# Question-1

**A circular hole of radius 1m is cut off from a disc of radius 6m. The centre of the hole is 3m from the centre of the disc. Find the centre of mass of the remaining disc.**

**Solution:**

Let O be the centre of the disc and O′ that of the hole.

To find the centre of mass, we use the fact that a body balances at this point, i.e. the algebraic sum of the moments of the weights about the centre of gravity is zero. The weight W_{1} of the disc acts at point O. The hole can be regarded as negative weight W_{2} acting at O′ . If X is the distance of the centre of gravity of the combination from point O, then

Also, W_{1} = ρ π × 6^{2} = 36 ρ π

W_{2} = ρ π × 1^{2} = ρ π (where ρ is the mass per unit area of the disc)

Substituting the values of W_{1} and W_{2}, we get

= 0.086 m.

The negative sign indicates that the centre of gravity is to the left of point O.

# Question-2

**Calculate the centre of mass of a non-uniform rod, whose mass per unit length**

**ρ varies as ρ = ρ**

where ρ

_{0}where ρ

_{0}is a constant, L is the length of rod and x is the distance of any point on rod measured from one end.**Solution:**

Let the rod OA be placed along the x-axis and its end O be chosen as the origin of the x-axis.

Consider a small element PQ of the rod lying between x and x + dx. Let dm be the mass of this element. Obviously,

dm = ρ dx

= x^{2} dx

If X is the distance of the centre of mass end O, then by definition,

X =

=

= .

# Question-3

**Calculate the moment of inertia of a uniform circular disc of mass 500 g, radius 10 cm about**

(i) the diameter of the disc

(ii) the axis tangent to the disc and parallel to its diameter and

(iii) the axis through the centre of the disc and perpendicular to its plane.

(i) the diameter of the disc

(ii) the axis tangent to the disc and parallel to its diameter and

(iii) the axis through the centre of the disc and perpendicular to its plane.

**Solution:**

Mass of the disc, M = 500 g

Radius of the disc, R = 10 cm

(i) Moment of inertia of disc about its diameter, I

_{d}=

= g cm

^{2}

= 12500 g cm

^{2}.

(ii) Applying the theorem of parallel axes. Moment of inertia of the disc about a tangent to the disc and parallel to the diameter of the disc

g cm

^{2}

= 62500 g cm

^{2}.

(iii) Moment of inertia of the disc about an axis, passing through the centre of disc and perpendicular to the plane of the disc

= g cm

^{2}

= 25000 g cm

^{2}.

# Question-4

**What will be the duration of the day, if the earth suddenly shrinks to of its original volume, mass remaining unchanged? Moment of inertia of sphere =**

**× mass ×**

**(radius)**

^{2}.**Solution:**

Let T

_{1}and ω

_{1}be the period of revolution and angular velocity respectively of the earth before contraction. Let T

_{1}and ω

_{2}be the corresponding quantities after contraction. Let I

_{1}and I

_{2}be the moments of inertia of the earth before contraction and after contraction respectively.

Applying the law of conservation of angular momentum,

I

_{1}

_{ω 1}= I

_{2ω 2}

∴

T

_{2}=

Volume after contraction =

∴

If n = 1, T

_{2 }= hour

= 1.5 hour.

# Question-5

**Solution:**

The given system can be written as,

The position vector of the centre of mass of a system of particles is given by

# Question-6

**Use second law of motion and derive the relation for a particle (capable of rotating about an axis) on which torque is acting.**

**Solution:**

Consider a particle of mass m and radius vector with respect to some origin is acted upon by a force . Suppose F

_{x}and F

_{y}be the components of the force and P

_{x}, P

_{y}be the components of momentum, then from Newton's 2

^{nd}law of motion,

Where

We know,

Let us substitute

Therefore

where is the angular momentum.

Thus, The rate of change of angular momentum of a particle is equal to the torque applied to it.

It is in accordance with Newton's second law of motion in which torque is replaced by and angular momentum is replaced by linear momentum .

# Question-7

**Three point masses m**

_{1}, m_{2}and m_{3}are located at the vertices of**an**

**equilateral**

**triangle of length a. Calculate the moment of inertia of the system about an axis along the altitude of the triangle passing through m**

_{1}.**Solution:**

The moment of inertia of the system is to be calculated about an axis, which passes through a point mass m_{1} and lies along the altitude of the triangle. If r_{1}, r_{2} and r_{3} be the distance of point masses m_{1}, m_{2} and m_{3} respectively from the axis AB, then the moment of inertia of the system about axis AB is given by

I =

Here, r_{1} = 0, r_{2} = a/2, r_{3} = a/2

I = m_{1}(0)^{2} + m_{2}+ m_{3 }

# Question-8

**A light vertical chain is being used to haul up an object of mass M kg attached to its lower end. The vertical pull applied has a magnitude F Newton at t= 0 and it decreases uniformly at a rate of f Newton per metre over a distance 'h' through the object is raised. Show that the velocity of the object after it has been raised through h metre is given by**

v =

v =

**Solution:**

Let the body rise through a distance y metre in t second. The instantaneous pull in the chain is then

P = (F -fy) Newton

If a is the upward acceleration of the object at this instant, we have from Newton's second law,

P = Mg = Ma

Or,

If V is the velocity of the body when it is risen h metre, we have, on integration between the limits y =0 to y = h,

V

^{2}=

=

V = .

