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Manasi

When a body rolls down, it has linear acceleration in downward direction. The friction, therefore, acts upward to counter sliding tendency as shown in the figure. This friction constitutes an anticlockwise torque providing the corresponding angular acceleration as required for maintaining the condition of rolling (if linear velocity is increasing, then angular velocity should also increase according to equation of accelerated rolling).

Rolling down an incline

Friction here plays a dual role :

1. It decelerates translational motion.

2. It accelerates rotational motion.

As the body rolls down, linear velocity increases with time such that its angular velocity also increases simultaneously in accordance with equation of rolling,

V_c =

R ........................(1)

The linear acceleration of the COM of the rolling body is equal to the component of acceleration due to gravity in x direction,

a_c = g sin

= g R ........................(2)

here, we select an appropriate pair of rectangular coordinates such that motion is along the positive direction of the x-coordinate. The various forces acting on the rolling disk are

Force of gravity, Mg, acting downward.

Normal force, N, perpendicular to the incline in y-direction.
Static friction,f_s, acting upward

Rolling down an incline

The force/ force components are acting in mutually perpendicular directions. As such, we can analyze motion in x-direction independently.

Thus, confining force analysis in x-direction, we apply Newton's law of motion for linear motion as :

F_x = Mg sin

- f_s = Ma_c.........................(3)

applying Newton's Law for rotation,

torque T = I

here the force due to gravity and normal force pass through the center of mass. As such, they do not constitute torque on the rolling disk. It is only the friction that applies torque on the disk, which is given as

T = f_s R = I

..........................(4)

also for rolling without sliding, a_c =

R ......................... (5)

combining 4 and 5,

f_s R = I

= I a_c/ R

=> f_s = I a_c/ R²

thus using this in 3,

Mg sin

- (I a_c/ R²) = Ma_c

=> a_c(M + I/ R²) = Mg sin

=> a_c= Mg sin

(M + I/ R²)