The gravitational pull of the Sun passes through COM of the Earth, providing centripetal force required for the rotation of the Earth about it. Thus, there is no external torque on the Earth. It means that the angular momentum of the Earth remains constant.

The melting of ice cap will result in the rise of sea level. From the point of view of MI of the Earth, it means redistribution of mass. The water equivalent of ice moves away from the axis of rotation, which passes through the poles. This results in an increase in the MI of the Earth.

Increase in the MI of the Earth, in turn, will decrease angular velocity. As the Earth rotate slowly, duration of a day on the Earth will increase.

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

Since no external torque is operating on the disk – objects system, the angular momentum of the system is conserved.

Let “ωf” be the common angular velocity of the composite system after objects are placed on the disk. Let subscripts “D”, “O” and “C” represent disk, objects and composite system respectively , then according to conservation of angular momentum,

I Di ω Di + I Oi ω Oi = I Ci ω f I Di ω Di + I Oi ω Oi = I Ci ω f

Here,

I Di = M R 2 2 ; ω Di = ω ; I Oi = 0 ; ω Oi = 0 I Di = M R 2 2 ; ω Di = ω ; I Oi = 0 ; ω Oi = 0

Also, the MI of the composite system is :

I Cf = ( M R 2 2 + 2 m R 2 ) = ( M 2 + 2 m ) R 2 I Cf = ( M R 2 2 + 2 m R 2 ) = ( M 2 + 2 m ) R 2

Putting in the equation of conservation law,

M R 2 ω 2 = ( M 2 + 2 m ) R 2 ω f M R 2 ω 2 = ( M 2 + 2 m ) R 2 ω f

ω f = M ω 2 ( M 2 + 2 m ) = M ω ( M + 4 m ) ω f = M ω 2 ( M 2 + 2 m ) = M ω ( M + 4 m )

Hence, options (a) is correct.

The Earth rotates around the Sun as gravitational pull provides the necessary centripetal force. This force passes through the center of mass of the Earth. As such, gravitational pull does not constitute torque on the Earth. Therefore, we can consider that angular momentum of the Earth remains unchanged.

L i = L f I i ω i = I f ω f L i = L f I i ω i = I f ω f

⇒ ω f = I i ω i I f ⇒ ω f = I i ω i I f

We know that time period of the Earth is :

T i = 24 hrs T i = 24 hrs

Hence, its angular velocity is :

⇒ ω i = 2 Π T i = 2 Π 24 = Π 12 rad / s ⇒ ω i = 2 Π T i = 2 Π 24 = Π 12 rad / s

Since data on mass and radius are missing, it is not possible to calculate MIs in two cases separately. However, we can find the ratio of MIs :

I i I f = 2 M R 2 5 2 5 2 M 3 x ( R 2 ) 2 = 1 x 3 x 4 2 = 6 I i I f = 2 M R 2 5 2 5 2 M 3 x ( R 2 ) 2 = 1 x 3 x 4 2 = 6

Putting in the equation,

ω f = 6 ω i = 6 x Π 12 = Π 2 rad / s ω f = 6 ω i = 6 x Π 12 = Π 2 rad / s

The new time period of rotation is :

T f = 2 Π ω f = 2 x 2 Π Π = 4 hrs T f = 2 Π ω f = 2 x 2 Π Π = 4 hrs

Hence, options (b) is correct.

The bodies acquire equal angular velocity due to friction operating between the surfaces in contact. However, friction here is internal to the system of two rotating disks. Thus, there is no external torque on the system and we can employ law of conservation of angular momentum :

Figure 2: Two disks are brought in contact along common axis of rotation.

Two rotating disks

L i = L f L i = L f

Here,

L i = I 1 ω 1 + I 2 ω 2 L i = I 1 ω 1 + I 2 ω 2

When disks come in tact and rotate about a common axis with equal angular velocity, they acquire common angular velocity, say ”ω”. Since, Two disks are rotating about a common axis of rotation, the MI of the combination is arithmetic sum of individual MIs. The common moment of inertia, “ I C I C ” is given as :

I C = I 1 + I 2 I C = I 1 + I 2

Thus, angular momentum of the system in this situation is :

L f = ( I 1 + I 2 ) ω L f = ( I 1 + I 2 ) ω

Putting values in the equation of conservation of angular momentum, we have :

⇒ I 1 ω 1 + I 2 ω 2 = ( I 1 + I 2 ) ω ⇒ I 1 ω 1 + I 2 ω 2 = ( I 1 + I 2 ) ω

⇒ ω = I 1 ω 1 + I 2 ω 2 I 1 + I 2 ⇒ ω = I 1 ω 1 + I 2 ω 2 I 1 + I 2

Hence, option (c) is correct.

