- #1

Ujjwal Basumatary

- 17

- 0

## Homework Statement

A uniform cylinder of mass ##M## and radius ##R## is released from rest on a rough inclined surface of inclined surface of inclination ##\theta## with the horizontal as shown in the figure. As the cylinder rolls down the inclined surface, what is the maximum elongation in the spring of stiffness ##k##?

**2. The attempt at the solution**

Here is what I have thought about the problem:

We choose the origin to coincide with the equilibrium position of the spring, the x-axis along the incline and the y-axis perpendicular to it.

Let ##a## be the linear acceleration at an arbitrary instant of time. Also let ##\alpha## be the angular acceleration of the block at the same. Since there is no slipping, we have ##a=R\alpha##.

The equation for tangential motion is, therefore $$Mg\sin\theta - kx-f=Ma$$

For angular motion, $$fR=I\alpha$$

Where,

##x =## displacement from the origin

##g =## acceleration due to gravity

##f =## force of friction

##I =## moment of inertia of the cylinder through its central axis

Substituting the value of ##f## from the second equation into the first, we obtain $$mg\sin\theta - kx- \frac{I\alpha}{R} = Ma$$

But at the extreme position, linear as well as tangential acceleration is zero. Substituting, ##a=\alpha=0## in the above we get $$x=\frac{Mg\sin\theta}{k}$$

But this is the elongation from the origin, and the spring was initially at ##-x##. Therefore maximum elongation is given by$$x_{max} = \frac{2Mg\sin\theta}{k}$$

Am I correct in my reasoning? The answer is given to be this but I suspect I might have missed something in the middle as I am new to rotational mechanics. Any help would be appreciated. Thanks a lot.