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Course by: Sunil Kumar Singh. E-mail the author

# Role of friction in rolling (check your understanding)

Module by: Sunil Kumar Singh. E-mail the author

Summary: 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 ". 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 difficult, 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.

## Understanding level (Role of friction in rolling motion)

### Exercise 1

In the front wheel driven car,

(a) friction at the rear wheel is in the direction of motion (b) friction at the rear wheel is in the opposite direction of motion (c) friction at the front wheel is in the direction of motion (d) friction at the front wheel is in the opposite direction of motion (a) friction at the rear wheel is in the direction of motion (b) friction at the rear wheel is in the opposite direction of motion (c) friction at the front wheel is in the direction of motion (d) friction at the front wheel is in the opposite direction of motion

#### Solution

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 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.

Hence, options (b) and (c) are correct.

### Exercise 2

A sphere can roll on

(a) a smooth horizontal plane (b) a smooth incline plane (c) a rough horizontal plane (d) a rough incline plane (a) a smooth horizontal plane (b) a smooth incline plane (c) a rough horizontal plane (d) a rough incline plane

#### Solution

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.

## Application level (Role of friction in rolling)

### Exercise 3

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 figure. If the disk rolls along a straight line as a result, then find the acceleration of the disk.

(a) 2 F 3 M (b) F 2 M (c) 4 F 3 M (d) 2 F 5 M (a) 2 F 3 M (b) F 2 M (c) 4 F 3 M (d) 2 F 5 M

#### Solution

This question is same as discussed in the course. The only difference 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.

The net force on the disk, on the other hand, imparts linear acceleration. Considering translation of the disk, the second law yields :

F + f S = M a C F + f S = M a C
(1)

Similarly, considering rotation of the disk, the second law yields :

- F R + f S R = I α - F R + f S R = I α

For rolling motion, a = - αR (negative for clockwise rotation). Substituting for “α”, we have :

- F R + f S R = - M R 2 x a C R - F + f S = - M a C 2 - F R + f S R = - M R 2 x a C R - F + f S = - M a C 2
(2)

Eliminating friction by solving two resulting (numbered) equations,

2 F = M a C + M a C 2 = 3 M a C 2 F = 3 M a C 4 a C = 4 F 3 M 2 F = M a C + M a C 2 = 3 M a C 2 F = 3 M a C 4 a C = 4 F 3 M

Hence, option (c) is correct.

### Exercise 4

A horizontal force,”F”, acts through the center of mass of a disk. If coefficient 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.

(a) μ M g (b) 3 μ M g (c) 3 μ M g 2 (d) μ M g 3 (a) μ M g (b) 3 μ M g (c) 3 μ M g 2 (d) μ M g 3

#### Solution

For sliding to start, the static friction between disk and the surface should be limiting stating friction.

f S = F S f S = F S

Now, limiting friction is given by :

F S = μ N = μ M g F S = μ N = μ M g

In order to find 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 - f S = M a C F - f S = M a C
(3)

Similarly, for rotation :

τ = f S R = I α = 1 2 x M R 2 a C R = M R a C 2 τ = f S R = I α = 1 2 x M R 2 a C R = M R a C 2

f S = M a C 2 f S = M a C 2
(4)

Substituting in the equation of motion for translation, we have :

f S = F 3 f S = F 3

The disk starts sliding, when static friction becomes equal to the limiting friction,

f S = F S f S = F S

F 3 = μ M g F 3 = μ M g

F = 3 μ M g F = 3 μ M g

Hence, option (b) is correct.

### Exercise 5

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 figure. Here,

(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 (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

#### Solution

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 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.


1. (b) and (c)    2. (a), (c) and (d)  3. (c)    4. (b) 5. (c) and (d)


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