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The statement is true.

Explanation : Rolling on a surface (horizontal or inclined) without slipping may be viewed as pure rotation about an horizontal axis through the point of contact, when viewed in the inertial frame of reference in which the surface is at rest. The point of contact of the wheel with the surface will be instantaneously at rest, resulting in a rolling motion, provided the wheel is able to ‘grip’ the surface, i.e., friction is necessary. With little or no friction, the wheel will slip at the point of contact. On an inclined plane, this will result in pure translation along the plane. On a horizontal surface, the wheel will simply rotate about its axis through the centre without translation.

When a ballet dancer stretches her arms while pirouetting, her moment of inertia increases, and consequently her angular speed decreases to conserve angular momentum.

If the Earth suddenly shrinks, mass remaining constant, the moment of inertia of the Earth will decrease, and consequently the angular velocity of rotation ω about its axis will increase. Since period T \(\propto\) \(\frac{1}{\omega}\), the duration of the day T will decrease.

When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Friction is necessary for any form of locomotion. Without friction, a vehicle cannot move. The banking angle for a road at a bend is calculated for optimum speed at which every vehicle can negotiate the bend without depending on friction to provide the necessary lateral centripetal force.

When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Data : r = 500 m, v = 72 kmph = 72 × \(\frac{5}{18}\) m/s

= 20 m/s, g = 10 m/s² , l = 1 m

tan θ =  \(\frac{v^2}{rg}\)

= \(\frac{(20)^2}{500\times 10}\) = 0.08

The required angle of banking,

θ = tan^-1 (0.08) = 4°4′

θ = tan^-1 (0.08) = 4°4′ 

The elevation of the outer rail relative to the inner rail, 

h = l sin θ 

= (1)(sin 4°4′) = 0.0709 m = 7.09 cm

Data : r = 280 m, v₁ = 10 m/s at t₁ = 0, v₂ = 14 m/s at t₂= 10 s, v₃ = 19 m/s at t₃ = 18 s

(i)  At t = t₂ , the radial acceleration is

Since the tangential acceleration is constant, it may be found from the data for any two times.

(ii)  If θ is the angle between the resultant linear acceleration and the radial acceleration,

tan θ = \(\frac{a_t}{a_r}\) = \(\frac{0.5}{0.7}\) = 0.7142

∴ θ = tan^-1 0.7142 = 35°32′

(iii) a_t = αr 

The angular acceleration,

α = \(\frac{α_t}r\) = \(\frac{0.5}{280}\)

= 1.785 × 10^-3 rad/s²

= 1.785 mrad/s²

Data : d = 40 km, /= 1 rot/s

∴ r = \(\frac{d}2\) = \(\frac{40\,km}2\) = 20 km = 2 × 10⁴ m

Linear speed, v = ωr = (2πf)r 

= (2 × 3.142 × 1)(2 × 10⁴) 

= 6.284 × 2 × 10⁴

= 1.257 × 10⁵ m/s (or 125.6 km/s)

Data : T = 24 hours = 24 × 60 × 60 s

Angular speed, \(\omega\) = \(\frac{2\pi}T\) = \(\frac{2\times 3.142}{24\times 60\times 60}\) = \(\frac{3.142}{43200}\)

= 7.273 x 10^-5 rad/s

The angular speed of the Earth due to its spin (rotational motion) is 7.273 × 10^-5 rad/s.

The motion of a particle along a complete circle or a part of it is called circular motion.

For a particle performing circular motion, its position vector with respect to the centre of the circle is called the radius vector. 

[Note : The radius vector has a constant magnitude, equal to the radius of the circle. However, its direction changes as the position of the particle changes along the circumference.]

There is no physical difference between them. It is just a question of usage. Circular motion of a body about an axis passing through the body is called rotation. Circular motion of a body around an axis outside the body is called revolution.

Linear velocity, \(\vec{v}\) = \(\vec{\omega}\) × \(\vec{r}\) where ω is the angular velocity and r is the radius vector.

At every instant,\(\vec{v}\), \(\vec{\omega}\) and \(\vec{r}\) are mutually perpendicular, so that in magnitude v = ωr.

1. It is an accelerated motion : As the direction of velocity changes at every instant, it is an accelerated motion. 

2. It is a periodic motion : During the motion, the particle repeats its path along the same trajectory. Thus, the motion is periodic.

In a uniform motion, a particle covers equal distances in equal intervals of time. Any motion which repeats itself in equal intervals of time is called a periodic motion. In a uniform circular motion (UCM), the particle takes the same time to complete each revolution, a distance equal to the circumference of the circle. Therefore, it is a periodic motion.

1. Circular motion of every particle of the blades of a fan or the dryer drum of a washing machine when the fan or the drum is rotating with a constant angular speed. 

2. Motion of the hands of a clock. 

3. Motion of an Earth-satellite in a circular orbit.

The periods of rotation of an hour hand and the Earth are T_h = 12 h and T_E = 24 h, respectively, so that their angular speeds are

\(\omega_h\) = \(\frac{2 \pi}{12}\) rad/h and \(\omega_E\) = \(\frac{2 \pi}{24}\) rad/h.

