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## Karnataka 1st PUC Physics Question Bank Chapter 7 System of Particles and Rotational Motion

Question 1.

What is a meant by rigid body?

Answer:

A body which does not change its shape and size when external forces are applied on it is called rigid body.

In a rigid body the distance between any pair of particles does not change.

Question 2.

Explain the different kinds of motion the rigid body can have.

Answer:

A rigid body can have

- translational motion
- rotational motion

1. Translational motion: In translational motion, at any instant of time, all the particles of the body have same velocity.

2. Rotational motion: In rotational motion, the rigid body rotates about a fixed axis. Every particle of the body moves in a circle in a plane perpendicular to the axis.

Question 3.

What kind of motion the rigid body can have when it is rotating about an axis?

Answer:

Rotational motion.

Question 4.

Mention few examples where the body rotates about a fixed axis.

Answer:

Ceiling fan, potters wheel, giant wheel, merry go round etc.

Question 5.

Mention few examples of rotational motion without a fixed axis.

Answer:

pinning top, pedestal fan oscillating table fan etc.

Question 6.

What is meant by precession of a spinning top?

Answer:

The movement of the axis of the top around the vertical axis is called precession of the spinning top.

Question 7.

What type motion is associated with a rigid body which is (a) fixed at open point (b) not fixed?

Answer:

(a) A rigid body fixed at one point or along a line can have only rotational motion.

(b) A rigid body which is not fixed can have either pure translation or a combination of translation and rotation.

Question 8.

Define centre of mass of a rigid body.

Answer:

Centre of mass of a rigid body is defined as a point where the entire mass of the body is assumed to be concentrated.

Question 9.

Explain how to locate the co-ordinates for centre of mass for a two particle system in one dimension.

Answer:

Consider a system of two particles of masses m_{1} and m_{2} which are at distance x_{1} and x_{2} respectively from the origin.

The centre of mass C is at a distance X from O which is given by

X = \(\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}\)

Question 10.

Find the co-ordinates of centre of mass of two identical masses which are distances x_{1} and x_{1} from the origin.

Answer:

The coordinate for the centre of mass is

Question 11.

Find the centre of mass of three particles of masses mj, m2 and m3 at distances x_{1}, x_{2} and x_{3} from the origin. What will be the position of centre of mass if three masses are identical?

Answer:

Consider a system of three particles of masses m_{1}, m_{2} and m_{3} which are at distances xu x2 and x3 from the origin O.

The centre of mass C is at a distance X which is given by

Question 12.

Write the expression for centre of mass of n particle system of masses m_{1}, m_{2}, m_{3}….m_{n} which are distances x_{1}, x_{2}, x_{3}….x_{n} from the origin.

Answer:

Consider a system of n particles of masses m_{1}, m_{2}, m_{3}….m_{n} which are at distances x_{1}, x_{2}, x_{3}….x_{n} from the

origin. The centre of mass C is at a distance X given by

Question 13.

Write the expression for centre of mass of n particle system of masses m_{1}, m_{2}, m_{3}….m_{n }with coordinates (x_{1},y_{1}), (x_{2},y_{2}) (x_{n},y_{n}).

Answer:

The co-ordinates of centre of mass (X, Y) of n particle system is

Question 14.

Mention the expression for position of centre of mass of n particles of masses m_{1}, m_{2}, m_{3}….m_{n } co-ordinates (x_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2}) (x_{n},y_{n},z_{n})

Answer:

The co-ordinates of centre of mass (X, Y, Z) of n particle system is

Question 15.

Mention the expression for centre of mass interms of position vector.

Answer:

Consider a system of n particles of masses m_{1}, m_{2}, m_{3}….m_{n} of position vectors r_{1}, r_{2}, r_{3}….r_{n} Then the

position vector R for centre of mass is

Answer:

Question 16.

Write the expression for the co-ordinate system of centre of mass in the vector form.

Answer:

The co-ordinate system for centre of mass in the vector form is

\(\vec{R}=\frac{1}{M} \int \vec{r} d m\)

Question 17.

Give the location of the centre of mass of (i) sphere (ii) cylinder (iii) ring (iv) cube each of uniform mass density.

Answer:

Centre of mass is

(i) at the centre of the sphere

(ii) at the centre of the ring

(iii) at the mid point and on the axis of the cylinder

(iv) at the point of the intersection of the diagonals.

Question 18.

Where does the centre of mass of a rectangular lamina lie?

Answer:

At the point of intersection of the diagonals of the rectangular lamina.

Question 19.

What is the location of the centre of mass of a triangular lamina? .

Answer:

Centre of mass is at the centroid of the triangle i.e., at the point of intersection of the medians.

Question 20.

Does the centre of mass of a body necessarily lie inside the body?

Answer:

No, it may lie outside the body.

Question 21.

Give an example for a body where the centre of mass lie inside the body.

Answer:

Solid sphere or circular disc.