# Question-9

**A solid cylinder of mass M and radius R has a light flexible rope wound round it. The rope carries a mass m at its free end. The mass is rest at a height h above the floor. Find the angular velocity of the cylinder at the instant the mass m, after release, strikes the floor. Assume friction to be absent and the cylinder to rotate about its own axis.**

**Solution:**

As the mass 'm' descends, its initial potential energy gets converted into kinetic energy of the falling mass M itself and kinetic energy of the cylinder set into rotation. We thus have

mgh =

Now I, the moment of inertia of the cylinder about its own axis is . Also ω = v/R. We thus have

mgh =

v^{2} =

# Question-10

**A body of mass 2 kg is tied at one end of a string 1m long. The other end is fixed and the body revolves in a horizontal circle. The maximum tension, which the string can withstand, is 2000 N. find the maximum number of revolutions per minute the body will make and its linear velocity when the string just breaks.**

**Solution:**

The centripetal force is provided by the tension in the string.

T = mω

^{2}

The tension increases as the speed of revolution increases. If n is the number of revolutions completed when the string just breaks, then

2 × (2π n)

^{2}× 1 = 2000

n

^{2}=

n = 5.03 rev/s

Let v be the maximum linear velocity when the string breaks, then

v = rω

= r(2 π n)

= 1 × 2π × 5.03

= 31.6 m/s.

After the string breaks, the body will move at a speed of 31.6 m/s tangential to the circle at that point.

# Question-11

**A small body slides down a frictionless track after being released from rest from a point at a height h. The track ends in a circular loop of diameter D in the vertical plane. The body remains in contact with the loop as it moves along it. Establish a relation between height h in terms of the diameter D of the loop.**

**Solution:**

The body is released at point P from a height h above the base of the circle.

At any point on the track, the forces acting on the body are its weight mg and the normal reaction F due to the track. At the highest point on the loop, both mg and F point towards the centre C. Thus we must have

F + mg = or v =

If the velocity is less than the downward pull of the weight is greater than the required centripetal force and the body will separate from the circle before reaching point B. To obtain the corresponding height h, consider the total energy at point P

E(P) = KE + PE = 0 + mgh (Q V = 0)

At point B, h = 2r and v^{2} = gr

E(B) =

Since the energy is conserved, we have

E(P) = E(B)

mgh = .

# Question-12

**Write the SI unit of force and Torque.**

**Solution:**

The SI unit of force is N, and that of torque is Nm. The correct unit of torque is Newton-meter, and not joule. This is done to create a difference between the scalar nature of work, and the vector nature of torque.

# Question-13

**Express torque j**

**→**

**in terms of the rate of change of linear momentum.**

**Solution:**

=

^{ }

^{= }

# Question-14

**Suppose a cyclist is negotiating a curve of radius r with speed**

**ν**

**. Write the conditions under which skidding will occur.**

**Solution:**

Skidding will occur in the event of:

(a) ‘ν ’ (the speed of the cyclist) being large

(b) ‘r’ being small

(c) the road surface being slippery or ‘m’ being small.

# Question-15

**If the angular momentum is conserved in a system whose moment of inertia is decreased, will it happen that rotational kinetic energy also be conserved?**

**Solution:**

According to the law of conservation of angular momentum,

I_{1 }_{ω 1 }= I_{2 ω 2 }

I_{1}^{2}_{ ω 1}^{2}_{ }= I_{2}^{2}_{ ω 2}^{2 }

I_{1} (I_{1 ω 1}^{2})= I_{2 }(I_{2 ω 2}^{2})

As I_{1 }< I_{2}, I_{1 ω 1}^{2} > I_{2 ω 2}^{2 }

I_{1 ω 1}^{2} > I_{2 ω 2}^{2}

Thus, the rotational kinetic energy of the system increases on decreasing its moment of inertia.

# Question-16

**We can write,**

**Σ**

**F**

That is, the resultant external force on the system of particles equals the total mass of the system multiplied by the acceleration of the centre of mass. Why do not the internal forces (from within the system) contribute to the motion of the system?

_{ext }= Ma_{c m }=That is, the resultant external force on the system of particles equals the total mass of the system multiplied by the acceleration of the centre of mass. Why do not the internal forces (from within the system) contribute to the motion of the system?

**Solution:**

The forces on any particle in the system may include both external and internal forces - external forces from outside the system, and internal forces from within the system. In accordance with Newton's third law, the forces of any arbitary particle 1 on any arbitrary particle 2, is equal and opposite to the force of particle 2 on particle 1. Therefore, when we sum up all the internal forces in the equation,

M a_{cm }= Σ m_{i }a_{i} = Σ F (where F is the force on particle t)

They cancel in pairs, and the net force on the system is due to only the external forces.

