the external force to cause linear acceleration.
For a smooth incline, there is no friction.
The external component of gravity along the direction of
incline pulls the sphere in translation only as there is no friction. As such, smooth incline can not support
accelerated rolling. On the other hand, rough horizontal plane and incline both support rolling by applying
friction.
Hence, options (a), (c) and (d) are correct.
Solution to Exercise (p. 2)
This question is same as discussed in the course. The only dierence is that we need to solve the equation
for the acceleration instead of friction.
First thing to note here is that external force acts tangentially and, therefore, constitutes a torque about
axis of rotation. The torque imparts angular acceleration to the disk. This is opposed by the friction. Its
direction is such that the torque due to friction opposes the torque due to external force. This means that
friction also acts in the direction of external force.
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A disk in rolling motion
Figure 6:
external force F acts tangentially at the highest point.
The net force on the disk, on the other hand, imparts linear acceleration. Considering translation of the
disk, the second law yields :
(1)
F + fS = M a C
Similarly, considering rotation of the disk, the second law yields :
− F R + fS R = Iα
For rolling motion, a = -
αR
(negative for clockwise rotation). Substituting for α, we have :
− F R + fS R = − M R2 x
− F + fS = −
aC
R
M aC
2
Eliminating friction by solving two resulting (numbered) equations,
2F = M aC +
⇒ F =
⇒ aC =
M aC
2
=
3M aC
2
3M aC
4
4F
3M
Hence, option (c) is correct.
Solution to Exercise (p. 2)
For sliding to start, the static friction between disk and the surface should be limiting stating friction.
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(2)
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A disk in rolling motion
Figure 7:
The external force F acts through center of mass of the disk.
fS = FS
Now, limiting friction is given by :
FS = µN = µM g
In order to nd the static friction for rolling, we make use of two Newton's laws of motion and equation
of accelerated rolling. For translation, we have :
For translation, we have :
F − fS = M a C
(3)
Similarly, for rotation :
τ = fS R = Iα =
1
2
fS =
x M R2 aRC =
M RaC
2
M aC
2
Substituting in the equation of motion for translation, we have :
fS =
F
3
The disk starts sliding, when static friction becomes equal to the limiting friction,
fS = FS
F
3
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= µM g
(4)
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⇒ F = 3µM g
Hence, option (b) is correct.
Solution to Exercise (p. 3)
As the plastic is pulled out from the bottom in the direction from left to right, the cylinder tends to slide
in the opposite direction to the motion of plastic sheet (the cylinder tends to stay where it was). Friction,
in turn, appears in the opposite direction i.e. from right to left to counter the sliding tendency.
A disk in rolling motion
Figure 8:
The plastic sheet is pulled out from beneath rapidly.
A tangential friction from right to left constitutes a clockwise torque. This imparts a clockwise angular
acceleration to the cylinder. This means that cylinder rotates in clockwise direction.
We should carefully consider the motion of the cylinder. The cylinder rotates clockwise. Since it is also
carried with the plastic sheet from right to left, the cylinder has the translational velocity with respect to
ground - that of the plastic sheet at that moment.
This observation, here, reveals that two motions are, as a matter of fact, opposite to the motions involved
in rolling. What it means that a clockwise rotation should actually involve a left to right translation not
right to left translation as in this case.
Hence, options (c) and (d) are correct.
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