Question 22.

Give an example for a body where the centre of mass lie outside the body.

Answer:

Circular spring or hollow cylinder.

Question 23.

Show that the product of total mass of a system of particles and the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.

Consider a system of particles of masses m_{1}, m_{2}, m_{3}….m_{n} having position vectors r_{1}, r_{2}, r_{3}….r_{n} If R is the

position vectors of centre of mass we have

differentiating with respect to time t

\(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\cdots \cdot \cdot \cdot \vec{F}_{n}=\vec{F}_{e x}\) , it is the sum of all external forces acting on the system of particles.

\(M \vec{A}=\vec{F}_{e x t}\)

Thus the centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

Question 24.

What is the path followed by the centre of mass of a projectile if it explodes midway in air?

Answer:

The centre of mass of the fragments of the projectile continues along the same parabolic path which it would have followed if there is no explosion.

Question 25.

Define linear momentum of system of particles.

Answer:

The linear momentum of a system of a system of particles is defined as vector sum of linear momentum of all the particles of the system.

OR

Total linear momentum of a system of particles is equal to the product of total mass of the system of particle and the velocity of centre of mass.

\(\vec{P}=M \vec{v}\)

Question 26.

Arrive at Newtons second law extended to a system of particles.

Answer:

Consider a system of particles of masses m_{1}, m_{2}, m_{3}….m_{n} having linear velocities \(\vec{v}_{1}, \vec{v}_{2}, \vec{v}_{3} \cdots \cdots \cdot \vec{v}_{n}\)

Linear momentum of particle of mass m_{1} is, \(\vec{p}_{1}=m_{1} \vec{v}_{1}\) linear velocity of momentum of mass m_{1} is \(\vec{p}_{2}=m_{2} \vec{v}_{2}\)

The vector sum of linear momentum of individual particles is the linear momentum of the system.

But linear momentum of the system of particles is equal to the product of total mass of the system and velocity of centre of mass.

i.e., \(\vec{P}=M \vec{v}\)

Differentiating both sides with respect to time t

∴ rate of change of linear momentum is net force acting on the system of particle. This is Newton’s second law extended to system of particles.

Question 27.

What is the total external force on a system of particles when its total momentum is constant?

Answer:

The total external force is zero.

Question 28.

What happens to total momentum of a system of particles when total external forces acting it is zero?

Answer:

The total momentum of the system of particles remains constant.

Question 29.

Give few examples where the motion of centre of mass of a body is not affected by internal forces.

Answer:

(i) When a projectile following a parabolic path explodes into two fragments in mid air, the forces leading to explosion forces i.e., force of gravity is the same before and after explosion. The centre of mass of the fragments follows the same path which it would have followed if there is no explosion.

(ii) When heavy nucleus (Radium, Ra) which is at rest, explodes into two lighter nuclei (Radon, Ra) and a-particle (He) because of internal forces. The fragments fly in the same line but in opposite direction. The centre of mass of the system continues to be at rest.

(iii) When the moving radium nucleus disintegrates into Radon and a-particle, the fragments fly in two different directions such that the centre of mass moves along the same path along which the original radium nucleus was moving.

Question 30.

Define cross product of two vectors.

Answer:

Cross product of two vectors is defined as product of the magnitude of the two vectors and sine of the angle between them.

If a and b are the two vectors then \(\vec{a} \times \vec{b}=a b \sin \theta \hat{n}\), where \(\hat{n}\) is the unit vector along the direction of the resultant.

Question 31.

Name the rule to find the direction of resultant of cross product of two vectors.

Answer:

The direction of the resultant is given by right hand screw rule or right hand thumb rule.

Question 32.

State right handed screw rule to find the direction of the vector product of two vectors.

Answer:

If head of a screw is turned in the plane of the two vectors a and b then the direction in which the tip of the screw advances represents the direction of the resultant c.

Question 33.

Distinguish between scalar product and vector product of two vectors.

Answer:

Scalar product

- Scalar product of two vectors is defined as product of magnitude of two vectors and cosine of the angle between them \(\vec{a} \cdot \vec{b}=a b \cos \theta\).
- The resultant is scalar.
- The scalar product of two vectors is commutative \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)

Vector product:

- Vector product of two vectors is defined as product of magnitude of two vectors and sine of the angle between them \(\vec{a} \times \vec{b}=a b \sin \theta \hat{n}\)
- The resultant is vector.
- The vector product of two vectors is not commutative \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\)

Question 34.

What is the magnitude of cross product of two identical vectors?

Answer:

For two identical vectors θ = 0.

∴ \(\vec{a} \times \vec{a}\) = a ^{2} sin0 = 0

Question 35.

What is the value of cross product of two unit vectors in the same plane?

Answer:

Zero.

i.e., î × î = 0 ĵ × ĵ = 0, k̂ × k̂ = 0

Question 36.

