A solid cylinder with base radius jis placed on top of an inclined plane of length land slope angle `alpha` (Fig). The cylinder rolls down without slipping. Find the speed of the centre of mass of the cylinder at the bottom of the plane, if the coefficient of rolling friction is k. Can rolling friction be neglected? Do the calculation for the following conditions: `l= 1 m, alpha = 30^@, r = 10 cm, k = 5 xx 10^(-4) m` . What would be the speed if, in the absence of friction, the cylinder slides down?

Answer

`2.7 m//s;3.4m//s`

Explanation

From the law of conservation of energy
`U- M_(tr)=K, " or " mg(l sin alpha+ r cos alpha)- T _("rol")l=(mv^2)/2+(Iomega^2)/2`
The force of rolling friction is `T_("rol") = kmg cos alpha//r` , the moment of inertia of a solid cylinder is `I = 1//2mr^2` , and the angular velocity is `omega=v//r` . We have
`mg(l sin alpha+r cos alpha)-(lkmgcos alpha)/r=(mv^2)/2+(mv^2)/4`
from which
`v=sqrt((4g)/2(l sin alpha+ r cos alpha -(lk cos alpha)/r))`
In principle the friction may not be neglected since in the absence of friction the disk will not roll down, but will slide down, and in this cage the kinetic energy of rolling should not be taken into account. But if the friction is small enough, the work of the force of friction may be neglected. The necessary condition for this is
`(lk)/rltltl tan alpha+r`
For the numerical example contained in the problem we have
`(lk)/r=(1xx5 xx10^(-4))/(10^(-1))=5 xx10^(-3), l tan alpha +r(sqrt3)/3+0.1=0.68`
i.e. the work of the force of rolling friction may be neglected in the calculation
When the cylinder slides down
`mg(l sin alpha+ r cos alpha)=1/2mv^2,v = sqrt(2g (l sin alpha + r cos alpha))`