Chapter 12 Rotational Motion Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes

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Introduction.

Translation is motion along a straight line but rotation is the motion of wheels, gears, motors, planets, the hands of a clock, the rotor of jet engines and the blades of

helicopters. First figure shows a skater gliding across the ice in a straight line with constant speed. Her motion is called translation but second figure shows her spinning at a constant rate about a vertical axis. Here motion is called rotation.

Up to now we have studied translatory motion of a point mass. In this chapter we will study the rotatory motion of rigid body about a fixed axis.

Rigid body : A rigid body is a body that can rotate with all the parts locked together and without any change in its
System : A collection of any number of particles interacting with one another and are under consideration during analysis of a situation are said to form a
Internal forces : All the forces exerted by various particles of the system on one another are called internal These forces are alone enable the particles to form a well defined system. Internal forces between two particles are mutual (equal and opposite).
External forces : To move or stop an object of finite size, we have to apply a force on the object from This force exerted on a given system is called an external force.

Centre of Mass.

Centre of mass of a system (body) is a point that moves as though all the mass were concentrated there and all external forces were applied there.

(1) Position vector of centre of mass for n particle system : If a system

consists of n particles of masses

m1, m2, m3 ……mn , whose positions vectors are

® ® ® ®

r1, r2, r3………… rn

respectively then position vector of centre of mass

® ® ® ®

® m1 r1 + m2 r2 + m3 r3 +…………….. mn rn

r = m + m + m +…………. m

3 n

Hence the centre of mass of n particles is a weighted average of the position vectors of n particles making up the system.

® ®

(2) Position vector of centre of mass for two particle system :

® = m1 r1 + m2 r2

and the centre of mass lies between the particles on the line joining them.

m1 + m2

® ®

® r + r

If two masses are equal i.e. m = m , then position vector of centre of mass = ¹ ²

1 2 r 2

(3) Important points about centre of mass

The position of centre of mass is independent of the co-ordinate system
The position of centre of mass depends upon the shape of the body and distribution of

Example : The centre of mass of a circular disc is within the material of the body while that of a circular ring is outside the material of the body.

In symmetrical bodies in which the distribution of mass is homogenous, the centre of mass coincides with the geometrical centre or centre of symmetry of the
Position of centre of mass for different bodies

The centre of mass changes its position only under the translatory There is no effect of rotatory motion on centre of mass of the body.
If the origin is at the centre of mass, then the sum of the moments of the masses of the system about the

centre of mass is zero i.e. S mi ri = 0 .

If a system of particles of masses m1 , m2 , m3 ,…… move with velocities v1 , v2 , v3 ,……

then the velocity of centre of mass vcm

= å mivi .

å m

If a system of particles of masses m1, m2, m3,…… move with accelerations a1 , a2 , a3 ,……

then the acceleration of centre of mass

Acm

= å miai

å m

If r is a position vector of centre of mass of a system

® ® æ ® ® ® ö

then velocity of centre of mass

vcm = d r

= d ç m₁ r₁ + m₂ r₂ + m₃ r 3 +…. ÷

dt dt ç

m1 + m2 + m3 +……….. ÷

® ® 2 ®

2 æ ® ® ö

Acceleration of centre of mass

Acm = d vcm = d r = d ç m₁ r₁ + m₂ r₂ +………… ÷

dt dt ² dt ² ç m₁ + m₂ + m₃ +…… ÷

è ø

® ® 2 ®

Force on a rigid body F =M Acm = M d r

dt ²

For an isolated system external force on the body is zero

® = d æ®

ö = 0 Þ ®

= constant .

F M dt ç v cm ÷

v cm

è ø

i.e., centre of mass of an isolated system moves with uniform velocity along a straight-line path.

