3. ROTATIONAL MOTION
1.
A circular disc of mass 10 kg and radius
0.2 m is set into rotation about an axis
passing through its centre and
perpendicular to its plane by applying
torque 10 Nm. Calculate the angular
velocity of the disc at the end of 6 s from
the rest.
Given :
τ
=
10 Nm
M
=
10 kg
R
=
0.2 m,
t
=
6s
=
0
ω1
To Find :
ω2
=
?
Formula :
i)
I
=
ii)
τ
=
iii) ω2
=
Solution :
Since M.I of disc,
I
∴
=
I
=
I
Now, τ
=
=
∴
α
=
∴
α
ω2
ω2
ω2
=
=
=
=
2.
1
MR2
2
Iα
ω1 + αt
1
MR2
2
Given :
D
=
∴
=
M
ω1
ω2
τ
To Find :
i)
ii)
Formula :
i)
∴
A solid sphere of diameter 25cm and
mass 25 kg rotates about an axis through
its centre. Calculate its moment of
inertia, if its angular velocity changes
from 2 rad/s to 12 rad/s in 5 second. Also
calculate the torque applied.
=
=
=
=
I
τ
=
=
=
0.25 m
?
?
M.I of sphere, I
=
2
MR2
5
Iα
2
MR2
5
I
=
I
=
I
=
2
 0.25 
× 25 × 

5
 2 
0.1562 kg m2
τ
=
Iα
τ
=
 ω2 – ω1 
I 

t


=
τ
3.
=
25
cm
2
25 kg
2 rad/s
12 rad/s
5s
ii)
τ
Solution :
1
× 10 × (0.2)2
2
0.2 kg m2
Iα
τ
10
=
I
0.2
50 rad/s
ω1 + αt
0 + 50 × 6
300 rad/sec
R
25 cm
=
2
ω1 – ω2 

∵ α =

t


 12 – 2 
0.1562 × 

 5 
0.3124 Nm
Calculate moment of inertia of a ring of
mass 500 g and radius 0.5 m about an
axis of rotation coinciding with its
diameter and tangent perpendicular to
its plane.
Given :
M
=
500 g
=
0.5 kg
R
=
0.5 m
Rotational Motion
MAHESH TUTORIALS SCIENCE
.. 14
?
?
i)
Id
=
ii)
IT
Solution :
=
∴
∴
4.
Id
=
Id
=
Id
IT
IT
IT
=
=
=
=
=
MR 2
2
2MR2
2
0.0625 kg m2
6.25 × 10–2 kg m2
2MR2
2 × 0.5 × (0.5)2
0.25 kg m2
2
R. Show that the
5
radius of gyration about a tangential
its diameter is
axis of rotation is
=
=
Kr
=
∴
Kd
=
Id + MR2
∴ MKT2
=
MK2d + MR2 ∵ I T = MK 2T 


 I = MK 2 
d 
 d
KT2
=
K2d + R2
2
T
K
=
 2 
R  + R2

5


KT2
=
2
R2 + R2
5
∴
KT
=
7 2
R
5
∴
KT
=
∴
2
2
A solid sphere has a radius ‘R’. If the
radius of gyration of this sphere about
Given :
M
Kd
=
MR 2
2
0.5 × ( 0.5 )
7
R.
5
mass of uniform solid sphere
Radius of gyration about
diameter
Radius of gyration about
tangent
2
R
5
=
A ballet dancer spins about a vertical axis
at 90 r.p.m with arms outstretched. With
the arms folded, the moment of inertia
about the same axis of rotation changes
to 75 %. Calculate the new speed of
rotation.
Given :
n1
=
90 r.p.m,
I1
=
M.I with arms out stretched
=
0.75I1
I2
To Find :
n2
=
?
Formula :
I 1 ω1
=
I 2 ω2
Solution :
I1 (2π
π n1)
=
I2 (2π
π n2)
[... ω = 2π
πn]
∴
n2
7
R
5
Proof :
From the theorem of parallel axes,
IO
=
I c + Mh2
Rotational Motion
7
R
5
5.
To Prove :
KT
[ . . . h = R]
IT
∴
To Find :
Id
=
IT
=
Formula :
=
 I1 
  n1
 I2 
=
I1
× 90
0.75I 1
=
∴
n2
=
100
× 90
75
120 r.p.m
MAHESH TUTORIALS SCIENCE
6.
.. 15
A torque of 400 Nm acting on a body Solution :
of mass 40 kg produces an angular
2
acceleration of 20 rad/s2. Calculate the
I1
=
MR21
5
moment of inertia and radius of gyration
of the body.
2
I2
=
MR22
Given :
5
τ
=
400 Nm
2
2
M
=
40 kg
 R1 
 R1 
I1
2
∴
=
=




α
=
20 rad/s
I2
 R2 
 2R 1 
To Find :
I
=
?
2
1
1
K
=
?
=
=
 
4
2
Formula :
i)
τ
=
Iα
I1
∴
=
1:4
I
I2
ii)
K
=
M
8.
A flywheel in the form of a disc, rotating
Solution :
about an axis passing through its centre
τ
400
and perpendicular to its plane, looses
I
=
=
α
20
100 J of energy, when slowing down
2
from 60 r.p.m to 30 r.p.m. Find its
I
= 20 kg m
moment of inertia about the same axis
I
and change in its angular momentum
Now, K =
M
Given :
n1
=
60 r.p.m.
20
1
∴
K
=
=
60
40
2
=
=
1 r.p.s,
60
∴
K
=
0.5
n2
=
30 r.p.m.
∴
K
=
0.707 m
30
1
=
=
r.p.s,
60
2
7.
If the radius of solid sphere is doubled
∆E =
100 J
by keeping its mass constant, compare
To Find :
the moment of inertia about any
i)
I
=
?
diameter.
ii)
∆L =
?
Given :
Formula :
R2
=
2R1
1
M
=
constant
i)
K.E. =
Iω2
To Find :
2
ii)
L
=
Iω
I1
=
?
Solution
:
I2
1
Formula :
i)
KE1 =
× I × (2πn1)2
2
2
I
=
MR2
= 2π2n21I
5
Similarly,
K.E2
=
2π2n22I
Rotational Motion
MAHESH TUTORIALS SCIENCE
.. 16
∴
∆E
∴
∆E
=
=
=
∴
I
=
=
=
=
∴
∴
∴
9.
. E
2
2 2
2
2
2
2
∆E

