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Rotation (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 modules on "Rotation" and "Rotation of rig body".

The questions have been selected to enhance understanding of the topics covered in the modules on " Rotation ". All questions are multiple choice questions with one or more correct answers. There are two sets of the questions. 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 (Rotation)

Exercise 1

A force F = 2i + j + 3k Newton acts at a point of a rigid body, defined by the position vector r = i + j + k meters. If the body rotates about z-axis of the coordinate system, then the magnitude of torque (in N-m) accelerating the rigid body is :

(a) 1 (b) 2 (c) 3 (d) 4 (a) 1 (b) 2 (c) 3 (d) 4

Solution

This question is included here with the purpose to highlight conceptual and visualization aspects of the calculation of torque. The torque about the axis of rotation i.e. z-axis is given as :

τ = r x F τ = r x F

where “r” and “F” are in the plane perpendicular to the axis of rotation i.e. in “xy”-plane. Explicitly, we can write :

τ z = r xy x F xy τ z = r xy x F xy

Figure 1: Component of force in “xy”-plane. .
Rotation
 Rotation   (rq1.gif)

The figure above shows the force component in “xy”-plane. Have a careful look at the force component in relation to the point of application, "P". The rigid body is deliberately not shown so that we can visualize vector components without any congestion in the figure.

Note here that body is rotating about z-axis. Thus, we need to calculate torque about z-axis only. The z-component of force does not constitute a torque about z-axis. Hence,

F xy = 2 i + j F xy = 2 i + j

The component of radius vector in xy - plane is shown in the figure below. Note that it is measured in the plane of rotation of the point of application of force about the axis. Again, we drop z-component. The “xy” components of force and radius vectors are shown in the red colors in the figure.

Figure 2: Component of position vector in “xy”-plane. .
Rotation
 Rotation   (rq2a.gif)

r xy = i + j r xy = i + j

Therefore, torque about the axis of rotation is :

τ z = r xy x F xy = ( i + j ) x ( 2 i + j ) = k - 2 k = - k τ z = r xy x F xy = ( i + j ) x ( 2 i + j ) = k - 2 k = - k

The magnitude of torque is :

τ z = 1 N - m τ z = 1 N - m

Note that the torque is in the direction of “-z”-axis and the body is rotating clockwise.

Alternatively, we can also work this problem without restricting evaluation of vector product in any plane as :

τ = r x F τ = r x F

τ =    | i         j        k |
       | 1         1        1 |
       | 2         1        3 |

τ = ( 1 x 3 - 1 x 1 ) i + ( 1 x 2 - 1 x 3 ) j + ( 1 x 1 - 2 x 1 ) k τ = ( 1 x 3 - 1 x 1 ) i + ( 1 x 2 - 1 x 3 ) j + ( 1 x 1 - 2 x 1 ) k

τ = 2 i - j - k τ = 2 i - j - k

The component of torque in z-direction is :

τ z = - k τ z = - k

The magnitude of torque is :

τ z = 1 N - m τ z = 1 N - m

Hence, option (a) is correct.

Exercise 2

The overview of two horizontal forces acting at the ends of a rod, pivoted at the middle point, is shown in the figure. The rod is free to rotate in horizontal plane. However, the magnitudes of forces, acting perpendicular to the rod, are such that rod is stationary. If the angle of force F 1 F 1 with respect to rod is changed, then rod can be held stationary if :

Figure 3: A rod is acted upon by a pair of forces in horizontal plane. .
Rotation
 Rotation   (rq3.gif)

(a) force F 1 is increased (b) force F 2 is decreased (c) point of application of force F 1 is brought closure to pivot (d) point of application of force F 2 is brought closure to pivot (a) force F 1 is increased (b) force F 2 is decreased (c) point of application of force F 1 is brought closure to pivot (d) point of application of force F 2 is brought closure to pivot

Solution

The rod will not rotate when torques due to the forces are equal in magnitude and opposite in direction :

Figure 4: A rod is acted upon by a pair of forces in horizontal plane. .
Rotation
 Rotation   (rq4.gif)

τ 1 = τ 2 τ 1 = τ 2

F 1 x r 1 = F 2 x r 2 F 1 x r 1 = F 2 x r 2

Since r 1 < r 2 r 1 < r 2 , we can conclude F 1 > F 2 F 1 > F 2 . The magnitude of torque due to force F 1 F 1 is given as :

τ 1 = r 1 x F 1 sin θ τ 1 = r 1 x F 1 sin θ

Evidently, torque due to F 1 F 1 is reduced when angle of application of force with respect to rod changes. In order to maintain the balance, the force F 1 F 1 needs to be increased or the force F 2 F 2 is decreased or the force F 2 F 2 is brought closure to the pivot.

Hence, options (a), (b) and (d) are correct.

Exercise 3

A flywheel with moment of inertia of 2 kg - m 2 kg - m 2 , is rotating at 10 rad/s about a perpendicular axis. The flywheel is brought to rest uniformly in 10 seconds due to friction at the axle. The magnitude of torque due to friction is (in N-m) :

(a) 0 (b) 5 (c) 10 (d) 15 (a) 0 (b) 5 (c) 10 (d) 15

Solution

The angular deceleration is uniform i.e. constant as given in the question. The torque due to friction at the axle is, therefore, also constant. Let “α” be the deceleration.