Problem 1. The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is 1.1 Å. Given, mass of carbon atom is 12 a.m.u. and mass of oxygen atom is 16 a.m.u., calculate the position of the center of mass of the carbon monoxide molecule [Kerala (Engg.) 2001]

3 Å from the carbon atom (b) 1 Å from the oxygen atom

Solution : (c) Let carbon atom is at the origin and the oxygen atom is placed at x-axis

^® ˆ ˆ ^® ˆ ˆ

m₁ = 12 , m₂ = 16 ,

r 1 = 0i + 0 j and r 2 = 1.1i + 0 j

® ®

® m r + m r

16 ´ 1.1 ˆ

r = 1 1 2 2 =

m1 + m2

28 i

r = 0.63 i

i.e. 0.63 Å from carbon atom.

Problem 2. The velocities of three particles of masses 20g, 30g and 50 g are r r

respectively. The velocity

10i ,10 j, and 10k

of the centre of mass of the three particles is [EAMCET 2001]

r r r

(a)

2i + 3 j + 5k

(b)

10(i + j + k)

(c)

20i + 30 j + 5k

(d)

2i + 30 j + 50k

Solution : (a) Velocity of centre of mass vcm

= m1v1 + m2v2 + m3v3 m1 + m2 + m3

= 20 ´ 10ˆi + 30 ´ 10ˆj + 50 ´ 10kˆ = 2ˆi + 3ˆj + 5kˆ.

100

Problem 3. Masses

8, 2, 4, 2 kg

are placed at the corners A, B, C, D respectively of a square ABCD of diagonal

80 cm . The distance of centre of mass from A will be [MP PMT 1999]

(a)

20 cm

(b)

30 cm

(c)

40 cm

(d)

60 cm

Solution : (b) Let corner A of square ABCD is at the origin and the mass 8 kg is placed at this corner (given in problem)

Diagonal of square d = a

= 80 cm Þ a = 40

2cm

m₁ = 8kg,

m₂ = 2kg,

m₃ = 4kg,

m4 = 2kg

Let

r 1 , r 2 , r 3 , r 4

are the position vectors of respective masses

r 1 = 0ˆi + 0ˆj ,

r2 = aˆi + 0ˆj , r3 = aˆi + aˆj , r4

= 0ˆi + aˆj

From the formula of centre of mass

r = m1 r1 + m2 r2 + m3 r3 + m4 r4 m1 + m2 + m3 + m4

= 15

2i + 15

2ˆj

\ co-ordinates of centre of mass = (15

of the corner = (0, 0)

2,15

2) and co-ordination

From the formula of distance between two points (x1, y1) and (x2, y2)

distance = =

Problem 4. The coordinates of the positions of particles of mass 7, 4

and 10 gm

= 30cm

are (1, 5, – 3), (2, 5,7)

and (3, 3, – 1) cm

respectively. The position of the centre of mass of the system would be

(a)

æ- 15 , 85 , 1 ö cm

(b)

æ 15 ,- 85 , 1 ö cm

(c)

æ 15 , 85 ,- 1 ö cm

(d)

æ 15 , 85 , 7 ö cm

ç 17 ÷

ç 21 ÷

Solution: (c)

m1 = 7gm , m2 = 4 gm , m3 = 10gm and

r1 = (ˆi + 5ˆj – 3kˆ), r2 = (2i + 5 j + 7k),

r3 = (3ˆi + 3ˆj – kˆ)

Position vector of center mass

r = 7(ˆi + 5ˆj – 3kˆ) + 4(2ˆi + 5ˆj + 7kˆ) + 10(3ˆi + 3ˆj – kˆ) = (45ˆi + 85ˆj – 3kˆ)

7 + 4 + 10 21

Þ r = 15 ˆi + 85 ˆj – 1 kˆ . So coordinates of centre of mass é15 , 85 , – 1ù .

7 21 7

Angular Displacement.

êë 7

21 7 úû

It is the angle described by the position vector r about the axis of rotation.

Angular displacement (q ) = Linear displacement (s)

Radius (r)

Unit : radian
Dimension : [M ⁰ L⁰ T ⁰ ]

® ® ®

Vector form S = q ´ r

i.e., angular displacement is a vector quantity whose direction is given by right hand rule. It is also known as axial vector. For anti-clockwise sense of rotation direction of q is perpendicular to the plane, outward and along the axis of rotation and vice-versa.

