1
$\begingroup$

enter image description here

Consider the system in the above diagram. The surface has enough friction to initiate pure rolling.

Questions: 1)If we stretch the spring from the equilibrium by about some distance x. The spring will gain potential energy. When the cylinder is released from this state the spring will do some work and increase the translational and rotational kinetic energy of the cylinder. I am able to understand the increase in translational KE. However, the increase in rotational KE doesn’t seem right to me since the spring applies a force on the center of the cylinder and therefor has 0 torque about the rotational axis (Passing through center of mass). Then why does the rotational kinetic energy of the cylinder increase?

(EDIT: New Question)

2)If the point of application of force changes will the rotational work done change ? Consider the case of a rolling ball on a rough surface. The point of application of friction (Point of contact) keeps changing however friction still does some rotational work. If friction did not displace the initial point of contact by an angle how can it still do work.

$\endgroup$
5
  • $\begingroup$ Hint: Look at the second sentence of your post. $\endgroup$ Sep 29 '19 at 17:55
  • $\begingroup$ I don’t get it.....does it have something to with friction ? $\endgroup$ Sep 29 '19 at 18:01
  • $\begingroup$ But isn’t friction a min conservative force. So why is energy conservation even valid ? $\endgroup$ Sep 29 '19 at 18:02
  • $\begingroup$ Yes. It has everything to do with it actually :) $\endgroup$ Sep 29 '19 at 18:02
  • $\begingroup$ Friction still does no net work on the cylinder. But friction is what causes the cylinder to roll. $\endgroup$ Sep 29 '19 at 18:03
1
$\begingroup$

It is because of friction. Friction has a torque about the center of the cylinder, so this force causes the cylinder to rotate and gain rotational kinetic energy.

But isn’t friction a non-conservative force. So why is energy conservation even valid?

Friction still does no net work. If the cylinder of radius $R$ moves a horizontal distance $\Delta x$, then it rotates by an angle $\Delta\theta=\Delta x/R$. Therefore, the "translational work" done by the constant friction force $f$ is $$W_t=-f\cdot\Delta x$$ and the "rotational work" done by friction is $$W_r=\tau\cdot\Delta\theta=fR\cdot\frac{\Delta x}{R}=f\cdot\Delta x$$

Therefore, we have the net work done by friction as $$W=W_t+W_r=-f\cdot\Delta x+f\cdot\Delta x=0$$

So the overall change in the cylinder's total kinetic energy just comes from the spring force, but the friction force still plays a role here. No net work does not mean nothing happens :)

$\endgroup$
6
  • $\begingroup$ Okay so are you trying to say that friction is converting the translational KE to rotational KE without any loss. $\endgroup$ Sep 29 '19 at 18:15
  • $\begingroup$ Also isn’t the work done by torque equal to torque multiplied to the angular displacement of the point of contact. In this case isn’t the point of contact at rest? $\endgroup$ Sep 29 '19 at 18:18
  • $\begingroup$ @AdityaAhuja Kind of, but not really. Really kinetic energy is just kinetic energy. You can break it down to translational and rotational to simplify calculations, visualizations, etc., but it's all just energy of motion at the end of the day. $\endgroup$ Sep 29 '19 at 18:26
  • $\begingroup$ What about the Rotational work done by friction (Check my comment above) ? $\endgroup$ Sep 29 '19 at 18:32
  • $\begingroup$ @AdityaAhuja "Rotationally" it's not at rest. $\endgroup$ Sep 29 '19 at 19:00
1
$\begingroup$

However, the increase in rotational KE doesn’t seem right to me since the spring applies a force on the center of the cylinder and therefor has 0 torque about the rotational axis (Passing through center of mass). Then why does the rotational kinetic energy of the cylinder increases.

There are at least two ways to look at this. One is that the motion of a wheel rolling is actually not a rotation about the axle, but rather it is a rotation about the point of contact with the ground. The spring does provide a torque about that axis.

The other way to look at it is that the contact force with the ground provides a torque about the axle of the wheel. This seems straightforward, except that this force does no work and yet the rotational KE increases. All of the KE comes from the PE of the spring. The contact force provides no energy itself but does function as a constraint which splits the energy from the spring between rotational and linear kinetic energy.

