(^634) „„„„„ A Textbook of Engineering Mechanics
Example 31.6. A homogeneous solid cylinder of mass 100 kg and 1 m diameter, whose axis
is horizontal, rotates about its axis, in frictionless bearings under the action of a falling block of
mass 10 kg, which is carried by a thin rope wrapped around the cylinder.
What will be the angular velocity of the cylinder two seconds after the motion? Neglect the
weight of the rope.
Solution. Given: Mass of cylinder (M) = 100 kg; Diameter of the cylinder (D) = 1 m or
radius (r) = 0.5 m; Mass of the block (m) = 10 kg and time (t) = 2 s.
We know that linear acceleration of the solid cylinder,
a=
2 2 10 9.8 1.63 m/s 2
2 (2 10) 100
mg
mM
××

+×+
and angular acceleration, α=
1.63 3.26 rad/s 2
0.5
a
r

∴ Angular velocity of the cylinder 2 seconds after the motion.
ω=ω 0 + αt = 0 + (3.26 × 2) = 6.52 rad/s Ans.
Example 31.7. A body of mass 6 kg is suspended by a light rope wound round a solid disc
of 60 kg and diameter 50 cm, the other end of the rope being fixed to the periphery of the pulley.
Find (i) acceleration of the descending mass, (ii) pull in the rope, and (iii) velocity after the
mass has descended 15 m. Take g as 9.8 m/s^2.
Solution. Given: Mass of the body (m) = 6 kg; Mass of the solid disc (M) = 60 kg; *Diam-
eter of the disc (D) = 50 cm and distance (s) = 15 m.
(i) Acceleration of the descending mass
We know that acceleration of the descending mass,
a=
2269.81.63 m/s 2
2(26)60
mg
mM
××

+×+
Ans.
(ii) Pull in the rope
We know that pull in the rope,
P=
60 6 9.8
49 N
2(26)60
Mmg
mM
××

+×+
Ans.
(iii) Velocity after the mass has descended 15 m
Let v= Velocity after the mass has descended 15 m.
We know that v^2 =u^2 + 2 a.s = 0 + (2 × 1.63 × 15) = 48.9
∴ v= 7.0 m/s Ans.
Example 31.8. A wheel has a 5.4 m long string wrapped round its shaft. The string is
pulled with a constant force of 98 N and it is observed that the wheel is rotating at 3 revolutions per
second, when the string leaves the axle. Find the moment of inertia of the wheel about its axis.
Solution. Given: Length of the string (l) = 5.4 m; Force (P) = 98 N and angular velocity
(ω) = 3 r.p.s. = 6π rad/s.
Let I= Moment of inertia of the wheel about its axis
We know that work done in pulling the string
= Force × Distance = 98 × 5.4 = 529.2 N-m ...(i)

• Superfluous data