2p radian = 360° = 1
If a body rotates about a fixed axis then all the particles will have same angular displacement (although linear displacement will differ from particle to particle in accordance with the distance of particles from the axis of rotation).

Angular Velocity.

The angular displacement per unit time is defined as angular velocity.

If a particle moves from P to Q in time Dt , w = Dq

where Dq

is the angular displacement.

Instantaneous angular velocity w = lim Dq

Dt ®0 Dt

= dq

Average angular velocity w _av

= total angular displacement

total time

= q ₂ – q₁

t – t

2 1

Unit : Radian/sec
Dimension : [M ⁰ L⁰T ^-1 ] which is same as that of

® ® ® ® ®

Vector form

v = w ´ r [where v = linear velocity, r = radius vector]

w is a axial vector, whose direction is normal to the rotational plane and its direction is given by right hand

screw rule.

(6) w = 2p

= 2pn

[where T = time period, n = frequency]

(7) The magnitude of an angular velocity is called the angular speed which is also represented by w .

Angular Acceleration.

The rate of change of angular velocity is defined as angular acceleration.

If particle has angular velocity w₁

at time t1 and angular velocity w ₂

at time t ₂

then,

® ®

Angular acceleration ® = w 2 – w 1

t 2 – t1

Dw d

d 2 ®

Instantaneous angular acceleration a = lim

Dt ®0 Dt

= w =

q .

dt ²

Unit : rad/sec ²
Dimension : [M ⁰ L⁰ T ^-2 ] .
If a = 0 , circular or rotational motion is said to be

Average angular acceleration a

= w ₂ – w₁ .

® ® ®

Relation between angular acceleration and linear acceleration a = a ´ r .
It is an axial vector whose direction is along the change in direction of angular velocity e. normal to the rotational plane, outward or inward along the axis of rotation (depends upon the sense of rotation).

Equations of Linear Motion and Rotational Motion.

Problem 5. The angular velocity of seconds hand of a watch will be [CPMT 2003]

(a)

p rad / sec

(b)

p rad / sec

(c)

60 p

rad / sec

(d)

30 p

rad / sec

Solution : (b) We know that second’s hand completes its revolution (2p) in 60 sec \ w = q

= 2p 60

= p rad/sec

Problem 6. The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the accelerator for 10 sec

it starts rotating at 4500 revolutions per minute. The angular acceleration of the wheel is [MP PET 2001]

(a) 30 radians/sec² (b) 1880 degrees/sec² (c) 40 radians/sec² (d) 1980 degrees/sec² Solution: (d) Angular acceleration (a) = rate of change of angular speed

2p æ 4500 – 1200 ö

3300

2p (n – n )

ç 60 ÷ 2p

360 degree

= 2 1

= è ø

= 60 ´

10 2p

sec ²

= 1980 degree / sec ² .

Problem 7. Angular displacement (q ) of a flywheel varies with time as q = at + bt ² + ct ³

then angular acceleration is given

by [BHU 2000]

(a)

a + 2bt – 3ct ²

(b)

2b – 6t

(c)

a + 2b – 6t

(d)

2b + 6ct

Solution: (d) Angular acceleration a = d2q = d2 (at + bt ² + ct³ ) = 2b + 6ct

dt² dt²

Problem 8. A wheel completes 2000 rotations in covering a distance of 9.5 km . The diameter of the wheel is [RPMT 1999]

(a)

1.5 m

(b)

1.5 cm

(c)

7.5 m

(d)

7.5 cm

Solution: (a) Distance covered by wheel in 1 rotation = 2pr = pD (Where D= 2r = diameter of wheel)

\ Distance covered in 2000 rotation = 2000 pD = 9.5 ´ 10³m (given)

\ D = 1.5 meter

Problem 9. A wheel is at rest. Its angular velocity increases uniformly and becomes 60 rad/sec after 5 sec. The total angular displacement is

(a)

600 rad

(b)

75 rad

(c)

300 rad

(d)

150 rad

Solution: (d) Angular acceleration a = w 2 – w1

= 60 – 0 = 12rad / sec ²

Now from q = w₁t + 1 a t ² = 0 + 1 (12)(5)² = 150 rad.