What is the value of cross product of any two units vectors in two mutually perpendicular planes?

Answer:

î × ĵ = k̂ and ĵ × î = -k̂

ĵ × k̂= i and k̂ × ĵ = -î

k̂ × î = ĵ and î × k̂= – ĵ

Question 37.

Express cross productors in the determinant form.

Answer:

Consider two vectors a and b.

This can be represented in the vector form as

a x b = \(\left|\begin{array}{ccc}

{\hat{i}} & {\hat{j}} & {\hat{k}} \\

{a_{x}} & {a_{y}} & {a_{z}} \\

{b_{x}} & {b_{y}} & {b_{z}}

\end{array}\right|\)

Question 38.

Mention any two examples for the cross product of two vectors.

Answer:

(i) Angular momentum is the cross product of position vector and linear momentum

\(\vec{L}=\vec{r} \times \vec{p}\)

(ii) torque is the cross product of position vector and force

\(\vec{\tau}=\vec{r} \times \vec{F}\)

Question 39.

Define angular displacement.

Answer:

The angle swept by the radius vector in a given interval of time is called angular displacement.

Question 40.

Mention the unit of angular displacement.

Answer:

The unit of angular displacement in S.I. system is radian (rad).

Question 41.

Define angular velocity.

Answer:

The time rate of change of angular displacement is called angular velocity. If A0 is angle described in a time interval ∆t then angular velocity ω = ∆θ/∆t

Question 42.

Mention the unit of angular velocity.

Answer:

The unit of angular velocity in S.I. system is rad s^{-1}.

Question 43.

Define uniform angular velocity.

Answer:

The angular velocity of a body is said to be uniform if it describes equal angles in equal intervals of time, however small the intervals may be.

Question 44.

Obtain the relation between linear velocity (v) and angular velocity (ω) OR Show that v = rω.

Answer:

Consider a rigid body rotating about an axis passing through the point O. Let P be a particle of mass m at a distance r from O. As the body rotates, the particle moves in a circular path.

Let the particle move from P to P’ covering an angle θ in a time interval t.

Question 45.

Define angular acceleration.

Answer:

Angular acceleration is defined as time rate of change of angular velocity.

If angular velocity changes by a small amount d.o in a time interval dt then

Question 46.

Mention the unit of angular acceleration.

Answer:

Unit of angular acceleration is rad s^{-2}.

Question 47.

Define angular momentum of a particle.

Answer:

Angular momentum of a particle is defined as moment of linear momentum of the particle.

Angular momentum of a particle is the vector product of position vector and linear momentum of the particle.

\(\vec{L}=\vec{r} \times \vec{p}\)

Question 48.

Show that torque is equal to time rate of change of angular momentum. OR Show that the time rate of

change of angular momentum of the particle is equal to the torque acting on it. [Show that \(\tau=\frac{d L}{d t}\)

Answer:

Consider a particle of ma.ss m rotating in a circle of radius r about a fixed point O. Then \(\overrightarrow{O P}=\vec{r}\) represents

the position vector of the particle. If y is the linear velocity of the particle, then linear momentum of the particle p = mV

∴ time rate of change of the angular momentum of a particle is equal to the torque acting on it.

Question 49.

Show that the time rate of total angular momentum of a system of particles about a point is equal to sum of the external torques acting on the system about the same point.

Answer:

The total angular momentum of a system of n particles is

Thus time rate of total angular momentum of a system of particles about a point is equal to sum of the external torque acting on the system about the same point.

Question 50.

State the condition for equilibrium of rigid bodies.

Answer:

A rigid body is said to be in mechanical equilibrium when it is in (i) translational equilibrium and (ii) rotational equilibrium.

(i) A body is said to be in translational equilibrium when vector sum of the forces is zero.

i.e„ \(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3} \cdots \cdots \cdot \vec{F}_{n}=\sum_{i=1}^{i=n} \vec{F}_{i}=0\)

In this case the linear momentum of the body does not change with time.

(ii) A body is said to be in rotational equilibrium when the vector sum of the torques is zero.

\(\vec{\tau}_{1}+\vec{\tau}_{2}+\vec{\tau}_{3}+\cdots \cdots \vec{\tau}_{n}=\sum_{i=1}^{i=n} \vec{\tau}_{i}=0\)

In this case the total angular momentum of the body does not change with time.

Question 51.

When a rigid body is said to be in mechanical equilibrium?

Answer:

A rigid body is said to be in mechanical equilibrium if both linear momentum and angular momentum are not changing with time, i.e., the body has neither linear acceleration nor angular acceleration.

Question 52.

What is a couple?

Answer:

Two equal unlike parallel forces acting at different points on a rigid body is called couple.

Question 53.

When the body is said to be n translational equilibrium?

Answer:

The body is said to be in translational equilibrium if the total force on the body is zero i.e., when the total linear momentum of the body does not change with time.

Question 54.

