### Exercise 1

Two particles of same mass, move with same speed along two parallel lines in opposite directions as shown in the figure. Then, magnitude of net angular moment of two particles about the origin of coordinate system :

Two moving particles |
---|

#### Solution

Here, moment arms of the particles are

Since angular momentum are along the same direction, we can find the net angular momentum by arithmetic sum as :

We see that the magnitude of net angular momentum is directly proportional to the perpendicular distance between the parallel lines. Further, the perpendicular distance is fixed for a given pair of parallel paths. Hence, net angular momentum is constant for a given pair.

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

##### Note:

### Exercise 2

A particle of mass “m” is moving with linear velocity v = (2i – j + k) m/s. At an instant, the particle is situated at a position (in meters) given by the coordinates 0,1,1. What is the angular momentum of the particle about the origin of coordinate system at that instant?

#### Solution

This question can be evaluated using cross product in component form as :

```
ℓ = | i j k |
| 0 1 1 |
| 2 -1 1 |
```

Hence, option (b) is correct.

### Exercise 3

A projectile of mass “m” is projected with a velocity “v”, making and angle “θ” with the horizontal. The angular momentum of the projectile about the point of projection, when it reaches the highest point is :

Projectile |
---|

#### Solution

The velocity of the projectile at the highest point is equal to the horizontal component of velocity as vertical component of velocity is zero. Further, there is no acceleration in the horizontal direction. Therefore, horizontal component of velocity is same as the horizontal component of the velocity at projection. It means that velocity of projection at the highest point is :

Projectile motion |
---|

Here, the moment arm i.e. the perpendicular drawn from the point of projection on the extended line of velocity is equal to the height attained by the projectile. Thus,

The angular momentum as required is :

Hence, option (a) is correct.

### Exercise 4

If “m”,”p” and “ℓ” denote the mass, linear momentum and angular momentum of the particle executing uniform circular motion along a circle of radius “r”, then kinetic energy of the particle is :

#### Solution

From the alternatives given, it is evident that we need to find kinetic energy in terms of linear and angular momentum. Now, the kinetic energy of the particle is expressed as :

The magnitude of linear momentum is given by :

Squaring both sides, we have :

Rearranging,

Combining equations,

Now, the magnitude of angular momentum is given by :

Squaring both sides, we have :

Combining equations,

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

### Exercise 5

Dimensional formula of angular momentum is :

#### Solution

In order to find dimensional formula, we use the expression of angular momentum :

The dimension of angular momentum is :

Hence, option (c) is correct.

### Exercise 6

Two point masses “m” and “2m” are attached to a mass-less rod of length “d”, pivoted at point “O” as shown in the figure. The rod is free to move in vertical direction. If “ω” be the angular velocity, then the angular momentum of two particles about the axis of rotation is :

Angular momentum of particle |
---|

#### Solution

This is a case of rotation. We can solve this problem using either the general expression or specific expression involving moment of inertia. When we use general expression, we should measure moment arm from the axis of rotation.

Since rod is rotating in anticlockwise direction, angular momentum of each particle is positive.

The angular momentum of first particle of mass “m” is :

The angular momentum of second particle of mass “2m” is :

Hence, option (b) is correct.