Physics
Session
Rotational Mechanics - 5
Session Objectives
Session Objective
1. Rolling of a body – Description
2. Rolling of a body - Mathematical
expression
3. Instantaneous center of rotation
Rolling of a body
Rolling without slipping : Motion is
combined rotation and translation.
There is no relative motion between
body and surface
P
Q
V
surface
Point of contact has no relative
motion with surface
Rolling of a body
Friction (f) is responsible for rolling
without slipping
No work is done by f when body
rolls without slipping
P
f=0
slipping
1
KE  MV2
2
spinning
no translation
1 2
KE  I
2
f0
rolling
1
1
2
KE  MV  Icm2
2
2
Relationship between linear
& angular velocity
The point in contact with surface is
momentarily at rest. Linear velocity
is velocity of center of mass (o).
Linear distance moved by centre of mass = S
It is also the distance PQ along the
edge of circular body
P
O
r

P
r
Vcm
r
S
Q
VCM
Relationship between linear
& angular velocity
 S  r (r : radius of circular
cross section)
 vCM 
ds
d
r
 r
dt
dt
VCM  r
Dynamics of rolling
Rolling without slipping =
Linear motion of center of mass
+ circular motion about center of mass.
VCM
VCM
VCM
r
+
r
r
2VCM

O
r
VCM
P
V=0
Point P is at rest with respect to the surface at this instant.
It is the instantaneous center of rotation .
Kinetic Energy
When considered with respect to P
1
KE  Ip2
2
 KE 
But Ip  ICM  Mr2
1
1
ICM2  MV2CM
2
2
(
vCM  r  )
Q
r
O
ICM 
1 2 
 vCM  M  2 
2
r 

VCM
r
rp
P
Kinetic Energy with respect to
instantaneous axis of rotation
Velocity vQ of the point Q, with respect
to the instantaneous centre rotation
vp  vCM  r    rp  
and a p  rp  
So torque (external) = I
(I measured from P)
KE 
1 2
I and L  I
2
are valid about instantaneous
point of rotation.
r
Vp
VCM
O
r
rp
P
Class Test
Class Exercise - 1
A disc, a ring and a sphere of same radius
and same mass are placed on a rough
horizontal table so that they can roll over
the table. An equal impulsive force is
imparted to each through their centres of
mass and they start rolling and slipping.
When the slipping stops the linear velocity
is greatest for
(a) ring
(b) disc
(c) sphere
(d) All are equal
Solution
1 2 1 2 1 2
I 
KE  I  mv  v  m  
2
2
2 
r2 
Ring: I =
mr2
1 2
KE  v (2m)  mvr 2
2
1 2
Disc : I  mr
2
1 2
1  3
KE  v  m  m   mv d2
2 
2  4
2 2
Sphere : I  mr
5
1 2
2  7
KE  v  m  m  
mv s2
2 
5  10
For same mass and KE, sphere has maximum linear
velocity.
Hence, answer is (c).
Class Exercise - 2
A solid sphere rolls on a horizontal
plane without slipping, percentage
of KE which is rotational is
approximately
(a) 28
(b) 3
(c) 72
(d) 100
Solution
For a sphere,
1 2 1
I  mv 2
2
2
1 2 2 2
2
  mr   mv 
25

KE 
1
mv 2 (0.4  1)
2
0.4
 Rotational KE percentage 
 100%
1.4
 28%

Hence, answer is (a).
Class Exercise - 3
A hoop of radius r rolls over a horizontal
plane with constant velocity v without
slipping. The velocity of any point t
seconds after it passes the top position is
vt
(a) v cos
r
vt
(c) 2v sin
r
vt
(b) 2v cos
r
 vt 
(d) 2v cos  
 2r 

v
Solution

is constant.
A
2V
B
r

C
r

–
2
v´
–
2
r´
P
Speed at B = v’
 v '  r ', v  r
v


 1  cos  

 r 
 cos  


2 

Solution contd..

r '








r  r cos 
r(1  cos 




cos(CPB)

cos
2




2

r  1  2cos
 1
2

  2r cos  

2 
cos

2


2
t
 2v cos
2
vt
 2v cos
2r
 v '  2r cos
   t 
Hence, answer is (d).
Class Exercise - 4
A cord is wound round the circumference
of a wheel of radius r. The axis of the
wheel is horizontal and moment of
inertia about its axis is I . A weight mg
is attached to the end of the cord and
falls from rest. After falling through a
distance h, the angular velocity of the
wheel will be
2gh
(a) r
  mr
(c)
 2 mgh 


