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!!
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.
!!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)