### 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 :

#### 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 :

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

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,

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.

Rotation |
---|

Therefore, torque about the axis of rotation is :

The magnitude of torque is :

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

The component of torque in z-direction is :

The magnitude of torque is :

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

Rotation |
---|

#### Solution

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

Rotation |
---|

Since

Evidently, torque due to

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

### Exercise 3

A flywheel with moment of inertia of 2

#### 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 :

The magnitude of torque is :

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 :

#### 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,

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

Here,

Hence,

Now, torque is also given as :

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) :

Rotation |
---|

#### 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.