A uniform cylinder (mass `M`) of radius `R` is kept on an accelerating platform (mass `M`) as shown in figure. If the cylinder rolls withut slipping on the platform, determine the magnitude of acceleration of the centre of mass of the cylinder. Assuming th coefficient of friction `mu=0.40`, determinete maximum acceleration the platform may have without slip between the clyinder and the platform.

Explanation

Figure shows the free body diagram of the cylinder. We have assumed the angular acceleration of the cylinder to be clockwise. In case of pure rolling the contact point of cylinder with surface does not slide. Therefore, the acceleration of th point P must be equal to that of the platform.

But `a=a_(0)-Ralpha` ........i
Which yields `a_(0)=Ralpha+a`
the equation of motion are
`SigmaF_(x)=f=(M)a_(0)`
`Sigmatau=-fxxR=(1/2MR^(2))alpha`..........iii
Negative sign shows that the torque of friction is opposite to the assume `alpha`.
From eqn i and ii we get
`f=M(Ralpha)`..........iv
Sunstituting the expression for `f` in eqn ii we obtain
`Ralpha=-2/3a`
From eqn i `a_(0)=-2/3+=1/3a`
The maximum value of frictional force `f_("max")=muN` where `N=Mg,` from eqn ii
`0.40N=M(1/3a)`
`=0.0Mg=M(1/3a)`
`a=(0.40)(3g)=12m//s^2`
Which is the maximum acceleration the platform may have without slipping between the cylinder and the platform.