Chapter 31 : Kinetics of Motion of Rotation 647
31.21.MOTION OF A BODY ROLLING DOWN A ROUGH INCLINED PLANE
WITHOUT SLIPPING
Fig. 31.17. Body rolling down on a rough inclined plane without slipping.
Consider a circular body rolling down an inclined rough plane without slipping as shown in
Fig. 31.16 (a).
Let M= Mass of the body,
I= Moment of inertia of the body,
k= Radius of gyration of the body,
r= Radius of the body,
θ= Inclination of the plane with the horizontal,
μ= Coefficient of friction between the plane and the body,
a= Linear acceleration of the body, and
α= Angular acceleration of the body.
We know that the normal reaction,
R=Mg cos θ
force of friction, F=μR = μMg cos θ
Since the body tends to roll downwards, therefore the force of friction will act upwards as
shown in Fig. 31.16 (b). Let us introduce two equal and opposite forces (each equal to the force of
friction F), through the centre of the body as shown in Fig. 31.16 (b).
A little consideration will show, that these two forces will not affect the motion of the system.
Now the rolling body is subjected to the following two forces :
- A force equal to Mg sin θ – F (acting downwards)
- A couple whose moment is equal to F × r (responsible for rolling down the body).
First of all, consider downward motion (neglecting rolling for the time being) of the body due
to force (Mg sin θ – F) acting on it. Since the body is moving with an acceleration (a), therefore
force acting on it
=Ma
We know that the force acting on it is responsible for this motion. Therefore
Mg sin θ – F=Ma ...(i)
Now consider the circular motion (i.e., rolling) of the body due to the couple (equal to P × r)
which is responsible for the rolling of the body. We know that linear acceleration of the body is
equal to its angular acceleration.
∴ a=rαand torque on the body, T=Iα
Now equating the couple (responsible for rolling) and torque acting on the body
F × r=Iα
F × r^2 =Iαr ...(Multiplying both sides by r) (Q a = rα)
=Ia