### Exercise 1

A uniform circular disc of mass “M” and radius “R” rotates about one of its diameters at an angular speed “ω”. What is the angular momentum of the disc about the axis of rotation?

Circular disk |
---|

#### Solution

The angular momentum of a rigid body is given by the equation,

In order to find MI about diameter, we first calculate moment of inertia, "

Applying theorem of perpendicular axes, the moment of inertia about the axis of rotation (i.e. about one of the diameters) is :

Therefore, the angular momentum of the disk about one of its diameter is :

Hence, option (d) is correct.

### Exercise 2

A thin uniform rod of mass "M" and length "x" rotates in a horizontal plane about a vertical axis. If the free end of the rod has a constant tangential velocity "v", then angular momentum of the rod about the axis of rotation is :

A rotating rod |
---|

#### Solution

The angular momentum of a rigid body is given by the equation,

Here, the rod rotates about a perpendicular axis at its end. Thus, we need to find MI of the rod, "I", about this axis. Using theorem of parallel axes, the MI about the axis of rotation is :"

A rotating rod |
---|

Putting in the expression of angular momentum :

The angular velocity is expressed in terms of tangential velocity as :

Substituting this value in the expression of angular momentum, we have

Hence, option (b) is correct.

### Exercise 3

A thin uniform rod of mass "M" and length "3x" rotates in a horizontal plane with an angular velocity “ω” about a perpendicular vertical axis as shown in the figure. Then, the angular momentum of the rod about the axis of rotation is :

A rotating rod |
---|

#### Solution

We know that angular momentum of a rigid body is given by :

We need to calculate the MI of the rod about the axis of rotation. Now, the MI of the rod about central axis is given by :

A rotating rod |
---|

Applying theorem of parallel axis, the MI about the axis of rotation passing through point “O” is :

Here,

Therefore, angular momentum about the axis of rotation is :

Hence, option (a) is correct.

Note 1 : Alternatively, we can also calculate the angular momentum of the rod in terms of angular momentum of the parts on either side of the axis of rotation. Since each part rotates about the same axis of rotation, the net angular momentum of the rod about the axis of rotation is algebraic sum of the angular momentum of the parts.

The MI of the left rod, about a perpendicular axis at one of the ends, is :

A rotating rod |
---|

Similarly, the MI of the right rod is :

Note 2 : It must be understood here that we can not treat the rod as a particle at its center of mass and then find the angular momentum. For example, in this case, the center of mass is at distance of "3x/2" from either end of the rod and at a distance of "x/2" from the point "O" through which axis of rotation passes. The MI of this particle equivalent rod about the axis of rotation is :

and

Obviously, this approach is conceptually wrong. The concept of center of mass by definition is valid for translation - not rotation.

### Exercise 4

A circular ring of mass “M” and radius “R” rolls on a horizontal surface without sliding at a velocity “v”. What is the magnitude of angular momentum of the ring about point of contact?

Rolling of a circular ring |
---|

#### Solution

Here, the point of contact and center of mass are in the plane of motion. Hence,

Now, the magnitude of angular momentum of the body, considering it as a particle at center of mass, with respect to point of contact is :

This angular momentum is clockwise and into the page.

The magnitude of angular momentum about center of mass is same as that about its central axis i.e. perpendicular axis passing through center of mass :

For rolling motion,

This angular momentum is also clockwise and into the page. Therefore, the magnitude of net angular momentum is :

Hence, option (b) is correct.

### Exercise 5

Q A solid sphere of mass “M” and radius “R” rolls on a horizontal surface without sliding at a velocity “v”. What is the magnitude of angular momentum of the ring about point of contact?

Rolling of a solid sphere |
---|

#### Solution

This question is similar to the previous one - with one exception that sphere is a three dimensional body as against the disk, which is two dimensional. However, the situation is same as far as the point about angular momentum is calculated and the center of mass. The point of contact and center of mass are in the same middle section of the sphere, along which motion takes place. Thus, the point of contact and center of mass of the solid sphere are in the plane of motion. Hence,

Now, the magnitude of angular momentum of the body as particle at center of mass with respect to point of contact is :

This angular momentum is clockwise and into the page.

The magnitude of angular momentum about center of mass is same as that about its central axis i.e. perpendicular axis passing through center of mass :

For rolling motion,

This angular momentum is also clockwise and into the page. The magnitude of net angular momentum, therefore, is :

Hence, option (d) is correct.