∴ \(\omega_h\) = 2\(\omega_E\)

In a nonu niform circular motion, the angular acceleration is an axial vector, perpendicular to the plane of the motion. The linear acceleration is in the plane of the motion. Hence, the angle between them is 90°.

The period of UCM of the particle,

T = \(\frac{2 \pi}{\omega}\) = \(\frac{2 \pi}{20 \pi}\) = 0.1 s

Consider a particle moving in a plane along a circular path of constant radius. If the particle is speeding up or slowing down, its angular speed ω and linear speed v both change with time. Then, the particle is said to be in a non uniform circular motion.

Linear speed and angular speed. (Also, kinetic energy, angular speed and angular momentum.)

A particle in uniform circular motion (UCM) moves in a circle or circular arc at constant linear speed v. The instantaneous linear velocity \(\vec{v}\) of the particle is along the tangent to the path in the sense of motion of the particle. Since \(\vec{v}\) changes in direction, without change in its magnitude, there must be an acceleration that must be

1. perpendicular to \(\vec{v}\)

2. constant in magnitude 

3. at every instant directed radially inward, 

i.e., towards the centre of the circular path.

Such a radially inward acceleration is called a centripetal acceleration.

∴ \(\vec{a}\) = \(\frac{d \vec{v}}{dt}\) = \(\vec{a_r}\)

If \(\vec{\omega}\) is the constant angular velocity of the particle and r is the radius of the circle,

\(\vec{a_r}\) = -\(\omega^2\vec r\)

where ω = | \(\vec{\omega}\) | and the minus sign shows that the direction of \(\vec{a_r}\) is at every instant opposite to that of the radius vector \(\vec{r}\). In magnitude,

\(a_r\) = \(\omega^2r\) = \(\frac{v^2}r\) = \(\omega v\)

[Note : The word centripetal comes from Latin for ‘centre-seeking’.]

Velocity and acceleration are nonuniform in UCM. 

(Also, centripetal force.)

(1) Period of revolution : The time taken by a particle performing UCM to complete one revolution is called the period of revolution or the period (T) of UCM.

T = \(\frac{2\pi r}v\) = \(\frac{2 \pi}{\omega}\)

where v and ω are the linear and angular speeds, respectively. 

SI unit: the second (s) 

Dimensions : [M°L°T¹].

(2) Frequency of revolution : The number of revolutions per unit time made by a particle in UCM is called the frequency of revolution (f).

The particle completes 1 revolution in periodic time T. Therefore, it completes 1/T revolutions per unit time

∴ Frequency f = \(\frac1T\) = \(\frac{v}{2 \pi r}\) = \(\frac{\omega}{2 \pi }\)

SI unit : the hertz (Hz), 1 Hz = 1 s^-1 

Dimensions : [M°L°T^-1]

A particle is said to perform uniform circular motion if it moves in a circle or a circular arc at constant linear speed or constant angular velocity.

Angular speed, ω = \(\frac{v\,cm/s}{r\,cm}\) = \(\frac{v}r\) rad/s.

In a circular motion, the instantaneous velocity \(\vec{v}\) is always tangential, in the sense of the motion. Hence, an inertial observer will see the stone fly off tangentially, in the direction of \(\vec{v}\) at the instant the string breaks.

1. Total linear momentum is conserved when \(\vec F_{extwernal}\) is Zer0

2. Angular momentum is conserved when \(\vec \tau_{extwernal}\) is zero.

  \(\vec F_{extwernal}\) = \(\frac{d\vec L}{dt}\)

Examples of centrifugal force :

1. A person in a merry-go-round experiences a radially outward force. 

2. Passengers of a car taking a turn on a level road experience a force radially away from the centre of the circular road. 

3. A coin on a rotating turntable flies off for some high enough angular speed of the turntable. 

4. As the Earth rotates about its axis, the centrifugal force on its particles is directed away from the axis. The force increases as one goes from the poles towards the equator. This leads to the bulging of the Earth at the equator.

A force which arises from gravitational, electromagnetic or nuclear interaction between matter is called a real force. The centrifugal force does not arise due to any of these interactions. Therefore, it is not a real force.

The centrifugal force in the noninertial frame of reference of a particle in circular motion is the effect of the acceleration of the frame of reference. Therefore, it is called a pseudo or fictitious force.

Correct answer is: (A) Angular velocity upwards, angular acceleration downwards.

Data : r = 25m, v = 36 km/h = 36 × \(\frac{5}{18}\) m/s = 10 m/s, m = 150 kg, g = 10 m/s²

1. Centripetal force, F = \(\frac{mv^2}r\) = \(\frac{150\times(10)^2}{25}\) = 600 N

2. Upward force = normal reaction of the road surface = mg = 150 × 10 = 1500 N

Correct answer is: (B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.

Correct answer is: (B) 2.5

Consider a thin uniform disc of mass M and radius R in the xy plane. Let Ix, ly and Iz be the moments of inertia of the disc about the x, y and z axes respectively.