Define moment of a couple or torque.

Answer:

The rotating effect produced by a couple on a body which is free to rotate about an axis is called moment of couple or torque. It is measured by product of magnitude of either force and the arm of the couple.

Torque = force × arm of the couple.

Question 55.

Mention the unit and dimension of torque.

Ansi Unit of torque is newton – metre (Nm)

The dimensions of torque is [ML^{2}T^{-2}]

Question 56.

Is torque is scalar or vector.

Answer:

Torque is a vector quantity.

Question 57.

Give few examples where the application of couple to rotatory effect.

Answer:

The applications of couple to produce rotatory effect are

- peddling of a bicycle
- opening or closing of a tap
- oscillations of a magnetic needle in earth’s magnetic field.

Question 58.

Show that moment of a couple does not depend on the point about which you take the moments.

Answer:

Consider two equal and opposite force F and – F acting at points A and B. Let r\ and r2 be the position vectors of these two forces about the origin O.

The moment of the couple F

= sum of the moments of the two forces forming the couple

Hence moment of a couple does not depend on the point about which the moment is taken.

Question 59.

What is an ideal lever?

Answer:

Ideal level is a light rod pivoted at a point along its length.

Question 60.

What is fulcrum?

Answer:

The point about which an ideal lever is pivoted is called fulcrum.

Question 61.

Mention few example for lever.

Answer:

- A see-saw on the children playground.
- A beam of a balance.

Question 62.

Explain the terms load and load arm.

Answer:

In case of a lever, the weight to be lifted is called load.

The load arm is the distance between the load and the fulcrum.

Question 63.

Explain the terms effort and effort arm.

Answer:

The force applied to lift a load is called effort.

The distance between effort and the fulcrum is called effort arm.

Question 64.

Define mechanical advantage of a lever.

Answer:

The mechanical advantage of a lever is defined as ratio of load to the effort.

Question 65.

Explain the principle of moments for a lever.

Answer:

Consider two like parallel forces F_{1} and F_{2} acting at A and B on a uniform bar. Let the forces act at distances d_{1} and d_{2} from the fulcrum 0. Let R be the reaction of the support at the fulcrum.

For translation equilibrium,

sum of the upward forces = sum of downward forces.

R = F_{1} + F_{2}

R_{1} – F_{1} – F_{2 }= 0

For rotational equilibrium

sum of the clockwise moment = sum of anticlockwise moments

F_{1}d = F_{2}d_{2}

i.e., load × load arm = effort × effort arm

but mechanical advantage load

M.a = \(\frac{\text { load }}{\text { effort }}\) = \(\frac{F_{1}}{F_{2}}=\frac{a_{2}}{d_{1}} \)

Question 66.

What is the mechanical advantage of a lever?

Answer:

Mechanical advantage of a lever signifies that a large load can be lifted using a small effort.

Question 67.

What happens when mechanical advantage of a lever is greater than one?

Answer:

When the mechanical advantage is greater than one, a small effort is enough to lift a large load.

Question 68.

Define centre of gravity of a body.

Answer:The centre of gravity of a body is defined as a fixed point about which the total gravitational torque on the body is zero.

Question 69.

What is the analogue of mass in rotational motion?

Answer:

Moment of inertia.

Question 70.

Obtain an expression for the kinetic energy of a rotating body.

Answer:

Consider a body a rotating about an axis passing through OO’. Let m_{1}, m_{2}, m_{3}…. are masses of the particles of the body which are at distances r_{1}, r_{2}, r_{3}…. from the axis. Let v_{1}, v_{2}, v_{3}…. are the linear velocities of the particles.

The kinetic energy of the particle of m_{1} linear velocity V_{1} is

K_{1} = \(\frac{1}{2} m_{1} v_{1}^{2}\)

The kinetic energy of the particle of mass m_{2}, linear velocity v_{2} is

K_{2} = \(\frac{1}{2} m_{2} v_{2}^{2}\)

Then the total kinetic energy of the body is the sum of the kinetic energy of the individual particles of the body. K = K_{1} + K_{2} + K_{3} + …….

where w is the angular velocity of all the particles of the body.

Question 71.

Define moment of inertia of a particle.

Answer:

Moment of inertia of a particle about an axis is the product of its mass and square of the distance of the particle from the axis of rotation.

I = mr^{2}

Question 72.

Define moment of inertia of a rigid body.

Answer:

The moment of inertia of: rigid body about an axis is defined as the algebraic sum of the product of the mass and the square of the distance of the particle from the axis of rotation

I = \(\sum_{i=1}^{i=n} m_{i} r_{i}^{2}\)

Question 73.

Mention the factors on which moment of inertia of a body depends.

Answer:

The moment of inertia of a rigid body rotating about an axis depends on

- mass of the body
- shape and size of the body
- distribution of mass from the axis of rotation
- position and orientation of the axis of rotation.

Question 74.

