8 Rotation
V1
rks pyks 'kq: djrs gS ,d ,slk chapter ftlls cPpks dks
lcls T;knk Mj yxrk gS... bcz this chapter says V2
Rigid body
m1
q1 q2
A V1 cos q1 B V2 cos q2 m2
Not Rigid body
Component of velocity along line joining A & B is same.
In a Rigid Body → V2cosq2 = V1cosq1
Types of Motion of a Rigid Body
HkkbZ lp cksyw rks ;s chapter cgqr gh etsnkj gS cl blls
Mjuk ugh gS... eSus youtube channel JEE Wallah ij
rotation dh 12 hr non-stop ,d class ys j[kh gS...
For more better understanding you can watch it
(or search saleem sir rotation motion on youtube)
Mains vkSj Advance nksuks osQ pov ls ;s cgqr important
chapter gS vkSj ftruk content eS ;g¡k cover dj jgk 1. Pure translation: Each points have same velocity.
g¡w that sufficient for you to build your concept... 2. Fixed axis rotation: Body rotates about a fixed axis.
almost gj izdkj osQ question eSus ;g¡k cover dj fn, gS 3. CRTM: Body rotates as well as translate.
pyks start djrs gSA
w
V2 = r 2 w
r1 r2
90-q2 q2
1
-q
90
RIGID BODY A q1 B
A rigid body is an object where the gap between any V1 = r1w
two points on the body do not change w.r.t time. V1 cos q1 = V2 cos q2
V2 sin q2 V1 sin q1
A wB/A
AB
Not Rigid body
B V2 sin q2 V1 sin q1
Rigid body m
r1 sin q1 r2 sin q2
Here, Distance AB not r2 w sin q2 r1 w sin q1 r
w
not changing with time. r1 sin q1 r2 sin q2
,Hence, we can say that in rigid body w of any point Q. Find moment of inertia of following system of
about any general point is same. particles about x-axis, y-axis and z-axis.
y 4kg
1kg
Rotation 'kq: djus ls igys ges ,d (3, 4)
(–6, 8) (0, 4) 3kg
article i<+uk iM+sxk Moment of inertia
(MOI) ftlus vkidks dkiQh maths ns[kus
dks feysxh.... tSls igys inertia gksrk Fkk 2kg
(tendency of an object to resist x
(3, 0)
changes in its state of motion) mlh
izdkj rotation esa property to oppose 5kg
to change in rotation motion is
called MOI. (–4, –4)
Sol. Ix =0 + 4 × 42 + 3 × 42 + 5 × 42 + 1 × 82 = 256
Iy =2 × 32 + 4 × 32 + 0 + 5 × 42 + 1 × 62 = 170
MOMENT OF INERTIA (MOI)
Iz =2 × 32 + 4 × 52 + 3 × 42 + 1 × 102 + 5 × (4 2)2
= 18 + 100 + 48 + 100 + 160
It is a property of a body by virtue of which it opposes
= 426
to change in its rotation motion.
Also, Iz = Ix + Iy = 256 + 170 = 426
For a point mass
MOI of a point mass about vjs ;s ns[kk rqeus.... ns[kk dh ugh fd Iz = Ix + Iy vkx;k....
an axis is calculated as m
2 ar
I = mr where r is the ⊥
distance from axis. r
For system of discrete
particle (point mass)
D;ksa vk;k D;k ges'kk vk,xk.... ?
I= mi ri 2
For a continuous body JEE Mains Favorite Question
I ∫ dm . x
2
Q. Three particles of mass m each are kept at an
Q. Find moment of inertia of following system of equilateral triangle of side L. Find MOI about x,
particles about y-axis. y and z axis.
y L L 3
m ,
Iy 2 2
vcs ;g¡k 5
kg osQ MOI
5kg 0 2kg osQ fy, L
x L
(3, 0) minus er
(–4, 0)
yxk nsuk ge
MOI fudky
jgs gS uk dh 60°
centre of m
Sol. Iy = 2 × 3 + 5 × 42 2
mass Ix
(0, 0) m (L, 0)
= 18 + 80 = 98
195
Rotation
, 2 L
L 3 3mL2
Sol. Ix = 0 + 0 + m
2 4
I= ∫ ldx.x2
0
L
2
L 5mL ∫
2
I= l0x.x2dx
Iy = 0 + mL2 + m
2 4 0
l 0 L4
2 2
=
L L 3 4
Iz = 0 + mL2 + m = 2mL2
2 2
Q. Find moment of inertia of uniform rod about
axis as shown in figure passing through its
Q. Find moment of inertia of uniform rod about axis centre
as shown in figure passing through one of its end I
m
(l = = linear mass density). (Uniform m, L)
L
m
l Centre
L
(m, L, Uniform) mL2
I=
m 12
Sol. We can consider the above rod as two rods of
Axis m L
mass and length .
