Engineering Mechanics

Chapter 31 : Kinetics of Motion of Rotation „„„„„ 639

Let P= Tension in the rope
a= Linear acceleration of the train, and
α= Angular acceleration of the drum.
First of all, consider the motion of the cage, which is being pulled by the rope due to the
torque on the drum.

Fig. 31.11.

We know that component of the weight of the cage along the plane

=

1

sinθ= 6 × 9.8 × 1.96 kN

30

Mg =

As the cage loaded with cars is moving upwards, therefore resultant force along the plane
=P – 1.96 ...(i)
Since the cage is moving with an acceleration of (a), therefore force acting on it
= Ma = 6 a ...(ii)
Equating equations (i) and (ii),
P – 1.96 = 6 a ...(iii)
Now consider motion of the drum, which is rotating about its axis due to torque. We know
that moment of inertia of the drum,
I=mk^2 = 1 × (0.45)^2 = 0.2 kg-m^2
We know that linear acceleration of the train is equal to the angular acceleration of the drum.

∴ a=rα = 0.5 α or α =

0·5

a

= 2a

Now accelerating torque,

T 1 =Iα = 0.2 × 2 a = 0.4 a kN-m

and torque due to tension in the rope,
T 2 =P × r = P × 0.5 = 0.5 P
∴ Total torque, T=T 1 + T 2 = 0.4 a + 0.5 P
We know that this total torque is equal to the torque applied to the drum.
3 = 0.4 a + 0.5 P

or P=

3–0.4

6–0.8

0.5

a

= a

Substituting this value of P in equation (iii),

6 – 0.8 a – 1.96 = 6 a or 6.8 a = 4.04

∴ a=

4.04=0.59m/s 2

6.8

Now substituting this value of a in equation (iii),

P – 1.96 = 6 × 0.59 = 3.54

or P= 3.54 + 1.96 = 5.5 kN