Now, I_x = I_y

since each represents the moment of inertia (MI) of the disc about its diameter and, by symmetry, the MI of the disc about any diameter is the same.

∴ I_x = I_y= 1/4 MR² (Given)

According to the theorem of perpendicular axes,

I_z = I_x + I_y = 2(1/4MR²) = 1/2 MR²

Let I be the MI of the disc about a tangent normal to the disc and passing through a point on its edge (i.e., a tangent perpendicular to its plane). According to the theorem of parallel axis,

I = I_CM + Mh²

Here, I_CM = I_z = 1/2 MR² and h = R.

∴ I = 1/2 MR² + MR² = 3/2 MR²

which is the required expression.

Correct Option is (C) \(\frac{3 L^3 \lambda}{8 \pi ^2}\)

Moment of inertia of the coil about the diameter

\(I = \frac{1}{2}MR^2\)

\(\because M = V \times \lambda\)

\(L = 2 \pi R\)

\(R = \frac{L}{2 \pi}\)

\(I = \frac{1}{2} L \lambda \frac{L^2}{4 \pi ^2}\)

\(I = \frac{1}{2} \times L \times \lambda \times \frac{L^2}{4 \pi ^2}\)

\(I = \frac{L^3 \lambda}{8 \pi ^2}\)

Using parallel axis theorem

\(I = I_{cm} + MR^2\)

\(I = \frac{L^3 \lambda}{8 \pi ^2} + L \lambda (\frac{L}{2 \pi})^2\)

\(I = \frac{L^3 \lambda}{8 \pi ^2} + \frac{L \lambda L^2}{4 \pi^2}\)

\(I = \frac{3 L^3 \lambda}{8 \pi ^2}\)

 (C) \(\frac{3\lambda L^3}{8\pi^2}\)

Moment of inertia of the sphere about a diameter

= \(\frac{2}5\)MR² = \(\frac{2}5\) x 15 x (0.1)² = 6 x 10^-2 kg.m²

Data : L = 120 cm = 1.2 m, m = 150 g = 0.15 kg,

r = 0.2 m, g = 9.8 m/s²

sin \(\theta\) = \(\frac{r}L\) = \(\frac{0.2}{1.2}\) = \(\frac{1}6\)

(B) centrifugal force

Data : Radius of the Earth = r = 6400 km 

= 6.4 × 10⁶ m, g = 9.8 m/s²

As the Earth rotates, the bodies on the equator revolve in circles of radius r. 

These bodies would not feel any weight if their centripetal acceleration (ωr) is equal to the acceleration due to gravity (g).

∴ ω²r = g 

The angular speed of the Earth’s rotation,

Correct option is (C) 9.42 N.m

The kinetic energy of a body of moment of inertia I and rotating with a constant angular velocity ω is

E = \(\frac12 I\omega^2\)

The angular momentum of the body, L = Iω.

∴E = \(\frac12 (I\omega)\)ω = \(\frac12 L\omega\)

This is the required relation.

Correct option is (A) \(E_2 > E_1\)

Angular Momentum                     \(L = I \omega\)

Given,

\(I_1 = I_2\)

\(I_1 \omega_1 = I_2 \omega_2\)

\(\frac{\omega_1}{\omega_2} = \frac{I_2}{I_1}\)

Kinetic Energy    K.E = \(\frac{1}{2} I\omega^2\)

\(\frac{E_1}{E_2} = \frac{I_1}{I_2} (\frac{\omega_1}{\omega_2})^2\)

\(\frac{E_1}{E_2} = (\frac{I_1}{I_2}) (\frac{\omega_1}{\omega_2})^2\)

\(\frac{E_1}{E_2} = \frac{I_1}{I_2} \times \frac{I_2{^2}}{I_1{}^2}\)

\(\frac{E_1}{E_2} = \frac{I_2}{I_1}\)

Given, \(I_1 > I_2\)

then,

\(E_2 > E_1\)

Correct option is (A) E₂ > E₁ 

(D) The planet’s angular momentum about the Sun

Data: f₁ = 10Hz, f₂ = 0 Hz, t = 6.28s 

The angular acceleration,

Examples of centripetal force :

1. For an Earth-satellite in a circular orbit, the centripetal force is the gravitational force exerted by the Earth on the satellite. 

2. In the Bohr atom, the centripetal force on an electron in circular orbit around the nucleus is the attractive Coulomb force of the nucleus.

3. When an object tied at the end of a string is revolved in a horizontal circle, the centripetal force is the tension in the string. 

4. When a car takes a turn in a circular arc on a horizontal road with constant speed, the force of static friction between the car tyres and road surfaces is the centripetal force.

Note : The tension in a string or the force of friction is electromagnetic in origin.

Data : ω₀ = 1200 rpm, ω = 600 rpm, f = 5 s 

Since the flywheel slows down uniformly, its angular acceleration is constant. Then, its average angular speed,

Its angular displacement in time t,

θ = ω_av .t = 15 × 5 = 75 revolutions