Me.ition the dimension and SI. Unit of moment of inertia.

Answer:

The dimension of moment of inertia is [M^{1}L^{2}] and its unit in SI system is kgm^{2}.

Question 75.

Is moment of inertia a scalar or vector quantity?

Answer:

Moment of inertia is a scalar

Question 76.

Moment of inertia is the measure of rotational inertia. Explain.

Answer:

Moment of inertia of a rotating body about an axis resist any charge in its rotational motion.

Hence it is the measure of rotational inertia of the body.

Question 77.

Define radius of gyration.

Answer:

Radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the centre of mass of the body.

Question 78.

Mention the expression for the moment of inertia in terms of radius of gyration.

Answer:

I = MK^{2}

where M is the mass of the body K is the radius of gyration.

Question 79.

Mention the dimension of radius of gyration.

Answer:

The dimension of radius of gyration is [L],

Question 80.

Calculate moment of inertia of a circular ring about an axis passing through its centre.

Answer:

Consider a circular ring of mass M and radius R rotating about an axis passing through its centre as shown in the diagram.

Each element of the ring is at a distance R from the axis and moves with a speed v = Rw

∴ Moment of inertia of a ring about an axis passing through its centre is I = MR^{2}

Question 81.

Which physical quantities are represented by the following?

(i) Rate of change of angular momentum.

(ii) Moment of linear momentum.

Answer:

(i) Torque

(ii) Angular momentum.

Question 82.

Why fly wheel is used automobile engine?

Answer:

Flywheel has larger moment of inertia. Hence it resist sudden increase or decrease of the speed of the vehicle. It allows a gradual change in the speed and prevents jerky motion and thereby ensure a smooth ride for the passengers.

Torque and work are both defined as four times distance. Explain how do they differ.

Answer:

(i) Work is a scalar quantity, but torque is a vector quantity

(ii) Work done is measured as the product of the applied force and the distance, which the body covers along the direction of the four. Whereas torque is measured as the product of the force and its perpendicular distance from the axis of notation.

Question 83.

State and explain theorem of parallel axes.

Answer:

Statement: The moment of inertia of a rigid body about an axis is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between two parallel axes.

Explanation: Consider a rigid body of mass M.

Let l be the moment of inertia of the body about an axis passing through the point O. Let 7C be the moment of inertia of the body about a parallel axis passing through the centre of mass C. If r be the perpendicular distance between the two parallel axes then

I = I_{C} +Mr^{2}

Question 84.

State and explain theorem of perpendicular axes theorem.

Answer:

The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to sum of the moment of inertia of the lamina about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

Explanation: Let Ix and Iy be the moment of inertia about two perpendicular axes X and Y which are in the plane of the lamina and intersecting at O. If Iz is the moment of inertia about Z axis which perpendicular to the plane of the lamina and is passing through the point O then

I_{z} = I_{x} + I_{y}

Question 85.

Using parallel axes theorem. Calculate the moment of inertia of the rod about an axis perpendicular to its length and passing through its one end. Given the moment of inertia about an axis passing through its centre and perpendicular to the length is \(\frac{M l^{2}}{12}\)

Answer:

Using parallel axes theorem

I’ = I + Mr^{2}

Question 86.

Using perpendicular axis theorem, calculate the moment of inertia of a Assume the moment of inertia about an axis perpendicular to its plane as \(\left(\frac{M R^{2}}{2}\right)\)

Answer:

The disc can be considered as plane laminar body.

Consider three concurrent axes X, Y and Z which are perpendicular to one another such that X and Y are in the plane of the disc.

Using perpendicular axes theorem,

I_{z} = I_{x} + I_{y}

∴ The moment of inertia of the disc about any of its diameter is \(\frac{M R^{2}}{4}\)

Question 87.

Mention the expression for moment of inertia of a thin rod about an axis perpendicular to its length and passing through the centre.

Answer:

I = \(\frac{M L^{2}}{12}\)

L → length of the rod

M → mass of the rod

Question 88.

Mention the expression for moment of inertia of circular ring about an axis perpendicular the plane and passing through its centre.

Answer:

I = MR^{2}

M → mass of the ring R→ radius of the ring

Question 89.

Mention an expression for moment of inertia of a circular disc about an axis perpendicular the plane and passing through its centre.

Answer:

I = \(\frac{M R^{2}}{2}\)

where M is mass of the disc ;

R is the radius of the disc.

Question 90.

Mention an expression for moment of inertia of a hollow cylinder about its axis.

Answer:

I = MR^{2}

M → is the mass of the cylinder ; R → is the radius of the cylinder.

Question 91.

Mention the expression for moment of inertia of a solid cylinder about its axis.

Answer:

I = \(\frac{M R^{2}}{2}\)

M → is the mass of the cylinder.

R → is the radius of the cylinder.

Question 92.

Mention the expression for the moment of inertia of the solid sphere about an axis passing through its diameter.

