Question

A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is (Fig.). Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

Answer 1

Let us consider the given diagram,

Frictional force (f) is acting in the opposite direction of F.

Now, let ‘a’ and α be the linear and angular acceleration respectively.

Let the acceleration of the centre of mass of disc be ‘a’, then

F-f = Ma .........(1)

where m is mass of disc.

Force of friction applies torque about centre.

The angular acceleration of the disc is α = a/R. (pure rolling)

Then τ = Iα or \(f\times R = (\frac{1}{2}MR^2)\times a\)

\(f \times R = (\frac{1}{2}MR^2)\frac{a}{R}\)

or Ma = 2f

F – f = 2f or F = 3f ..........(2)

from (1) and (2), we get

f ≤ μN

Thus, f = F/3.

Since there is no sliding.

or f ≤ μ mg

or F ≤ 3 μ Mg.