$\endgroup$
6
  • $\begingroup$ If we analyse the system from the point of contact will the rotational kinetic energy change ? $\endgroup$ Sep 29 '19 at 18:28
  • $\begingroup$ Yes. And the spring provides that torque $\endgroup$
    – Dale
    Sep 29 '19 at 18:32
  • $\begingroup$ Will the total kinetic energy change? $\endgroup$ Sep 29 '19 at 18:33
  • $\begingroup$ Yes. As the spring does work on the wheel the total KE changes. $\endgroup$
    – Dale
    Sep 29 '19 at 18:49
  • $\begingroup$ Now let us say that we analyse the total KE at same time from both axes. Will they be equal ? $\endgroup$ Sep 29 '19 at 18:54
0
$\begingroup$

I am able to understand the increase in translational KE. However, the increase in rotational KE doesn’t seem right to me. The spring applies a force on the center of the cylinder and therefore has zero torque about the rotational axis (which passes through the center of mass). Why does the rotational kinetic energy of the cylinder increase?

I think the easiest way to look at this problem is to look at it as realistically as possible.

  • When the spring is stretched it applies a force on the center of mass of the cylinder.
  • If there was no friction, the cylinder would just slide, so you would get an increase in kinetic energy equal to the amount of work done by the spring over a distance. So: $$KE_{linear} = W = k\Delta x$$
  • But, since the cylinder is on a surface with friction, the cylinder will roll as well as translate. And, like @Aaron Stephens said, energy is energy, so the sum of the rotational kinetic energy plus the translational kinetic energy is still going to equal the work done with each increment of displacement. Therefore: $$KE_{linear}+KE_{angular} = k\Delta x$$

I think the confusion comes when trying to imagine that friction supplies a force that gives the cylinder rotational kinetic energy. Common sense dictates that the spring is the source of the force that causes the cylinder to roll and translate. It is not friction that supplies the force that causes the cylinder to roll.

This is obviously true (that the spring is the source of the force) since the spring is the entity in the problem that has potential energy stored within it. Static Friction doesn't supply energy to move anything. Rather, static friction is more like a bond that connects things together so they don't move.

(Note, if there is friction, and the two surfaces move in relation to each other, it's sliding friction, and that's a different/more complex problem where we have to compute the amount of heat lost too.)

Thus, the cylinder and the table are bonded together by static friction, so they can't slide over each other. So, when the spring pulls on the center of the cylinder, the table and cylinder don't slide. At that point (when the cylinder moves a tiny bit), the cylinder starts to translate and pick up some kinetic energy, but because the cylinder and table are bonded, the cylinder rolls incrementally also and picks up a little rotational kinetic energy too.

Therefore, the total amount of energy supplied by the spring contracting is divided into two compartments, 1) some rolling/angular kinetic energy, and 2) some linear/translational kinetic energy.

Rolling problems are made confusing by pretending (framing the problem in terms such) that friction supplies a motive force. Just look at the problem realistically. Identify the source of the force (which always comes from the conversion of one type of energy into another). Locate the lever arms and hinge points (the points around which motion rotates). The problem seems more intuitive when you model the problem to reflect reality.

$\endgroup$
0
$\begingroup$

That's my favourite type of question- entirely conceptual, with no maths!

The answer is that the cylinder undergoes a succession of tiny rotations about its successive contact points with the ground, which is equivalent to a lateral translation combined with a rotation about the cylinder's axis. That will come utterly clear if you follow this explanation.

Imagine replacing the cylinder with a tall rod balanced on one end. I hope you can easily imagine that if you apply a lateral force to its centre of mass it will topple over. It topples because the force creates a turning moment about its point of contact with the ground, with friction doing the job of restraining the foot of the rod. Note that the rod is now rotated- what was its vertical axis is now its horizontal one.

Now imagine replacing your circular cylinder with one with a pentagonal cross section. If you apply a lateral force the pentagon will roll over its contact point with the ground onto its next face. The effect is that the pentagon has shifted over and has been rotated about its axis, even though the cause was moment of force around a contact point with the ground.

Next imagine replacing the pentagonal cylinder with cylinders of increasingly many faces, a hexagonal one, then an octagonal one, and so on. In each case a force applied to their central axis will cause the cylinder to topple onto its next face.

The circular cylinder is simply the extreme case. If you think of it as having a very large number of very tiny faces, its rolling is actually toppling from one tiny face to the next.

$\endgroup$
1
  • $\begingroup$ @Macro Ocram even if I visualise rolling like this, I still cant see the point of contact being displaced at that particular instant....wont the point of contact be at rest and the cylinder perform pure rotational motion about an axis passing through the point of contact. How can we now say that friction will do work in this case ? ( there is a force but no displacement ) $\endgroup$ Oct 1 '19 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.