Engineering Mechanics

Chapter 31 : Kinetics of Motion of Rotation „„„„„ 645

Let m= Mass of the hanging body A,
P= Tension in the string,
M= Mass of the rolling body B,
I= Moment of inertia of the rolling body B,
k= Radius of gyration of the rolling body B,
r= Radius of the rolling body B,
μ= Coefficient of friction between the plane and the body,
a= Linear acceleration of the rolling body B, and
α= Angular acceleration of the rolling body B.
From the geometry of the motion, we find that the acceleration of the hanging body A will be
(2a). We know that the normal reaction on the horizontal plane for the body B,

R= Mg

and force of friction, F=μR = μΜg

Since the rolling body tends to roll towards right, therefore force of friction will act towards
left as shown in Fig. 31.14 (b). Let us introduce two equal and opposite forces (each equal to the
force of friction, F) through the centre of the rolling body as shown in Fig. 31.14. (b).

A little consideration will show, that these two forces will not effect the motion of the system.
Now the rolling body is subjected to the following forces :

1. A force equal to P – F (acting towards right)


2. A couple whose moment is equal to F × r (responsible for rolling the body).
   First of all, consider motion of the hanging body A, which is coming down. We know that
   the forces acting on it are mg (downwards) and P (upwards). As the body is moving downwards,
   therefore resultant force

=mg – P ...(i)
Since the body is moving downwards with an acceleration of (2a), therefore force acting on
this body

=m × 2 a = 2 ma ...(ii)

Equating equations (i) and (ii),

mg – P=2 ma

or P=mg – 2 ma ...(iii)

Now consider the linear motion (neglecting rolling for the time being) of the body B, on the
rough horizontal plane due to the force (P – F) acting on it. Since the body is moving with an
acceleration (a) towards right, therefore force acting on it.

=Ma (towards right)
We know that the force acting towards right is responsible for this motion. Therefore
P – F=Ma ...(iv)
Now consider the circular motion (i.e. rolling) of the rolling body due to the couple (equal
to P × r) which is responsible for rolling the body. We know that linear acceleration of the body is
equal to its angular acceleration.

∴ a=rα

and torque acting on the body,

T=Iα