Page No 180: - Chapter 7 System Of Particles & Rotational Motion class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 7.23:

A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m².

(a) What is his new angular speed? (Neglect friction.)

(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Answer:

(a) 58.88 rev/min (b) No

(a)Moment of inertia of the man-platform system = 7.6 kg m²

Moment of inertia when the man stretches his hands to a distance of 90 cm:

2 × m r²

= 2 × 5 × (0.9)²

= 8.1 kg m²

Initial moment of inertia of the system, 

Angular speed, 

Angular momentum, 

Moment of inertia when the man folds his hands to a distance of 20 cm:

2 × mr²

= 2 × 5 (0.2)² = 0.4 kg m²

Final moment of inertia, 

Final angular speed = 

Final angular momentum,   … (ii)

From the conservation of angular momentum, we have:

(b)Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.

Question 7.24:

A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

(Hint: The moment of inertia of the door about the vertical axis at one end is ML²/3.)

Answer:

Mass of the bullet, m = 10 g = 10 × 10^–3 kg

Velocity of the bullet, v = 500 m/s

Thickness of the door, L = 1 m

Radius of the door, r = 

Mass of the door, M = 12 kg

Angular momentum imparted by the bullet on the door:

α = mvr

Moment of inertia of the door:

Question 7.25:

Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω₁ ≠ ω₂.

Answer:

(a)

When the two discs are joined together, their moments of inertia get added up.

Moment of inertia of the system of two discs, 

Let ω be the angular speed of the system.

Total final angular momentum, 

Using the law of conservation of angular momentum, we have:

(b)Kinetic energy of disc I,

Kinetic energy of disc II,

Total initial kinetic energy, 

When the discs are joined, their moments of inertia get added up.

Moment of inertia of the system, 

Angular speed of the system = ω

Final kinetic energy E_f:

The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.

Question 7.26:

(a) Prove the theorem of perpendicular axes.

(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x² + y²).

(b) Prove the theorem of parallel axes.

(Hint: If the centre of mass is chosen to be the origin  ).

Answer:

(a)The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

A physical body with centre O and a point mass m,in the x–y plane at (x, y) is shown in the following figure.

Moment of inertia about x-axis, I_x = mx²

Moment of inertia about y-axis, I_y = my²

Moment of inertia about z-axis, I_z = 

I_x + I_y = mx² + my²

= m(x² + y²)

(b)The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

Suppose a rigid body is made up of n particles, having masses m₁, m₂, m₃, … , mn, at perpendicular distances r₁, r₂, r₃, … , rn respectively from the centre of mass O of the rigid body.

The moment of inertia about axis RS passing through the point O:

I_RS = 

The perpendicular distance of mass mi, from the axis QP = a + ri

Hence, the moment of inertia about axis QP:

Now, at the centre of mass, the moment of inertia of all the particles about the axis passing through the centre of mass is zero, that is,

Question 7.27:

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by  .

Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

Answer:

A body rolling on an inclined plane of height h,is shown in the following figure:

m = Mass of the body

R = Radius of the body

K = Radius of gyration of the body

v = Translational velocity of the body

h =Height of the inclined plane

g = Acceleration due to gravity

Total energy at the top of the plane, E­₁= mgh

Total energy at the bottom of the plane, 

But

From the law of conservation of energy, we have:

Hence, the given result is proved.

Question 7.28:

A disc rotating about its axis with angular speed ωois placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?

Answer:

v_A = Rω_o; v_B = Rω_o;  ; The disc will not roll

Angular speed of the disc = ω_o

Radius of the disc = R

Using the relation for linear velocity, v = ω_oR

For point A:

v_A = Rω_o; in the direction tangential to the right

For point B:

v_B = Rω_o; in the direction tangential to the left

For point C:

in the direction same as that of v_A

The directions of motion of points A, B, and C on the disc are shown in the following figure

Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.