Question

A solid cylinder of mass, m (=2kg) is kept on a rough horizontal rug of mass, M (=4kg). If the coefficient of friction between the cylinder and rug is μ and that between the rug and floor is zero, find the acceleration of the cylinder and rug respectively. (g=10 m/(s\^2))
(Assume cylinder does not slip on the rug)

a) 7/3 and 35/3
b) 10/7 and 30/7
c) 8/11 and 32/11
d) None

Solution

The free body diagram of the rug and the cylinder are as follows:

a) The rug:

b) The cylinder:

The equation for motion for the rug will be:

F - f = MA

Substituting the values of m, M and g, we get:

20 - 20μ = 4A

or, A = 5 - 5μ

The equation of motion for the cylinder are:

For translational motion:

f = ma

Substituting the values of m and g, we get:

a = 10μ

For rotational motion:

r * f = I * α

(where r is the radius of the cylinder, I is the moment of inertia, and α is the angular acceleration.)

We know that the moment of inertia of a solid cylinder is,

I = (mr\^2)/2

Therefore, we get:

r * α = (2f)/m = f = 20μ

Since the cylinder is rolling over the rug without slipping, we can use the no-slip condition, i.e., the tangential acceleration at point P of the cylinder is equal to that at point Q of the rug.

Therefore,

r * α + a = A

Substituting the values from the above equations, we get:

20μ + 10μ = 5 - 5μ

Therefore, μ = 1/7

Substituting in the expressions for a and A, we get:

a = 10/7 , and

A = 30/7