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

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

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.

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

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.

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.

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.

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.

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.