Chapter 31 : Kinetics of Motion of Rotation „„„„„ 637

Equating equations (viii) and (ix),

(P 1 – P 2 ) r=Iα

(P 1 – P 2 ) r^2 =Iαr ...(Multiplying both sides by r)

=Ia ...(∴ a = rα)

∴ (P 1 – P 2 )= 2

Ia

r

Adding equations (iii) and (vi)

m 1 g – P 1 + P 2 – m 2 g=m 1 a + m 2 a

g (m 1 – m 2 ) – (P 1 – P 2 )=a (m 1 + m 2 )

Substituting the value of (P 1 – P 2 ) in the above equation,

(^122)
.
(– )–
Ia
gm m
r
=a (m 1 + m 2 )
g (m 1 – m 2 )= 2 () 12
I
amm
r
⎡⎤
⎢⎥++
⎣⎦
∴ a=^12
2 12
(– )
()
gm m
I
mm
r
⎡⎤++
⎢⎥⎣⎦
Notes: 1. The values of pulls (or tensions) in the two strings (P 1 and P 2 ) may now be found out
from the equations (iii) and (vi) i.e.
P 1 =m 1 g – m 1 a = m 1 (g – a)
and P 2 =m 2 a + m 2 g = m 2 (a + g)

2. There will be some torque acting on the pulley due to the difference of tension in the
   two strings, such that
   T 1 =(P 1 – P 2 ) r
   where r= Radius of the pulley.


3. The net torque acting on the pulley,
   T=T 2 – T 1
   where T 2 is the torque obtained from the relation (T = Iα)
   Example 31.10. Two bodies of masses 15 kg and 5 kg are attached to the two ends of a
   flexible rope, which is passed over a pulley of mean radius 200 mm having a mass of 10 kg and
   radius of gyration 150 mm. Find the acceleration of the masses and pulls on either side of the rope.

Solution. Given: Mass of the first body (m 1 ) = 15 kg; Mass of the second body (m 2 ) = 5 kg;
Radius of the pulley (r) = 200 mm = 0.2 m; Mass of the pulley (M) = 10 kg and radius of gyration
(k) = 150 mm = 0.15 m.

Acceleration of the masses

We know that the mass moment of inertia of the pulley,
I=Mk^2 = 10 (0.15)^2 = 0.225 kg-m^2
and acceleration of the masses,

a=

12 2

(^2122)
(– ) 9.8(15–5)
m/s
()0.225
(15 5)
(0.2)
gm m
I
mm
r

⎡⎤⎡ ⎤
⎢⎥⎢ ⎥++ ++
⎣⎦⎣⎦

98
5·625+ 20 = 3.82 m/s
(^2) Ans.