Answer:

I = \(\frac{2}{5}\) MR^{2}

M → is the mass of the sphere.

R → is the radius of the sphere.

Question 93.

Write the kinematic equations of linear motion and hence the corresponding kinematic equations for rotational motion.

Answer:

The kinematic equations of linear motion with uniform acceleration are

(i) ω = ω_{0} + at

(ii) θ = θ_{0} + ω_{0}t + \(\frac{1}{2}\)at^{2}

(iii) ω^{2} = ω_{0}^{2} + 2α θ

The kinematic equations for rotational motion with uniform angular acceleration are

(i) v = v_{0} + at

(ii) x = x_{0} + v_{0}t + \(\frac{1}{2}\)at^{2}

(iii) v^{2} = v_{0}^{2} + 2ax

Question 94.

derive the equation ) ω = ω_{0} + at from the first principles..

Answer:

Question 95.

Obtain an expression for the work done by the torque.

Answer:

Consider a rigid body rotating about a fixed axis. Let F_{1} be the force acting on a particle P_{1}. The particle describes a circular path of radius r_{1}.

Work done by the force as the particle moves through a distance P_{1}P’_{1} = ds is

work done, dW_{1} = \(\vec{F}_{1} \cdot \overrightarrow{d s}_{1}\)

dW_{1} = F_{1}ds_{1} cosΦ _{1}

But ds_{1} = rdθ

dW_{1} = F_{1}(r_{1}dθ)cosΦ _{1}

from the diagram,

Φ _{1} + α = 90°

Φ _{1} – (90 – α)

dW_{1}= F_{1} (r_{1}dθ) cos (90 – α)

= F_{1}r_{1}θsinα

torque due to force F_{1} is

\(\tau_{1}=\vec{r}_{1} \times \vec{F}_{1}=r_{1} F_{1} \sin \alpha\)

work done by the torque dW_{1} = t_{1}dθ

total work done by all the forces acting on the body

dW = (t_{1} + t_{2} + …….. )dθ

But t_{1}+t_{2}+ = t

work done dW = tdθ

Question 96.

Show that the torque of a rotating body is given by the product of moment of inertia of body and its angular acceleration. OR Show that t = I α

Answer:

Consider a rigid body rotating about a fixed axis. Let Fi be the force acting on a particle Pressure. The particle describes a circular path of radius r_{1}

torque due to the force F_{1} is \(\tau_{1}=\vec{r}_{1} \times \vec{F}_{1}\)

= r_{1}F_{1}sinα

work done by the torque

dW = t_{1}dθ

total work done by all the forces acting on the body dW =(t_{1} + t_{2} + ……)dθ

dW = tdθ

dividing both sides by dt

\(\frac{d W}{d \tau}=\tau \frac{d \theta}{d \tau}\)

i.e., instantaneous power P = tw …. (1)

The workdone by the external torque is used to increase the kinetic energy of the body.

The rate of increase of kinetic energy \(\frac{d}{d t}\left(\frac{1}{2} I \omega^{2}\right)=\frac{1}{2} I \cdot 2 \cdot \omega \cdot \frac{d \omega}{d t}=I \omega \alpha\) …. (2)

equating (1) and (2),

tw = Iwα

t = Iα

Question 97.

State the law of conservation of angular momentum. Give two examples to illustrate the law.

Answer:

Statement:- In the absence of torque, the angular momentum of a rotating body remains constant.

(i) Spinning ballet dancer: A spinning ballet dancer uses the principle of conservation of angular momentum to change her angular speed.

When she folds her arms and bring her leg closer her body, the moment of inertia decreases and thereby her angular speed increases. When she wants to decrease her speed, she stretches her arms and one of her leg, the moment of inertia increases and thereby the angular speed decreases.

(ii) Spring board diver: A diver jumping from a spring board, folds is arms and legs and curls his body. Consequently the moment of inertia decrease hence can spin and perform somersault.

When he is about to enter water, he stretches his arms and legs consequently the moment of inertia increases and angular speed decreases. He can enter the water surface without splashing.

Question 98.

Obtain an expression for kinetic energy of a rolling body.

Answer:

A rolling body has kinetic energy due to both translatory and rotational motion.

Kinetic energy due to translatory motion i.e., kinetic energy about its centre of mass K_{t} = \(\frac{1}{2} m v_{c m}^{2}\)

where m is the mass of the body

v_{cm} is the centre of mass of the body.

Kinetic energy due to rotatory motion K_{t} = \(\frac{1}{2}\) Iw^{2}

where I is the moment of inertia

w is the angular velocity.

∴ total kinetic energy of the rolling body

Question 99.

A body of mass 0.2 kg executes circular motion in a circle of radius 1 m with an angular speed of 60 rpm. Calculate (i) angular speed in rad s^{-1} (ii) linear speed (iii) moment of inertia about the axis of rotation.

