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Angular mometum (check your understanding)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Objective questions, contained in this module with hidden solutions, help improve understanding of the topics covered under the module "Angular momentum".

The questions have been selected to enhance understanding of the topics covered in the module titled " Angular momentum ". All questions are multiple choice questions with one or more correct answers. There are two sets of the questions. The “understanding level” questions seek to unravel the fundamental concepts involved, whereas “application level” are relatively difficult, which may interlink concepts from other topics.

Each of the questions is provided with solution. However, it is recommended that solutions may be seen only when your answers do not match with the ones given at the end of this module.

Understanding level (Angular momentum)

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 :

Figure 1: Two particles of same mass, move with same speed
Two moving particles
 Two moving particles  (amq1.gif)

(a) is directly proportional to the perpendicular distance between the parallel lines (b) is inversely proportional to the perpendicular distance between the parallel lines (c) is constant for a given pair of parallel paths (d) is zero (a) is directly proportional to the perpendicular distance between the parallel lines (b) is inversely proportional to the perpendicular distance between the parallel lines (c) is constant for a given pair of parallel paths (d) is zero

Solution

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 :

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

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

Note:
Since we can always orient coordinates similar to the situation here, we can conclude that angular momentum about any point is constant for a given pair of parallel paths under similar conditions of mass and speed. Will the angular momentum be constant at a point between parallel paths? The answer is yes.

Exercise 2

A particle of mass “m” is moving with linear velocity v = (2ij + 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?

(a) i + j - 2 k (b) 2 i + 2 j - 2 k (c) i + 2 j - 2 k (d) 2 i + j - 2 k (a) i + j - 2 k (b) 2 i + 2 j - 2 k (c) i + 2 j - 2 k (d) 2 i + j - 2 k

Solution

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


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

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

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 :

Figure 2: A projectile of mass “m” is projected with a velocity “v”, making and angle “θ” with the horizontal.
Projectile
 Projectile  (amq2.gif)

(a) m v 3 sin 2 θ cos θ 2 g (b) m v 3 sin 2 θ 2 g (c) m v 2 sin 2 θ 2 g (d) m v 3 sin 2 θ cos θ g (a) m v 3 sin 2 θ cos θ 2 g (b) m v 3 sin 2 θ 2 g (c) m v 2 sin 2 θ 2 g (d) m v 3 sin 2 θ cos θ g

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 :

Figure 3: The projectile is at maximum vertical displacement.
Projectile motion
 Projectile motion  (amq3.gif)

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 :

= m r v = m v 2 sin 2 θ 2 g x v cos θ = m v 3 sin 2 θ cos θ g = m r v = m v 2 sin 2 θ 2 g x v cos θ = m v 3 sin 2 θ cos θ g

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 :

(a) p 2 2 m (b) p 2 2 m r (c) 2 2 m r 2 (d) 2 2 m r (a) p 2 2 m (b) p 2 2 m r (c) 2 2 m r 2 (d) 2 2 m r

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 :

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 :

= m v r = m v r

Squaring both sides, we have :

2 = m 2 v 2 r 2 2 = m 2 v 2 r 2

m v 2 = 2 m r 2 m v 2 = 2 m r 2

Combining equations,

K = 2 2 m r 2 K = 2 2 m r 2

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

Exercise 5

Dimensional formula of angular momentum is :

(a) M L 3 T - 1 (b) M L 2 T - 2 (c) M L 2 T - 1 (d) M L T - 1 (a) M L 3 T - 1 (b) M L 2 T - 2 (c) M L 2 T - 1 (d) M L T - 1

Solution

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

= m v r sin θ = 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.

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 :

Figure 4: The rod is free to move in vertical plane.
Angular momentum of particle
 Angular momentum of particle   (amq4.gif)

(a) 3 m ω d 2 4 (b) 2 m ω d 2 3 (c) 2 m ω d 2 5 (d) 5 m ω d 2 9 (a) 3 m ω d 2 4 (b) 2 m ω d 2 3 (c) 2 m ω d 2 5 (d) 5 m ω d 2 9

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 :

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

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

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

Hence, option (b) is correct.

Application level (Angular momentum)

Exercise 7

A particle of mass 100 gram is moving with a constant speed at √2 m/s along a straight line y = x + 2 in "xy" – plane. Then, the magnitude of the angular momentum (in kg - m 2 s kg - m 2 s ) about the origin of the coordinate system, if “x” and “y” are in meters, is :

(a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4

Solution

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

= m ( r x v ) = 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,

= m r v = 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
 Angular momentum of a particle   (amq5.gif)

r = OB = 2 sin 45 ° = 2 r = OB = 2 sin 45 ° = 2

Hence,

= 0.1 x 2 x 2 = 0.2 kg - m 2 s = 0.1 x 2 x 2 = 0.2 kg - m 2 s

Hence, option (b) is correct.

Exercise 8

A projectile of mass “m” is projected with a velocity “u”, making and angle “θ” with the horizontal. The angular momentum of the projectile about the point of projection at a given instant “t” during its flight is :

Figure 6: Angular momentum of a projectile at a given time.
Projectile motion
 Projectile motion   (amq6.gif)

(a) - m u g t 2 cos θ i (b) - 2 m u g t 2 cos θ j (c) - m u g t 2 cos θ k (d) - 1 2 x m u g t 2 cos θ k (a) - m u g t 2 cos θ i (b) - 2 m u g t 2 cos θ j (c) - m u g t 2 cos θ k (d) - 1 2 x m u g t 2 cos θ k

Solution

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
 Projectile motion   (amq7.gif)

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 :

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

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

= 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 = 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

= 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 = 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

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

Hence, option (d) is correct.

Answers

1. (a) and (c)   2. (b)    3. (a)    4. (a) and (c)   5.  (c)   6. (b)  
7. (b)           8. (d)

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