2 2

Problem 10. A wheel initially at rest, is rotated with a uniform angular acceleration. The wheel rotates through an angle q ₁

in first one second and through an additional angle q ₂ in the next one second. The ratio q 2 is

q₁

(a) 4 (b) 2 (c) 3 (d) 1

Solution: (c) Angular displacement in first one second q

= 1 a (1)² = a ……(i) [From q = w t + 1 a t ² ]

¹ 2 2 ¹ 2

Now again we will consider motion from the rest and angular displacement in total two seconds

q₁ + q ₂ = 1 a (2)² = 2a

Solving (i) and (ii) we get q = a

and q = 3a

……(ii)

\ q 2 = 3 .

¹ 2 ² 2 q₁

Problem 11. As a part of a maintenance inspection the compressor of a jet engine is made to spin according to the graph as shown. The number of revolutions made by the compressor during the test is

(a) 9000 (b) 16570 (c) 12750 (d) 11250

Solution: (d) Number of revolution = Area between the graph and time axis = Area of trapezium

= 1 ´(2.5 + 5)´ 3000 = 11250 revolution.

Problem 12. Figure shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If x and y be the distances travelled by A and B in the same time interval, then

x = 2y
x = y
y = 2x
None of these

Solution: (c) Linear displacement (S) = Radius (r) × Angular displacement (q)

\ S µ r

(if q =

constant)

Distance travelled by mass A (x) = Radius of pulley concerned with mass A (r) = 1

Þ y = 2x .

Distance travelled by mass B (y) Radius of pulley concerned with mass B (2r) 2

Problem 13. If the position vector of a particle is ® = (3ˆi + 4ˆj) meter and its angular velocity is ® = (ˆj + 2kˆ) rad/sec then its

r w

linear velocity is (in m/s)

(a)

(8ˆi – 6ˆj + 3kˆ)

(b)

(3ˆi + 6ˆj + 8kˆ)

ˆi ˆj

(c)

kˆ

– (3ˆi + 6ˆj + 6kˆ)

(d)

(6ˆi + 8ˆj + 3kˆ)

Solution: (a)

v = w ´ r

=(3ˆi + 4ˆj + 0kˆ) ´(0ˆi + ˆj + 2kˆ) = 3 4

0 1

0 = 8ˆi – 6ˆj + 3kˆ

Moment of Inertia.

Moment of inertia plays the same role in rotational motion as mass plays in linear motion. It is the property of a body due to which it opposes any change in its state of rest or of uniform rotation.

Moment of inertia of a particle I = mr ² ; where r is the perpendicular distance of particle from rotational
Moment of inertia of a body made up of number of particles (discrete distribution)

I = m r ² + m r ² + m r ² + …….

1 1 2 2 3 3

Moment of inertia of a continuous distribution of mass, treating the element of mass dm at position r as particle

dI = dmr ²

i.e.,

I = ò r ² dm

Dimension : [ML²T ⁰ ]
I. unit : kgm².
Moment of inertia depends on mass, distribution of mass and on the position of axis of

Moment of inertia does not depend on angular velocity, angular acceleration, torque, angular momentum and rotational kinetic
It is not a vector as direction (clockwise or anti-clockwise) is not to be specified and also not a scalar as it has different values in different Actually it is a tensor quantity.
In case of a hollow and solid body of same mass, radius and shape for a given axis, moment of inertia of hollow body is greater than that for the solid body because it depends upon the mass

Radius of Gyration.

Radius of gyration of a body about a given axis is the perpendicular distance of a point from the axis, where if whole mass of the body were concentrated, the body shall have the same moment of inertia as it has with the actual distribution of mass.

When square of radius of gyration is multiplied with the mass of the body gives the moment of inertia of the body about the given axis.

