# Solid cylinder over a rug problem

**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!!

Therefore, the answer is b)