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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 xaxis along the incline and the yaxis 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 $$Mg\sin\theta  kxf=Ma$$
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.
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