Question

A uniform cylinder of mass `M` and radius `R` rolls without slipping, down a slope of angle `theta`. The cylinder is connected to a spring of spring constant `K` while the other end is connected to a rigid support at `P`?

The cylinder is released when the spring is unstretched. The maximum distance that the cylinder travels is?

Answer 1

Let the uniform cylinder of mass `M` and outer radius `R` starts from rest and moves down the incline of an angle `theta` with the horizontal.

Let the center of mass of the cylinder move through a maximum distance `x` as the cylinder rolls without slipping down the incline, which is equal to drop through a vertical distance `h` when its movement is stopped by the stretched spring.

Let `g` be acceleration due to gravity.
Mechanical potential energy of stretched spring `=1/2Kx^2`
Loss in Gravitational Potential energy `=mgh=mg(xsintheta)`

Due the stated motion of cylinder, there is no loss of energy to friction. Using law of Conservation of Energy,

`:.1/2Kx^2=Mgxsintheta`
`=>x=(2Mgsintheta)/K`