A plank of mass `M`, is placed on a smooth surface over which a cylinder of mas `m(=M)` annd radius `R=1m` is placed as shown in figure. Now the plane is pulled towards the right wilth an external force `F(=2Mg)`. If the cyinder does not slip over the surfcace of the plank, find the linear acceleration of he plank and the cylinder and the angular acceleration of the cylinder. (Take `g=10 ms^(2)`)

Explanation

Here we are give that cylinder does not sip over the plank surface, it is the case of pure rolling, we can se firction on cylinder in any direction. Here we choose towards the right. As friction is acting on the cylinder towards the right, it must be towards the left on the plank as shown in the force diagram of figure.

Let the plank move towards the right with an acceleration `a_(1)` the cylinder will experience a pseudoforce `ma_(1)` in the left direction, due to which it will roll towards the left with respect to the plank with an acceleration `a_(2)`. As we have used pseudoforce, `a_(2)` must be with respect to the plank. Let it angular accelration during rolling be `alpha`, we have `a_(2)=Ralpha`.
For tanslationa motion of the plank, we have
`F-f=Ma_(1)`.............i
For translational motion of the cylinder with respect to the plank, we have
`ma_(1)-f=ma_(2)` ........ii
For rotational motion of the cylinder with respect to the plank we have
`fR=Ialpha`
`fR+(1/2mR^2)(a_2/R)`
or `f=1/2ma_(2)`........iii
From eqn i and ii we get
`ma_(1)-1/2ma_(2)=ma_(2)`
`a_(1)=3/2a_(2)`.........iv
Using eqn i, iii and iv we get
`F-1q/2ma_(2)=3/2Ma_(2)`
`a_(2)=(2F)/(3M+m)=10m//s^(2)`
From eqn iv `a_(1)=(3F)/(3M+m)=15 m//s^(2)`
As we have already discused that the value of `a_(2)` is relative to the plank, the net acceleration of the cylinder will be given as `a_(1)-a_(2)`.
Hence, the acceleration of the cylinder is
`a_("cylinder")=a_(1)-a_(2)=15-10=5m//s^(2)`
The angular accelertionn of the cylinder is
`alpha=(a_(2))/R=10 rad//s^(2)`