Answer:

M = 0.2 kg; R = 1 m w = 60 rpm

(i) Angular speed co = 60 rpm

= \(\frac{60 \times 2 \pi}{60}\)

= 2π rad s^{-1}

(ii) Linear speed v = Rw

= 1 x 2π

= 2 x 3.14 = 6-28 ms^{-1}

(iii) 1 = MR^{2} = 0.2 x 1^{2}

= 0.2 kgm^{2}

Question 100.

The moment of inertia of a solid sphere about a tangent is \(\frac{5}{3}\) MR^{2}, where M is mass and R is radius of the sphere. Find the M.I. of the sphere about its diameter.

Answer:

Here, I_{tan} = \(\frac{5}{3}\) MR^{2}

The diameter of sphere is the axis through the centre of gravity of the sphere. If IG is M.I. of the sphere about its diameter, then according to the theorem of parallel axes, we have

I_{tan} = I_{G} = MR^{2} (Here, h = R)

∴ I_{G} = I_{tan} – MR^{2} = \(\frac{5}{3}\) MR^{2} – MR^{2} = \(\frac{2}{3}\) MR^{2}

Question 101.

Two bodies of mass 2g and 3g having position vectors (3î + 5ĵ + 4k̂) and (1î + 2ĵ + 3k̂) cm respectively. Find the position vector of the centre of mass.

Answer:

m_{1} = 2g; m_{2} = 3g; r_{1} = (3î + 5ĵ + 4k̂) r_{2} = (1î + 2ĵ + 3k̂)

x_{1} = 3 cm; y_{1} = 5 cm; z_{1} = 4 cm

x_{2} = 1 cm; y_{2} =2 cm; z_{2} = 3 cm

the position vector of centre of mass is R = (Xî + yĵ + Zk̂)

Question 102.

A sphere of mass 1 kg radius 10 cm rolls on ground viith a speed 1 ms^{-1}. Calculate the kinetic energy of the sphere.

Answer:

A sphere rolling on the ground has both translation and rotating kinetic energy.

Question 103.

The linear speed of rotating body increases from 60 rpm to 120 rpm in 5 s. Find (I) angular acceleration.

Answer:

Question 104.

104. that with her arms extended, find her angular velocity when she folds her arms. Calculate the

percentage change in her kinetic energy.

Answer:

ω_{1} =6π rads^{-1} I_{1}=I

ω_{2} = ?

I_{2} = \(\frac{60}{100} I\) = 0.6 I

using the law of conservation of angular momentum

I_{1}ω_{1} = I_{2}ω_{2}

I × 6π = 0.6 I × ω_{2}

ω_{2} =10π rad ^{-1}

percentage change in kinetic energy

∴Kinetic energy increases by 66.67%

Question 105.

Given the location of the centre of mass of a (i) sphere (ii) cylinder (iii) ring and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Answer:

(i) Sphere at the centre of the sphere.

(ii) ring at the centre of the ring.

(iii) cylinder at the mid point on the axis of the cylinder

(iv) cube at the point of intersection of the diagonals.

The centre of mass of a body does not lie always inside the body. It can be outside the body also.

Question 106.

In the HC1 molecule, the separation between the nuclei of the two atoms is about 1.27 A (1A – 10-10 m). Find the approximate location of the CM of the molecule, given that the chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Answer:

If the hydrogen molecule is placed at the origin, the co-ordinate system for the centre of mass x is

Question 107.

A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is speed of the CM of the (trolley + child) system?

Answer:

Since there is no external forces acting on the system (trolley + child), the speed of centre of mass of the system remains the same.

Question 108.

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from the left end.

Answer:

Let T_{1} and T_{2} be the tension in the two strings.

For equilibrium of the bar in the horizontal direction.

Taking the moment about the point W,

sum of the clockwise moment = sum of the anticlockwise moment.

Question 109.

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is

1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each 1 back wheel.

Answer:

M= 1800 kg

distance between front and back axle = 1.8 m distance of front axle from the centre of gravity = 1.05 m

Let R_{1} and R_{2} be the reactional force exerted by the ground on front and back wheels. For translational equilibrium,

sum of the upward forces = sum of the down ward forces R_{1} + R_{2} =1800 × 9.8 = 17640 …(1)

For rotational equilibrium,

sum of the clockwise moments = sum of the anticlockwise moments

Question 110.

Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of

the sphere about any of its diameters to be \(\frac{2}{5}\) MR^{2} . Where M is the mass of the sphere and R is the radius of the sphere, (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be \(\frac{M R^{2}}{4}\) find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer:

(a) Using parallel axes theorem,

Moment of inertia about the tangent = Moment of inertia about the diameter + MR^{2}

I = \(\frac{2}{5}\) MR^{2}+ MR^{2}

I = \(\frac{7}{5}\) MR^{2}

(b) Using perpendicular axes theorem

I_{z} = I_{x} + I_{y}

I_{z} = \(\frac{M R^{2}}{4}+\frac{M R^{2}}{4}\)

I = \(\frac{M R^{2}}{2}\)

using parallel axes theorem,

Moment of inertia normal to the disc and passing through its edge = I_{z} + MR^{2}

I = \(\frac{M R^{2}}{2}\) + MR^{2} = \(\frac{3}{2}\) MR^{2}

Question 111.