Applying, equation of motion for constant angular acceleration, we have :

ω f = ω i + α t ω f = ω i + α t

Here, ω i = 10 rad / s ; ω f = 0 ; t = 2 s , Here, ω i = 10 rad / s ; ω f = 0 ; t = 2 s ,

0 = 10 + α x 2 0 = 10 + α x 2

α = - 5 rad / s 2 α = - 5 rad / s 2

The magnitude of torque is :

τ = I α = 2 x 5 = 10 N - m τ = I α = 2 x 5 = 10 N - m

Hence, option (c) is correct.

Exercise 4

A circular disk of mass 2 kg and radius 1 m is free to rotate about its central axis. The disk is initially at rest. Then, the constant tangential force (in Newton) required to rotate the disk with angular velocity 10 rad/s in 5 second is :

(a) 2 (b) 4 (c) 8 (d) 16 (a) 2 (b) 4 (c) 8 (d) 16

Solution

This question is similar to earlier one except that we need to find the tangential force instead of torque. Since torque is equal to the product of tangential force and moment arm, we need to find torque first and then find the required tangential force.

A constant tangential force results in a torque along the axis of rotation. The torque, in turn, produces a constant angular acceleration “α”. As such, we can apply equation of motion for constant angular acceleration,

ω f = ω i + α t ω f = ω i + α t

Here, ω i = 0 ; ω f = 10 rad / s ; t = 2 s , Here, ω i = 0 ; ω f = 10 rad / s ; t = 2 s ,

10 = 0 + α x 2 10 = 0 + α x 2

α = - 5 rad / s 2 α = - 5 rad / s 2

Applying Newton’s second law for rotation, we have :

τ = I α τ = I α

Here,

I = M R 2 = 2 x 1 2 = 2 Kg - m 2 I = M R 2 = 2 x 1 2 = 2 Kg - m 2

Hence,

τ = I α = 2 x 2 = 4 N - m τ = I α = 2 x 2 = 4 N - m

Now, torque is also given as :

τ = R F τ = R F

F = τ R = 4 1 = 4 N F = τ R = 4 1 = 4 N

Hence, option (b) is correct.

Exercise 5

A wheel of 10 kg, mounted on its central axis, is acted upon by four forces. The lines of action of forces are in the plane perpendicular to the axis of rotation as shown in the figure. If the magnitudes of forces and their directions during rotation are same, then (considering, R = 10 m) :

Figure 5: Rotation about central axis. .
Rotation
 Rotation  (rq5.gif)

(a) the magnitude of net torque is zero (b) the magnitude of net acceleration is 2 rad / s 2 (c) the magnitude of net acceleration is 4 rad / s 2 (d) the direction of rotation is anti-clockwise (a) the magnitude of net torque is zero (b) the magnitude of net acceleration is 2 rad / s 2 (c) the magnitude of net acceleration is 4 rad / s 2 (d) the direction of rotation is anti-clockwise

Solution

The forces of 3N and 5N act through the axis of rotation. They do not constitute torque. The force of 4N constitute a clockwise torque of 4 x 10 = 40 N-m. On the other hand, the force of 2N constitute an anticlockwise torque of 2 x 20 = 40 N-m.

The cylinder, therefore, is acted by two equal, but opposite torques. Therefore, the net torque on the body is zero.

Hence, option (a) is correct.

Application level (Rotation)

Exercise 6

Two flywheels of moments of inertia 1 and 2 kg - m 2 kg - m 2 and radii 1 and 2 meters respectively are connected by a belt as shown in the figure. The angular speed of larger wheel is increased at a constant rate 2 rad / s 2 rad / s 2 without any slippage between wheels and the belt. Then,

Figure 6: There is slippage between belt and flywheels .
Connected flywheels
 Connected flywheels   (rq6.gif)

(a) Tension in the belt is same (b) Linear velocity of the belt is same everywhere (c) Torques on the wheels are same (d) Angular accelerations of the wheels are same (a) Tension in the belt is same (b) Linear velocity of the belt is same everywhere (c) Torques on the wheels are same (d) Angular accelerations of the wheels are same

Solution

Since, no slippage between wheels and the belt is involved, the linear velocity of the belt is same through out.

The rotation of the larger wheel causes the smaller wheel to rotate as well. Since two wheels are accelerated from at constant rate, it means that they are acted upon by constant torques. On the other hand, a net tangential force on either of the wheels is required to constitute a torque. As such, tension in the belt needs to be different so that there is a net tangential force on the flywheel.

Let ω be the angular speed of the larger wheel. The corresponding linear speed of the belt is :

v = ω 1 x r 1 = ω x 2 = 2 ω v = ω 1 x r 1 = ω x 2 = 2 ω

The angular speed of the smaller wheel is :

ω 1 = v r 2 = 2 ω 1 = 2 ω ω 1 = v r 2 = 2 ω 1 = 2 ω

On the other hand, the linear acceleration of the belt is :

a = α 1 x r 1 = 2 x 2 = 4 m / s 2 a = α 1 x r 1 = 2 x 2 = 4 m / s 2

The angular acceleration of the smaller wheel is :

α 2 = a r 2 = 4 1 = 4 rad / s 2 α 2 = a r 2 = 4 1 = 4 rad / s 2

Thus, angular accelerations of the wheels are not same. Now, the torque on the larger wheel is :

τ 1 = I 1 x α 1 = 2 x 2 = 4 N - m τ 1 = I 1 x α 1 = 2 x 2 = 4 N - m

The torque on the smaller wheel is :

τ 2 = I 2 x α 2 = 1 x 4 = 4 N - m τ 2 = I 2 x α 2 = 1 x 4 = 4 N - m

Thus, the torques on the wheels is same.

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

Note:
Torques need not be always equal. This result is specific for the set of values given in the question.

Answers


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

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