A cylinder is released from rest from the top of an incline of inclination `theta` and length 'L'. If the cylinder rolls without slipping. What will be its speed when it reaches the bottom?

Explanation

Let the mass of the cylinder be m and its radilus r. Suippose the linear spee of the cylinder when it reaches the bottom is v. As the cylinder rolls without slilpping, its angular speed about its axis is `omega=v/r`. The kinetic energy at the bottom will be
`K=1/2Iomega^2+1/2mv^2`
`=1/2(1/2mr^2)omega^2+1/2mv^2=1/4+1/2mv^2=3/4mv^2`
This should be equal to the loss of potential energy mgl sintheta`. Thus, `3/4mv^2=mglsintheta`
`or v=sqrt(4/3gl sin theta)`