Question

(`a`) A rigid body of radius of gyration `k` and radius `R` rolls without slipping down a plane inclined at an angle `theta` with the horizontal. Calculate its acceleration and the frictional force acting on it.
(`b`) If the body be in the form of a disc and `theta=30^(@)`, what will be the acceleration and the frictional force acting on it.

Answer 1

(`a`) Due to component `mg sin theta`, linear velocity is increasing, for pure rolling (`v_(c.m.)=Romega`, at anytime), hence angular velocity should increase, for this friction should act upward to provide angular acceleration. Also at the point of contact `A`, relative acceleration should be zero. Since inclined plane is at rest, hence acceleration of sphere along the plane at `A` should be zero, i.e.
`a_(A)=a-Ralpha=0impliesa=Ralpha` (`i`)
Linear motion: `mg sintheta-f=ma` (`ii`)
Rotational motion: `tau_(0)=fR=I_(0)alpha=mk^(2)alpha` (`iii`)
`fR=mk^(2)(a)/(R)implies f=ma(K^(2))/(R^(2))` in (`ii`)
`mgsintheta-ma(k^(2))/(R^(2))=ma`
`a(1+(k^(2))/(R^(2)))=gsintheta`
`a=(gsintheta)/(1+(k^(2))/(R^(2)))`, `f=(mgsintheta)/(1+(R^(2))/(k^(2)))`
(`b`) For disc, `(k^(2))/(R^(2))=(1)/(2)`
`a=(gsin30^(@))/(1+(1)/(2))=(g)/(3)`
`f=(gsin30^(@))/(1+2)=(mg)/(6)`