I = Mk ² or k = .

Here k is called radius of gyration.

From the formula of discrete distribution

I = mr12 + mr2 2 + mr3 2 +……….. + mr ²

If m₁ = m₂ = m₃ =….. = m then

I = m(r12 + r2 2 + r3 2 +………….. r ² )

From the definition of Radius of gyration,

I = Mk ²

By equating (i) and (ii)

Mk ² = m(r12 + r2 2 + r3 2 +…………….. + r ² )

……..(i)

……..(ii)

nmk ² = m(r12 + r2 2 + r3 2 +…………… + r ² )

[As

M = nm ]

\ k =

Hence radius of gyration of a body about a given axis is equal to root mean square distance of the constituent particles of the body from the given axis.

Radius of gyration

(k)

depends on shape and size of the body, position and configuration of the axis of

rotation, distribution of mass of the body w.r.t. the axis of rotation.

Radius of gyration (k) does not depends on the mass of

Dimension [M ⁰ L¹T ⁰ ].
I. unit : Meter.

Significance of radius of gyration : Through this concept a real body (particularly irregular) is replaced by a point mass for dealing its rotational

Example : In case of a disc rotating about an axis through its centre of mass and perpendicular to its plane

k = = = R

So instead of disc we can assume a point mass M at a distance (R / the rotational motion of the disc.

2) from the axis of rotation for dealing

Note : @ For a given body inertia is constant whereas moment of inertia is variable.

Theorem of Parallel Axes.

Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of the body about an

axis parallel to given axis and passing through centre of mass of the body I_g and body and a is the perpendicular distance between the two axes.

I = I g + Ma ²

Ma²

where M is the mass of the

Example: Moment of inertia of a disc about an axis through its centre and

perpendicular to the plane is

1 MR ² , so moment of inertia about an axis through its

tangent and perpendicular to the plane will be

I = Ig + Ma²

I = 1 MR² + MR²

\ I = 3 MR²

Theorem of Perpendicular Axes.

According to this theorem the sum of moment of inertia of a plane lamina about two mutually perpendicular axes lying in its plane is equal to its moment of inertia about an axis perpendicular to the plane of lamina and passing through the point of intersection of first two axes.

I z = I x + I y

Example : Moment of inertia of a disc about an axis through its centre of mass and perpendicular to its plane is

1 MR² , so if the disc is in x–y plane then by theorem of perpendicular axes

i.e.

I z = I x + Iy

MR² = 2ID

I = 1 MR²

D 4

[As ring is symmetrical body so

I x = Iy = I D ]

Note : @In case of symmetrical two-dimensional bodies as moment of inertia for all axes passing through the centre of mass and in the plane of body will be same so the two axes in the plane of body need not be perpendicular to each other.

Moment of Inertia of Two Point Masses About Their Centre of Mass.

Let m1

and m2

be two masses distant r from each-other and

r1 and r2

be the distances of their centre of

mass from m₁ and m₂

(1) r₁ + r₂ = r

(2) m1r1 = m2r2

respectively, then

= m2

m + m

r and

r2 =

m1 r

m₁

r₁

Centre of mass

r₂

m₂

m + m

1 2 1 2

I = m r ² + m r ²

I = é

1 1

m1m2

2 2

ù r ² = m r ²

[where m =

m1m2

is known as reduced mass

m < m

and m < m .]

êm + m ú

m + m ¹ ²

ë 1 2 û 1 2

In diatomic molecules like above

H 2 , HCl

etc. moment of inertia about their centre of mass is derived from

Analogy Between Tranlatory Motion and Rotational Motion.

Translatory motion Rotatory motion
Mass	(m)	Moment of Inertia	(I)
Linear momentum	P = mv	Angular Momentum	L =	Iw
	P = 2mE		L =	2IE
Force	F = ma	Torque	t =	Ia
Kinetic energy	E = 1 mv² 2		E =	1 Iw ² 2
	E = P 2			L2
	2m		E =	2I

Moment of Inertia of Some Standard Bodies About Different Axes.

