Here, moment arms of the particles are y 1 y 1 and y 1 y 1 . Let the mass of each particle be “m” and speed of each particle is “v”. Then, the angular momentum of the individual particles about the origin of coordinate system are :

ℓ 1 = m y 1 j x v 1 i = - m y 1 v k ℓ 2 = m y 2 j x ( - v 2 ) i = m y 2 v k ℓ 1 = m y 1 j x v 1 i = - m y 1 v k ℓ 2 = m y 2 j x ( - v 2 ) i = m y 2 v k

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

ℓ = ℓ 1 + ℓ 2 = m v ( y 2 - y 1 ) k ℓ = ℓ 1 + ℓ 2 = m v ( y 2 - y 1 ) k

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.

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

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

⇒ ℓ = ( 1 x 1 - 1 x - 1 ) i + ( 1 x 2 - 0 ) j + ( 0 - 2 x 1 ) k ⇒ ℓ = 2 i + 2 j - 2 k ⇒ ℓ = ( 1 x 1 - 1 x - 1 ) i + ( 1 x 2 - 0 ) j + ( 0 - 2 x 1 ) k ⇒ ℓ = 2 i + 2 j - 2 k

Hence, option (b) is correct.

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 :

Figure 3: The projectile is at maximum vertical displacement.

Projectile motion

v x = v cos θ v x = v cos θ

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,

r ⊥ = H = v 2 sin 2 θ 2 g r ⊥ = H = v 2 sin 2 θ 2 g

The angular momentum as required is :

⇒ ℓ = m r ⊥ v = m v 2 sin 2 θ 2 g x v cos θ ⇒ ℓ = m v 3 sin 2 θ cos θ g ⇒ ℓ = m r ⊥ v = m v 2 sin 2 θ 2 g x v cos θ ⇒ ℓ = m v 3 sin 2 θ cos θ g

Hence, option (a) is correct.

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 :

K = 1 2 x m v 2 K = 1 2 x m v 2

The magnitude of linear momentum is given by :

p = m v p = m v

Squaring both sides, we have :

⇒ p 2 = m 2 v 2 ⇒ p 2 = m 2 v 2

Rearranging,

⇒ m v 2 = p 2 m ⇒ m v 2 = p 2 m

Combining equations,

K = p 2 2 m K = p 2 2 m

Now, the magnitude of angular momentum is given by :

ℓ = m v r ℓ = m v r

Squaring both sides, we have :

⇒ ℓ 2 = m 2 v 2 r 2 ⇒ ℓ 2 = m 2 v 2 r 2

⇒ m v 2 = ℓ 2 m r 2 ⇒ m v 2 = ℓ 2 m r 2

Combining equations,

K = ℓ 2 2 m r 2 K = ℓ 2 2 m r 2

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

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

ℓ = m v r sin θ ℓ = m v r sin θ

The dimension of angular momentum is :

[ K ] = [ m ] [ v ] [ r ] [ sin θ ] = [ M ] [ L T - 1 ] [ L ] = [ M L 2 T - 1 ] [ K ] = [ m ] [ v ] [ r ] [ sin θ ] = [ M ] [ L T - 1 ] [ L ] = [ M L 2 T - 1 ]

Hence, option (c) is correct.

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 :

ℓ 1 = m v r = m ω x 2 d 3 x 2 d 3 = 4 m ω d 2 9 ℓ 1 = m v r = m ω x 2 d 3 x 2 d 3 = 4 m ω d 2 9

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

ℓ 2 = m v r = 2 m ω x d 3 x d 3 = 2 m ω d 2 9 ℓ 2 = m v r = 2 m ω x d 3 x d 3 = 2 m ω d 2 9

ℓ = ℓ 1 + ℓ 2 = 4 m ω d 2 9 + 2 m ω d 2 9 = 2 m ω d 2 3 ℓ = ℓ 1 + ℓ 2 = 4 m ω d 2 9 + 2 m ω d 2 9 = 2 m ω d 2 3

Hence, option (b) is correct.

The angular momentum of a particle about a point is defined as :

ℓ = m ( r x v ) ℓ = m ( r x v )

The magnitude of angular momentum can be evaluated using any of the three methods discussed in the module on the subject. In this case, evaluation of angular momentum involving moment arm is most appropriate. Thus,

ℓ = m r ⊥ v ℓ = m r ⊥ v

We, now, analyze the situation by drawing the figure. The path of the motion is shown. The moment arm i.e. the perpendicular distance of velocity vector from the origin is :

Figure 5: The path of the moving particle.

Angular momentum of a particle

r ⊥ = OB = 2 sin 45 ° = √ 2 r ⊥ = OB = 2 sin 45 ° = √ 2

Hence,

⇒ ℓ = 0.1 x √ 2 x √ 2 = 0.2 kg - m 2 s ⇒ ℓ = 0.1 x √ 2 x √ 2 = 0.2 kg - m 2 s

Hence, option (b) is correct.

We need to find a general equation for angular momentum for a given time “t”. We observe here that projectile motion is a two dimensional motion for which we can have information in component form. It is, therefore, required to obtain component information for position and the velocity.

Figure 7: Angular momentum of a projectile at a given time.

Projectile motion

x = u cos θ x t y = u sin θ x t - 1 2 g t 2 v x = u cos θ v y = u sin θ - g t x = u cos θ x t y = u sin θ x t - 1 2 g t 2 v x = u cos θ v y = u sin θ - g t

Now, the angular momentum is :

ℓ = | i	                   j	               k |
    | utcosθ	   utsinθ – 1/2g t*2	       0 |
    | u cosθ       usinθ – gt	               0 |

⇒ ℓ = m { u cos θ x t ( u sin θ - g t ) - u cos θ ( u sin θ - 1 2 x g t 2 ) } k ⇒ ℓ = m { u cos θ x t ( u sin θ - g t ) - u cos θ ( u sin θ - 1 2 x g t 2 ) } k

⇒ ℓ = m ( u cos θ x t x u sin θ - u cos θ x t x g t - u cos θ x u sin θ x t + 1 2 x g t 2 x u cos θ ) k ⇒ ℓ = m ( u cos θ x t x u sin θ - u cos θ x t x g t - u cos θ x u sin θ x t + 1 2 x g t 2 x u cos θ ) k

⇒ ℓ = m ( u 2 t cos θ x sin θ - u g t 2 cos θ - u 2 t cos θ x sin θ + 1 2 x u g t 2 cos θ ) k ⇒ ℓ = m ( u 2 t cos θ x sin θ - u g t 2 cos θ - u 2 t cos θ x sin θ + 1 2 x u g t 2 cos θ ) k

⇒ ℓ = - 1 2 x m u g t 2 cos θ k ⇒ ℓ = - 1 2 x m u g t 2 cos θ k

Hence, option (d) is correct.