2 π n 22 – n 12 


2
100
2

2  1 
2 ( 3.14 )   – 12 
 2 

∆L
=
=
Formula :
L
=
Iω
= constant
Solution :
L′
=
L′ ′
′
′
I ω
=
I ′′ ω ′ ′
I′(ω2 – ω2 )
=
(I + I) ω′′
I × 2π (n2 – n1)
=
2I × 2πn′′
n2 – n1
=
2n ′′
240 – 120
2
n′ ′
–100
23
2 ( 3.14 )  
4
–200
3 ( 3.142 )
2
I
=
– 6.753 kg m2
Negative sign shows that energy is lost
I
=
6.759 kg m2
L =
Iω
L =
I (2πn)
(∵ ω = 2πn)
L =
2πIn
∆L =
L2 – L1
=
2π I (n2 – n1)
=
∴
– K.E1
(2π n – 2π2n21) I
2π (n – n21) I
K
1

2 × 3.14 × 6.753  – 1 
2

– 3.142 × 6.753
21.21 kg m2/s
=
n′ ′
=
60 rpm
10.
A uniform circular disc with its plane
horizontal is rotating about a vertical
axis passing through its centre at a speed
of 180 r.p.m. A small piece of wax of
mass 1.9 g falls vertically on the disc and
sticks to it at a distance of 25 cm from
the axis. If the speed of rotation is now
reduced by 60 r.p.m., calculate moment
of inertia of the disc.
Given :
n1
=
180 r.p.m
=
n2
=
=
m
h
Two wheels of moment of inertia 4 kg
m2 rotate side by side at the rate of
I1
120 rev/min and 240 rev/min and
respectively in the opposite directions.
I2
If now both the wheels are coupled by
means of weightless shaft so that both
To find :
the wheels now rotate with a common
I1
angular speed. Find the new speed of
Formula :
rotation.
I1ω1
Given :
2
I
=
4 kg m
I2
n1
=
120 rpm
n2
=
240 rpm
Solution :
To find :
I2
n′ ′ =
?
Rotational Motion
=
=
=
=
=
180
=
3 r.p.s,
60
(180 – 60) r.p.m
120
=
2 r.p.s,
60
1.9 g
=
1.9 × 10–3 kg,
r
25 cm =
0.25 m
M.I of the disc about vertical
axis
M.I of disc about same axis
with wax
=
?
=
I 2 ω2
=
I1 + Iwax
(Parallel axes theorem)
=
I1 + Iwax
MAHESH TUTORIALS SCIENCE
∴
∴
∴
I2
Consider,
I 1 ω1
I1 (2πn1)
I1n1
I1 (n1 – n2)
.. 17
I1 + mr2
=
I 2 ω2
(I1 + mr2) × (2πn2)
(I1 + mr2) n2
mr2n2
=
=
=
=
Ml 2
Ml 2
+
12
16
=
7
7
Ml2 =
(1) (1)2
48
48
0.1458 kg m2
=
∴
Io
=
K
=
K
=
2
∴
I1
=
mr n 2
n1 – n2
2
=
I1
=
1.9 × 10 – 3 × ( 0.25 ) × 2
3–2
2.375 × 10–4 kg m2
11.
A thin uniform rod of length 1 m and
mass 1 kg is rotating about an axis
passing through its centre and
perpendicular to its length. Calculate
moment of inertia and radius of gyration
of the rod about an axis passing through
a point mid way between the centre and
its edge, perpendicular to its length.
Given :
M
=
1 kg
l
=
1m
h
=
To find :
i)
I
ii)
K
Formula :
i)
I0
ii)
K
∴
?
?
=
Ic + Mh2
=
A homogenous rod XY of length L and
mass ‘M’ is pivoted at the centre ‘C’ such
that it can rotate freely in vertical plane.
Intially the rod is in the horizontal
position. A blob of wax of same mass ‘M’
that of the rod falls vertically with the
speed ‘V’ and sticks to the rod midway
between points C and Y. If the rod rotates
with angular speed ‘ω’ what will be
angular speed in terms of V and L ?
Solution :
ITotal =
Irod + Iwax
∴
I
M
Solution :
M. I of rod about an axis through its ∴
centre is given by
∴
0.1458
1
12.
l
4
=
=
I
=
M
0.3818 m
=
ML2
+ Mr2
12
=
ML2
ML2
+
12
4
L

∵ r = 
2

7ML2
48
Since angular momentum is conserved,
Initial angular = Final angular
momentum
momentum
ITotal =
L
MV   =
4
 7ML2

 48
Ic
=
Ml 2
12
∴
V
4
=
7L
ω
48
I0
=
Ic + Mh2
∴
ω
=
V 48
×
4 7L
I0
=
l
Ml 2
+M  
12
4
∴
ω
=
12 V
7 L
2

 ω

Rotational Motion