2 2
2
m L
Sol. dm = ldx w 2 2 mL2
2 I 2
dI = dmx dm 3 12
∫ dI = ∫ dm.x2 HkkbZ bls integration ls Hkh dj ldrs gS... rqe yksx
x dx
L try djks
I= ∫ ldx.x2 $ L /2 L /2
0 Hint: I ! % dm " x2 ! % &dx " x2
L # L /2 # L /2
= l ∫ x .dx2
0 I=0
0 Axis
lL3 m L3 mL2 I=0
I
3 L 3 3 Moment of inertia of rod about the given axis will
be 0.
Q. Find moment of inertia of rod whose linear mass
density is given by l = lox about axis passing 0
through one end as shown in figure. Axis is perpendicular to plane
ml2
Moment of inertia in the given case will be I =
3
I
Q. Find MOI for following where a particle of
l = l0 x mass m is attached at the end of rod of mass m
and length L.
x dx I
(m, L) Point mass
m
Sol. dI = dmx2
mL2 4 2
∫ dI = ∫ dm.x2 I
3
mL2
3
mL
196 Physics P
W
, Q. A rod makes an angle q with the axis of rotation.
vxj uniform ring dh txg m mass dk non-
Find MOI about the given axis.
uniform ring gksrk ;k m mass dk arc/semi
I=?
ring gksrk rks Hkh I = mr2 vkrk lkspks lkspks
q
0 MOI of Hollow cylinder (m, R) about axis as shown
in figure.
m, L
Sol. I=? (m, L) uniform rod Hollow cylinder
R
x I = mR2
q dx
dm
r = x sin q
Q. Find MOI of uniform Disc of mass m and radius
R about axis perpendicular to the plane of disc
dI = dm.r2 (r: perpendicular distance)
and passing through centre.
∫ dI = ∫ ldx (x sin q) 2
Top view Side view
L w mR2
I
= lsin q ∫ x dx
2 2
2
0
I mR2
m 2 L 3 I
= sin q 2
L 3
ML2 sin2 q
=
3 m
Sol. Surface mass density s
pR2
Method-2: SKC
Area of thin strip of width dr
vxj rod osQ lkjs mass dks eS uhps mrkj y¡w
dA = 2pr . dr
rks sin q length dh ,d rod cu tk,xh
Mass of element = dm = s dA
m (L sin q)2
hence I dm = s dA Disc
3
dr (m, R)
MOI = dI = dm.r2
0 MOI of a uniform ring (m, R) about axis as shown r
dI = sdA.r2
in figure.
R
2 dm
Top view Side view ∫ dI ∫ s2pr.drr
Ring (m, R) w 0
R
s2p∫ r3dr
dm
0
Axis dr
m R4 2pr
.2p.
pR2 4
I ∫ (dm)R R2 ∫ dm mR2
2
mR2
2
I = mR2
197
Rotation
V1
rks pyks 'kq: djrs gS ,d ,slk chapter ftlls cPpks dks
lcls T;knk Mj yxrk gS... bcz this chapter says V2
Rigid body
m1
q1 q2
A V1 cos q1 B V2 cos q2 m2
Not Rigid body
Component of velocity along line joining A & B is same.
In a Rigid Body → V2cosq2 = V1cosq1
Types of Motion of a Rigid Body
HkkbZ lp cksyw rks ;s chapter cgqr gh etsnkj gS cl blls
Mjuk ugh gS... eSus youtube channel JEE Wallah ij
rotation dh 12 hr non-stop ,d class ys j[kh gS...
For more better understanding you can watch it
(or search saleem sir rotation motion on youtube)
Mains vkSj Advance nksuks osQ pov ls ;s cgqr important
chapter gS vkSj ftruk content eS ;g¡k cover dj jgk 1. Pure translation: Each points have same velocity.
g¡w that sufficient for you to build your concept... 2. Fixed axis rotation: Body rotates about a fixed axis.
almost gj izdkj osQ question eSus ;g¡k cover dj fn, gS 3. CRTM: Body rotates as well as translate.
pyks start djrs gSA
w
V2 = r 2 w
r1 r2
90-q2 q2
1
-q
90
RIGID BODY A q1 B
A rigid body is an object where the gap between any V1 = r1w
two points on the body do not change w.r.t time. V1 cos q1 = V2 cos q2
V2 sin q2 V1 sin q1
A wB/A
AB
Not Rigid body
B V2 sin q2 V1 sin q1
Rigid body m
r1 sin q1 r2 sin q2
Here, Distance AB not r2 w sin q2 r1 w sin q1 r
w
not changing with time. r1 sin q1 r2 sin q2
,Hence, we can say that in rigid body w of any point Q. Find moment of inertia of following system of
about any general point is same. particles about x-axis, y-axis and z-axis.
y 4kg
1kg
Rotation 'kq: djus ls igys ges ,d (3, 4)
(–6, 8) (0, 4) 3kg
article i<+uk iM+sxk Moment of inertia
(MOI) ftlus vkidks dkiQh maths ns[kus
dks feysxh.... tSls igys inertia gksrk Fkk 2kg
(tendency of an object to resist x
(3, 0)
changes in its state of motion) mlh
izdkj rotation esa property to oppose 5kg
to change in rotation motion is
called MOI. (–4, –4)
Sol. Ix =0 + 4 × 42 + 3 × 42 + 5 × 42 + 1 × 82 = 256
Iy =2 × 32 + 4 × 32 + 0 + 5 × 42 + 1 × 62 = 170
MOMENT OF INERTIA (MOI)
Iz =2 × 32 + 4 × 52 + 3 × 42 + 1 × 102 + 5 × (4 2)2
= 18 + 100 + 48 + 100 + 160
It is a property of a body by virtue of which it opposes
= 426
to change in its rotation motion.