Body Axis of

Rotation

Ring About an axis passing through

C.G. and perpendicular to its plane

Ring About its diameter

Figure Moment of inertia

MR ²

k k²/R²

R 1

Ring About a tangential axis in its own plane

3 3

2 R 2

Ring About a tangential axis perpendicular to its own plane

2MR ² 2

Disc About an axis passing through

C.G. and perpendicular to its plane

Disc About its Diameter

R 1

2 4

Disc About a tangential axis in its own plane

5 R 5

2 4

Disc About a tangential axis perpendicular to its own plane

3 3

2 R 2

Annular disc inner radius = R₁ and outer radius = R₂

Passing through – –

the centre and

perpendicular to

the plane

Annular disc Diameter – –

Annular disc Tangential and – –

Parallel to the diameter

Annular disc Tangential and – –

perpendicular to the plane

Solid cylinder About its own axis 1

R 2

Solid cylinder Tangential (Generator)

3 3

2 R 2

Solid cylinder About an axis

L2 R 2

passing through its

C.G. and perpendicular to its own axis

12 4

Solid cylinder About the diameter

L2 R 2

of one of faces of 3 + 4

the cylinder

Cylindrical shell

About its own axis

MR² R 1

Cylindrical shell Tangential

(Generator)

2MR² 2

Cylindrical shell About an axis

L2 R 2

passing through its

C.G. and perpendicular to its own axis

12 2

Cylindrical shell About the diameter

L2 R 2

of one of faces of 3 + 2

the cylinder

Hollow cylinder with inner radius = R₁ and outer radius = R₂

Axis of cylinder

R2 R1

Hollow cylinder with inner radius = R₁ and outer radius = R₂

Tangential

Solid Sphere About its diametric

axis

2 2

5 R 5

Solid sphere About a tangential

axis

7 7

5 R 5

Spherical shell About its diametric

axis

2 2

3 R 3

Spherical shell About a tangential

axis

5 5

3 R 3

Hollow sphere of About its diametric

inner radius R₁

and

axis

outer radius R₂

Hollow sphere Tangential

Long thin rod About on axis

passing through its centre of mass and

perpendicular to

the rod.

Long thin rod About an axis

ML²

Rectangular lamina of length l and breadth b

passing through its 3

edge and L

perpendicular to the rod

Passing through the

centre of mass and b

perpendicular to

the plane l

Rectangular lamina Tangential

perpendicular to the plane and at the mid-point of breadth

Rectangular lamina Tangential

perpendicular to the plane and at the mid-point of length

ii iii

M[b ² + t ² ]

Rectangular

Passing through

i (i)

parallelopiped length l, breadth b, thickness t

centre of mass and t

parallel to l

Length (x)

Rectangular parallelepiped length l, breath b, thickness t

Elliptical disc of semimajor axis = a and semiminor axis = b

breadth (z)
thickness (y)

Tangential and parallel to

length (x)
breadth (y)
thickness(z)

Passing through CM and perpendicular to the plane

M[l ² + t ² ]

i ii

iii

(ii)

(iii)

(i)

(ii)

(iii)

M [l² + b² + 3t²] 12

Solid cone of radius R and height h

Axis joining the vertex and centre of the base

Equilateral triangular lamina with side a

Passing through CM and perpendicular to the plane

Ma ²

Right angled triangular lamina of sides a, b, c

Along the edges

Mb ²

b c

Ma ²

a 1 6

2 3

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Introduction.

Centre of Mass.

(1) Position vector of centre of mass for n particle system : If a system

(2) Position vector of centre of mass for two particle system :

(3) Important points about centre of mass

Angular Displacement.

Angular Velocity.

Angular Acceleration.

Equations of Linear Motion and Rotational Motion.

Moment of Inertia.

Radius of Gyration.

Theorem of Parallel Axes.

Theorem of Perpendicular Axes.

Moment of Inertia of Two Point Masses About Their Centre of Mass.

Analogy Between Tranlatory Motion and Rotational Motion.

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