# Question-17

**A solid cylinder of mass 20kg rotates about it is axis with angular speed 100s**

^{-1}. The radius of the cylinder is 0.25m.What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about it is axis?**Solution:**

Mass of the solid cylinder, M = 20 kg

Angular speed of the cylinder, ω = 100 s^{-1 }

Radius of the cylinder, R = 0.25 m

Moment of inertia of the cylinder about it is axis

=MR^{2 }= × 20 × (0.25)^{2 }= 0.625 kg m^{2 }

Kinetic energy of rotation = Iω ^{2}= × 0.0625 × (100)^{2}

= 3125 Joules

Angular momentum, L = Iω = 0.625 × 100 = 62.5 Joules.

# Question-18

**Define the term radius of gyration of a body about some axis of rotation.**

**Solution:**

The distance of a point from the axis of rotation, a which if whole of the mass of the body were supposed to be concentrated, its moment of inertia about the axis will be the same as that determined by the actual distribution of mass of the body is called radius of gyration.

# Question-19

**Why a helicopter must necessarily have two propellers?**

**Solution:**

If there were only one propeller in the helicopter then the helicopter would have to turn in the opposite direction to conserve angular momentum.

# Question-20

**Why two spheres of the same mass and same radius, one solid and the other hollow have unequal moment of inertia?**

**Solution:**

Moment of inertia of a body depends upon the distribution of mass of the body about the axis of rotation mass of solid sphere is distributed uniformly about the axis of rotation. On the other hand, mass of hollow sphere is distributed away from the axis of rotation. Hence moment of inertia of hollow sphere is larger than the moment o inertia of solid sphere.

# Question-21

**A disc is recast into hollow and thin cylinder of same radius, which will have large moment of inertia?**

**Solution:**

Hollow cylinder will have larger moment of inertia because most of its mass is located at comparatively larger distance from the axis of rotation.

# Question-22

**The cap of the pen can be easily opened with the help of two fingers than with one finger. Explain why?**

**Solution:**

We are able to apply a couple of force with the help of two fingers. This couple enables us to open the cap of the pen easily.

# Question-23

**Should the center of mass of a body necessarily lie inside the body? Explain.**

**Solution:**

No, it may lie outside the body. In case of a semicircular ring, it is at the center, which is outside the ring.

# Question-24

**A flywheel is revolving with constant angular velocity. A chip of its rim breaks and flies away, what will be the effect on the angular velocity?**

**Solution:**

On account of reduction in mass, there will be decrease in moment of inertia. Since angular momentum is to be conserved therefore angular velocity will be increased.

# Question-25

**A ladder is at rest with the upper end against a wall and the lower end on the ground. Is the ladder more likely to slip when a man stands on it at the bottom or at the top? Explain.**

**Solution:**

It is more likely to slip when a man stands at the top of the ladder. This is due to the fact that the main weight will provide an extra torque for the slipping of the ladder.

# Question-26

**If the torque acting on a particle about an arbitrary origin is zero, what can we say about the angular momentum of the particle about the origin?**

**Solution:**

We know = τ

if τ = 0, = 0

or L = constant.

# Question-27

**You are given two identical balls – one tied to one end of a smaller spring and the other tied to one end of a longer spring. Which one will you find it easy to rotate in a horizontal circle? Explain your answer.**

**Solution:**

The torque τ required to rotate the ball with an angular acceleration α is given by τ = 1α , where I is the moment of inertia of the system about the axis of rotation. The moment of inertia of the system will be less if the ball is tied to a shorter spring. Therefore, the torque required will also be less.

# Question-28

**Show that the angular momentum of a satellite of mass Ms revolving round the earth having mass Me in an orbit of radius r is equal to [G M**

_{e}M_{s}^{2}r]^{1/2}.**Solution:**

Mass of the satellite = M

_{s }

Mass of the earth = M_{e }

Radius of the orbit to the satellite = r

Suppose the satellite revolves around the earth with orbital speed v, the gravitational force between the earth and the satellite provides the necessary centripetal force required by the satellite to revolve in the circular orbit around the earth.

Now, the required centripetal force = .

# Question-29

**When there is no external torque acting on a rotating body, which of the following quantities can change?**

(i) angular acceleration

(ii) angular momentum

(iii) angular speed.

(i) angular acceleration

(ii) angular momentum

(iii) angular speed.

**Solution:**

(i) The relation between torque angular acceleration is τ = I αSince τ = 0 ∴ α = 0

Hence, angular acceleration will not change, when external torque acting on the body is zero

(ii) The relation between torque and angular momentum is given by τ = dL/dt

When τ = 0,

or L = constant

Hence, angular acceleration will not change, when external torque acting on the body is zero

(iii) Since L = I ω = constant, if τ = 0 then L is constant. Therefore, angular speed (ω) will change, if moment of inertia of the body changes.