Applying law of conservation of angular momentum, the expression of common angular velocity is given as (as derived in the earlier question) :

Figure 4: Two disks are brought in contact along common axis of rotation.

Two rotating disks

ω = I 1 ω 1 + I 2 ω 2 I 1 + I 2 ω = I 1 ω 1 + I 2 ω 2 I 1 + I 2

The energy dissipated due to the friction is equal to the change in the angular kinetic energy. Also as there is dissipation of energy, final angular kinetic energy is less than initial angular kinetic energy of the system. The change in angular kinetic energy, therefore, is :

Δ K = 1 2 x I i ω i 2 - 1 2 x I f ω f 2 Δ K = 1 2 x I i ω i 2 - 1 2 x I f ω f 2

Here, final angular speed and moment of inertia of the combined disk are :

ω f = ω ; I f = I 1 + I 2 ω f = ω ; I f = I 1 + I 2

Putting values, we have :

⇒ Δ K = 1 2 x I 1 ω 1 2 - 1 2 x I 2 ω 2 2 = 1 2 x ( I 1 + I 2 ) ω 2 ⇒ Δ K = 1 2 x I 1 ω 1 2 - 1 2 x I 2 ω 2 2 = 1 2 x ( I 1 + I 2 ) ω 2

Substituting value of common angular velocity and solving, we have :

⇒ Δ K = I 1 I 2 ( ω 1 - ω 2 ) 2 2 ( I 1 + I 2 ) ⇒ Δ K = I 1 I 2 ( ω 1 - ω 2 ) 2 2 ( I 1 + I 2 )

Hence, option (b) is correct.

The line of action of the force on the block due to obstruction passes through the axis of rotation about which block suddenly rotates i.e. about the obstruction for a brief period. This force does not constitute torque. We can, therefore, conclude that there is no torque on the block. Obviously, it is a single body system. According to law of conservation of angular momentum,

L i = L f L i = L f

The angular momentum of the block, treating it as a particle like translating body, about the point of obstruction on the surface is :

L i = - M v r ⊥ = - M v L 2 L i = - M v r ⊥ = - M v L 2

The angular momentum of the block, treating it as a rigid body rotating about the axis perpendicular to the plane of motion and passing through the point of obstruction, is :

Figure 6: The cubical block overturns at P.

A block hitting an obstruction

L f = I ω L f = I ω

where “ω” is the angular velocity of the block about the axis of rotation.

Thus,

⇒ - M v L 2 = I ω ⇒ - M v L 2 = I ω

For moment of inertia of the block, it can be treated like a rod. Its MI about the axis passing through COM is,

⇒ I COM = ( L 2 + L 2 ) 12 = M L 2 6 ⇒ I COM = ( L 2 + L 2 ) 12 = M L 2 6

Figure 7: Axis of rotation lies in the central plane as seen from any of the six sides.

Axis of rotation

Recall that the formula used for MI here is that of plate about an axis that passes through COM but which is in the plane of plate. It is a valid assumption as cube can be considered to be a cylinder with square face. Since MI of the plate and its corresponding cylinder are same, MI of the cube is same as that of corresponding square plate.

Here, the axis can be considered to lie in the central plane consisting of its COM. A central plane with perpendicular axis is shown for an enlarged cube. Since cube is symmetric about any such axis as seen from any of the six faces, we can conclude that MI as calculated above is about an axis, which is perpendicular to the plane of motion and passing through its COM. Now, using theorem of parallel axes, we have MI about the point of obstruction :

⇒ I = I COM + M L 2 ⇒ I = I COM + M L 2

From the geometry,

OP = r = L √ 2 OP = r = L √ 2

⇒ r 2 = L 2 2 ⇒ r 2 = L 2 2

Putting in the expression,

⇒ I = M L 2 6 + M L 2 2 = 2 M L 2 3 ⇒ I = M L 2 6 + M L 2 2 = 2 M L 2 3

Now, putting in the equation of conservation of angular momentum, we have :

⇒ - M v L 2 = 2 M L 2 3 x ω ⇒ - M v L 2 = 2 M L 2 3 x ω

⇒ ω = - 3 v 4 L ⇒ ω = - 3 v 4 L

The negative sign indicates that the cubical block overturns in clockwise rotation about point "P". The angular speed of the block about obstruction, therefore, is equal to the magnitude of angular velocity.

Hence, option (a) is correct.