7.21. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

From, 1/2 mv2 + 1/2 I ω2 = mgh
1/2mv2 + 1/2 (1/2 mr2) ω2 = mgh
3/4 mv2 = mgh
h = 3v2/4g = 3 × 52/4 × 9.8 = 1.913 m

If s is the distance up the inclined plane, then as
sin θ = h/s
s = h/sin θ  = 1.913/sin 30° = 3.826 m

Time taken to return to the bottom

(Hint: Consider the equilibrium of each side of the ladder separately.)

The given situation can be shown as:

(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, I_i = 7.6 + 8.1 = 15.7 kg m²
Angular speed, ω_i = 300 rev/min
Angular momentum, L_i = I_iω_i  =  15.7 × 30   ….(i)
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, I_f = 7.6 + 0.4 = 8 kg m²
Final angular speed = ω_f
Final angular momentum, L_f = I_fω_f = 0.79 ω_f …… (ii)
From the conservation of angular momentum, we have:
I_iω_i  =  I_fω_f
∴ ω_f = 15.7 × 30  / 8  =  58.88 rev/min

(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.

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.)

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 = m / 2
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door:
α = mvr
= (10 × 10^-3 ) × (500) × (1/2)  =  2.5 kg m² s^-1    …(i)
Moment of inertia of the door:
I = ML² / 3
= (1/3) × 12 × 1² = 4 kgm²
But α = Iω
∴ ω = α / I
= 2.5 / 4
= 0.625 rad s^-1

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.

Angular speed of disc I = ω₁

Moment of inertia of disc I = I₂
Angular speed of disc I = ω₂
Angular momentum of disc I, L₁ = I₁ω₁
Angular momentum of disc II, L₂ = I₂ω₂
Total initial angular momentum L_i = I₁ω₁ + I₂ω₂
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the system of two discs, I = I₁ + I₂
Let ω be the angular speed of the system.
Total final angular momentum, L_T = (I₁ + I₂) ω
Using the law of conservation of angular momentum, we have:
L_i = L_T
I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω
∴ ω = (I₁ω₁ + I₂ω₂) / (I₁ + I₂)

(b) Kinetic energy of disc I, E₁ = (1/2) I₁ω₁²
Kinetic energy of disc II, E₁ = (1/2) I₂ω₂²
Total initial kinetic energy, E_i = (1/2) ( I₁ω₁² + I₂ω₂²)
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system, I = I₁ + I₂
Angular speed of the system = ω
Final kinetic energy E_f:  =  (1/2) ( I₁ + I₂) ω²
=  (1/2) ( I₁ + I₂) [ (I₁ω₁ + I₂ω₂) / (I₁ + I₂) ]²
= (1/2) (I₁ω₁ + I₂ω₂)² / (I₁ + I₂)
∴ E_i – E_f
= (1/2) ( I₁ω₁² + I₂ω₂²) – (1/2) (I₁ω₁ + I₂ω₂)² / (I₁ + I₂)
Solving the equation, we get
= I₁ I₂ (ω₁ – ω₂)² / 2(I₁ + I₂)
All the quantities on RHS are positive
∴ E_i – E_f > 0
E_i > 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.

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 ∑ m_ir_i = 0).

(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 y-axis, I_y = my²
Moment of inertia about z-axis, I_z = m(x² + y²)^1/2

I_x + I_y = mx² + my²
= m(x² + y²)
= m [(x² + y²)^1/2]^1/2
I_x + I_y = I_z
Hence, the theorem is proved.

(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, … , m_n, at perpendicular distances r₁, r₂, r₃, … , r_n respectively from the centre of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O:

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 v² = 2gh / [1 + (k²/R²) ]
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.

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

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, E_b = KE_rot + KE_trans
= (1/2) I ω² + (1/2) mv²
But I = mk² and ω = v / R
∴ E_b = (1/2)(mk²)(v²/R²) + (1/2)mv²
= (1/2)mv² (1 + k² / R²)
From the law of conservation of energy, we have:
E_T = E_b
mgh = (1/2)mv² (1 + k² / R²)
∴ v = 2gh / (1 + k² / R²)
Hence, the given result is proved.

9.28. A disc rotating about its axis with angular speed ω_ois 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?

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:
v_c = (R/2)ω_o 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.

(b) What is the force of friction after perfect rolling begins?

A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.

Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, ω₀ =10 π rad s^–1
Coefficient of kinetic friction, μ_k = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma
μ_kmg= ma
Where,
a = Acceleration produced in the objects
m = Mass
∴ a = μ_kg … (i)
As per the first equation of motion, the final velocity of the objects can be obtained as:
v = u + at
= 0 + μ_kgt
= μ_kgt … (ii)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ= –Iα
α = Angular acceleration
μ_kmgr = –Iα
∴ α = –μ_kmgr / I     …..(iii)
Using the first equation of rotational motion to obtain the final angular speed:
ω = ω₀ + αt
= ω₀ + (–μ_kmgr / I)t    ….(iv)
Rolling starts when linear velocity, v = rω
∴ v = r (ω₀ – μ_kmgrt / I)    …(v)
Equating equations (ii) and (v), we get:
μ_kgt = r (ω₀ – μ_kmgrt / I)
= rω₀ – μ_kmgr²t / I    ….(vi)
For the ring:
I = mr²
∴ μ_kgt = rω₀ – μ_kmgr²t / mr²
= rω₀ – μ_kgt
2μ_kgt = rω₀
∴ t = rω₀ / 2μ_kg
= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8  =  0.80 s    ….(vii)
For the disc: I = (1/2)mr²
∴ μ_kgt = rω₀ – μ_kmgr²t / (1/2)mr²
= rω₀ – 2μ_kgt
3μ_kgt = rω₀
∴ t = rω₀ / 3μ_kg
= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8  =  0.53 s   …..(viii)
Since t_d > t_r, the disc will start rolling before the ring.

7.31. A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µ_s = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?

Mass of the cylinder, m = 10 kg
Radius of the cylinder, r = 15 cm = 0.15 m
Co-efficient of kinetic friction, µ_k = 0.25
Angle of inclination, θ = 30°
Moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr²
The various forces acting on the cylinder are shown in the following figure:

a = mg Sinθ / [m + (I/r²) ]
= mg Sinθ / [m + {(1/2)mr²/ r²} ]
= (2/3) g Sin 30°
= (2/3) × 9.8 × 0.5  =  3.27 ms^-2


(a) Using Newton’s second law of motion, we can write net force as:
f_net = ma
mg Sin 30° – f = ma
f = mg Sin 30° – ma
= 10 × 9.8 × 0.5 – 10 × 3.27
49 – 32.7 = 16.3 N

(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.

(c) For rolling without skid, we have the relation:
μ = (1/3) tan θ
tan θ = 3μ = 3 × 0.25
∴ θ = tan^-1 (0.75) = 36.87°.

7.32. Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

(a) False
Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.

(b) True
Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False
This is becausehen a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True
This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.

(e) True
This is because rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, the wheel will simply slip under the effect of its own weight.