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.

Answer:

Moment of inertia of the hollow cylinder about its axis I_{1} = MR^{2}

But t = αI

1/2

∴ α_{2} > α_{1}, the sphere will acquire greater speed than the cylinder after a given interval of time.

Question 112.

A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s“\ The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Answer:

Moment of inertia of a solid cylinder about its axis I = \(\frac{M R^{2}}{2}\)

= \(\frac{20 \times(0 \cdot 25)^{2}}{2}\)

I = 0.625 kg m^{2}

Kinetic energy of cylinder

K = \(-\frac{1}{2} I \omega^{2}\)

I = 0.625 kg m^{2}; co = 100 rad s^{-1}

.-. K =\(-\frac{1}{2}\) × 0.625 × (100)^{2}

K = 0.3125 × 10^{4}J = 3125 J

Angular momentum L = Iw

= 0.625 × 100 L = 62.5Js

Question 113.

(a) A child stands arbitrary the centre of a turntable with his two arms outstretched. The turn table is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia is \(\frac{2}{5}\) times the initial value? Assume that the turntable rotates without friction, (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Answer:

Initial moment of inertia I_{1} = I

Final moment of inertia I_{2} = \(\frac{2}{5}\) I

Initial angular momentum w_{1} = 40 rev/min using the law of conservation of momentum.

I_{1}w_{1} = I_{2}w_{2}

There is no supply of energy from outside. Therefore the increase in kinetic energy is due to internal energy of the child.

Question 114.

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the linear acceleration of the rope? Assume that there is no slipping.

Answer:

Moment of inertia of the hollow cylinder

I = MR^{2} = 3 × 0.42 = 0.48 kg m^{2}

torque acting on the cylinder

force = force × distance

T = FR = 30 × 0.4

= 12 Nm

Further

torque t = I α

12 = 0.48a

α = \(\frac{12}{0+48}\) = 25 rad s^{-2}

Linear acceleration a = Rα

= 0.4 × 25 = 10 ms^{-2}

Question 115.

To maintain a rotor at a uniform angular speed of 200 rad s^{-1} an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? Assume that the engine is 100% efficient.

Answer:

w = 200 rad s^{-1} t = 180 Nm

Power required by the engine P = tw

= 180 × 200

= 36000 W

= 30 × 10^{3} W = 30kW

Question 116.

From a uniform disc of radius R, a circular hole of radius \(\frac{R}{2}\) is cut out. The centre of the original disc.

Locate the centre of gravity of the resulting flat body.

Answer:

Let M be the mass of the disc of radius, then mass per unit area of the disc

When a circular hole of radius \(\frac{R}{2}\) is cut out, mass of the portion removed

Let C be the centre of mass of the disc of mass M and C be the centre of mass of the portion removed from the disc.

When the circular disc is removed, the remaining portion is considered to be two mass M concentrated at C (origin) and M’ concentrated at C’.

If X is the point through which the centre of mass of the remaining portion of the disc shifts C.

The negative sign indicates that the centre of mass shift by a distance \(\frac{R}{6}\) to the left of the point C.

Question 117.

A metre stick is balanced on a knife edge at its centre. When two coins of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm what is the mass of the meter stick?

Answer:

Using the principle of moments

sum of the clockwise moments = sum of the anticlockwise moments

W(5) = 10 × 33

w = \(\frac{10 \times 33}{5}\)

= 66 g

∴ mass of the meter stick is 66 g.

Question 118.

A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination, (a) Will it reach the bottom with the same speed in each case (b) Will it take longer to roll down on plane than the other? (c) if so, which one and why?

Answer:

Using the law of conservation of energy

Total kinetic energy at the bottom of the plane = Total potential energy at the maximum height

∵ h is same for both the planes, the solid sphere will reach the ground with the same speed in both the planes.

(b) and (c) the acceleration V for an object rolling down is given by

Question 119.

A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that it centre of mass has a

speed of 20 cm ^{-1}. How much work has to done to stop it?

Answer:

Total energy of the hoop is sum of kinetic energy due to translation motion and rotational motion

Question 120.

=The oxygen iolecule has a mass of 5.30 × 10^{26}kg and a moment of inertia of 1.94 × 10 kg m^{2} about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 ms’ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule

Answer:

Moment of inertia of the oxygen molecule

I = 2 (moment of inertia of each oxygen atom)

Given kinetic energy of rotation = \(\frac{2}{3}\) (kinetic energy of translation)