2
   2mr 
 2mgh
(b) r 
2
   mr
(d)
2gh



Solution
v = r
Loss in PE = Gain in KE

v
m
h
m
Solution contd..
1
2 1
 mgh  Iwheel  mv 2
2
2
1 v2 1
 I
 mv 2
2 r2 2
 2 1  I  mr 2  2
1 I
v
 
 m v  
2  r2
2  r2 




vr
2mgh
I  mr 2
Hence, answer is (b).
Class Exercise - 5
A mass M is supported by a massless
string wound round a uniform cylinder of
mass M and radius R . On releasing the
system from rest, the acceleration of
mass M is
(a) g
(c)
2
g
3
(b)
g
2
(d) depends on R
R
M
Solution
v = r
R
M
Loss in PE = Gain in KE
when M has dropped a height h.
M
Solution contd..
1
2 1 2
Mgh  Mv  I
2
2
2
1
1
1
v


 Mv 2   Mr 2 
2
22
 r2
3
gh  v 2
4
3v 2
h
4g

2
 v  0  2ah
v2 2
a
 g
2h 3
Hence, answer is (c).
Class Exercise - 6
A ring and a cylinder of the same mass
m and the same radius R are released
with the same velocity at the same time
on a flat horizontal surface such that
they start rolling immediately on
release towards a wall equidistant from
both. The rolling friction is finite. Which
of the two will reach the wall first?
Solution
For rolling,
KE 
1
1
mv2  I2
2
2
[v = r]
1
I 
2
 mv 1 
2
2
 mr 
As m and v are same, KE is same for both.
For the ring: I = mr2
 mv 2  KE  vr 
KE
m
Solution contd..
1 2
For the cylinder : I  mr
2
 KE 
3
mv2  vc 
4
4 KE
3 m
 vc  vr
So the cylinder will reach first.
Class Exercise - 7
A body of mass m and radius r is rolled
up on irregular surface with finite but
negligible friction so that it attains a
5 v2
maximum vertical height of
.
6 g
What is the moment of inertia of the
body and what may be its shape?
h
v
Solution
2
1
1
2 1 2
2 1 v
KE  mv  I  mv  I
2
2
2
2 r2
v


   r for rolling


1 2
I 
 v m 
2
2 
r 
As friction is negligible,
Loss in KE = Gain in PE
5 mgv 2
mgh 
6 g
5 mgv 2
1 2 
I 

 v m 1 
2
6 g
2
mr 

5
I
 I
3
mr 2
2 2
 I  mr
3
The body is probably a hollow
sphere.
Solution contd..
Hint/Formula:
1
I 
2
mv 1 
  mgh
2
2
 mr 
1 v2 
I  1 v2  2  5 v2
h
1  
1 


2
2 g  mr  2 g  3  6 g
Class exercise - 8
A cylinder AB of radius R, length L and
mass M has two cords A and B wound
around it. The ends of the cords are
attached to a fixed, horizontal support
and the cords are vertical. When the
cylinder is released, the cords unwind
and the cylinder moves down while
rotating. What is the angular
acceleration of the cylinder and the
tension in the cords as the cylinder falls?
A
B
R
L
Solution
The equations of motion:
Mg – 2T = Ma (Translational)
2TR = I  (Rotational)
a
MR2
I
and  
(Motion of cylinder is rotational)
2
R
Then 2TR 
 2T 
1
a 1
MR2  MRa
2
R 2
1
Ma
2
1
 Mg  Ma  Ma
2
a
2
2g
g 
3
3R
1
T  Mg
6
Class Exercise - 9
A disc of radius R is spin about its axis so
that its angular velocity is w. Then it is
slowly placed on a rough horizontal
surface of coefficient of friction m. How
long will the disc takes to stop? (Plane of
the disc is parallel to surface all through
the rotation.)
Solution
Take a concentric ring of radius r and
width dr (as one of a series of
concentric rings making up the disc).

r
dr
Its mass: dM 
M
R
2
 2rdr 
M
R
2
2rdr
Solution contd..
Torque on the ring due to friction:
d = dMrg
So total torque  on the disc opposing the motion:
R
R
2Mg 2
2
  d 
r dr  MgR
3
R2
0
0


2
 d 
I 
    MgR
3
 dt 
t
0
1

2
I

MR


2


3 R
3 R
  dt  
d


t


4

g
4 g
0

Thank you