Also, Iz = Ix + Iy = 256 + 170 = 426
For a point mass
MOI of a point mass about vjs ;s ns[kk rqeus.... ns[kk dh ugh fd Iz = Ix + Iy vkx;k....
an axis is calculated as m
2 ar
I = mr where r is the ⊥
distance from axis. r
For system of discrete
particle (point mass)
D;ksa vk;k D;k ges'kk vk,xk.... ?
I= mi ri 2
For a continuous body JEE Mains Favorite Question
I ∫ dm . x
2
Q. Three particles of mass m each are kept at an
Q. Find moment of inertia of following system of equilateral triangle of side L. Find MOI about x,
particles about y-axis. y and z axis.
y L L 3
m ,
Iy 2 2
vcs ;g¡k 5
kg osQ MOI
5kg 0 2kg osQ fy, L
x L
(3, 0) minus er
(–4, 0)
yxk nsuk ge
MOI fudky
jgs gS uk dh 60°
centre of m
Sol. Iy = 2 × 3 + 5 × 42 2
mass Ix
(0, 0) m (L, 0)
= 18 + 80 = 98
195
Rotation
, 2 L
L 3 3mL2
Sol. Ix = 0 + 0 + m
2 4
I= ∫ ldx.x2
0
L
2
L 5mL ∫
2
I= l0x.x2dx
Iy = 0 + mL2 + m
2 4 0
l 0 L4
2 2
=
L L 3 4
Iz = 0 + mL2 + m = 2mL2
2 2
Q. Find moment of inertia of uniform rod about
axis as shown in figure passing through its
Q. Find moment of inertia of uniform rod about axis centre
as shown in figure passing through one of its end I
m
(l = = linear mass density). (Uniform m, L)
L
m
l Centre
L
(m, L, Uniform) mL2
I=
m 12
Sol. We can consider the above rod as two rods of
Axis m L
mass and length .
2 2
2
m L
Sol. dm = ldx w 2 2 mL2
2 I 2
dI = dmx dm 3 12
∫ dI = ∫ dm.x2 HkkbZ bls integration ls Hkh dj ldrs gS... rqe yksx
x dx
L try djks
I= ∫ ldx.x2 $ L /2 L /2
0 Hint: I ! % dm " x2 ! % &dx " x2
L # L /2 # L /2
= l ∫ x .dx2
0 I=0
0 Axis
lL3 m L3 mL2 I=0
I
3 L 3 3 Moment of inertia of rod about the given axis will
be 0.
Q. Find moment of inertia of rod whose linear mass
density is given by l = lox about axis passing 0
through one end as shown in figure. Axis is perpendicular to plane
ml2
Moment of inertia in the given case will be I =
3
I
Q. Find MOI for following where a particle of
l = l0 x mass m is attached at the end of rod of mass m
and length L.
x dx I
(m, L) Point mass
m
Sol. dI = dmx2
mL2 4 2
∫ dI = ∫ dm.x2 I
3
mL2
3
mL
196 Physics P
W
, Q. A rod makes an angle q with the axis of rotation.
vxj uniform ring dh txg m mass dk non-
Find MOI about the given axis.
uniform ring gksrk ;k m mass dk arc/semi
I=?
ring gksrk rks Hkh I = mr2 vkrk lkspks lkspks
q
0 MOI of Hollow cylinder (m, R) about axis as shown
in figure.
m, L
Sol. I=? (m, L) uniform rod Hollow cylinder
R
x I = mR2
q dx
dm
r = x sin q
Q. Find MOI of uniform Disc of mass m and radius
R about axis perpendicular to the plane of disc
dI = dm.r2 (r: perpendicular distance)
and passing through centre.
∫ dI = ∫ ldx (x sin q) 2
Top view Side view
L w mR2
I
= lsin q ∫ x dx
2 2
2
0
I mR2
m 2 L 3 I
= sin q 2
L 3
ML2 sin2 q
=
3 m
Sol. Surface mass density s
pR2
Method-2: SKC
Area of thin strip of width dr
vxj rod osQ lkjs mass dks eS uhps mrkj y¡w
dA = 2pr . dr
rks sin q length dh ,d rod cu tk,xh
Mass of element = dm = s dA
m (L sin q)2
hence I dm = s dA Disc
3
dr (m, R)
MOI = dI = dm.r2
0 MOI of a uniform ring (m, R) about axis as shown r
dI = sdA.r2
in figure.
R
2 dm
Top view Side view ∫ dI ∫ s2pr.drr
Ring (m, R) w 0
R
s2p∫ r3dr
dm
0
Axis dr
m R4 2pr
.2p.
pR2 4
I ∫ (dm)R R2 ∫ dm mR2
2
mR2
2
I = mR2
197
Rotation
