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Rotational Mechanics question with spring

  • #1

Homework Statement


Untitled.png

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.
 

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  • #2
Orodruin
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It is a bit unclear to me what you mean by the spring's equilibrium position. You could mean one of two things:
  • Where the system is at equilibrium.
  • Where the spring is unstretched.
Your argumentation in the end seems to suggest the first, but your argumentation for the location of the zero-acceleration point seems to suggest the latter.

That being said, you could also just do an energy argument, which will avoid any rotational involvement altogether.
 
  • #3
It is a bit unclear to me what you mean by the spring's equilibrium position. You could mean one of two things:
  • Where the system is at equilibrium.
  • Where the spring is unstretched.
Your argumentation in the end seems to suggest the first, but your argumentation for the location of the zero-acceleration point seems to suggest the latter.

That being said, you could also just do an energy argument, which will avoid any rotational involvement altogether.
I'm sorry for being unclear. I meant the position where the spring is unstretched.
 
  • #4
Orodruin
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Then that is where the contraption was released so
But this is the elongation from the origin, and the spring was initially at −x.
is not true. The spring was initially at ##x = 0##. However, the point you have computed is not that where the system changes direction, it is where it has zero acceleration.
 
  • #5
Then that is where the contraption was released so

is not true. The spring was initially at ##x = 0##. However, the point you have computed is not that where the system changes direction, it is where it has zero acceleration.
I see that flaw. Thanks a lot for pointing it out. I found a better and neater solution.

The decrease in gravitional potential energy is equal to the increase in spring potential energy and hence we have $$mgx\sin\theta = \frac{1}{2}kx^2$$
Rearranging yields the answer. I hope this is correct :)
 

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