# Cubical block on a cylinder

1. Nov 17, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
A cubical block of side L rests on a fixed cylindrical drum of radius R. Find the largest value of L for which the block is stable.

2. Relevant equations

3. The attempt at a solution
There is only one contact point between the cube and cylinder. There are three forces acting on the cube. The normal reaction from the cylinder, the friction and weight. The normal reaction and the friction pass through the contact point. Now how do I form the equations here?

Any help is appreciated. Thanks!

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• ###### kleppner 6.35.png
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2. Nov 17, 2013

### voko

It might be easier to find the potential energy due to gravity as a function of the block's tilt angle and then look for its minima.

3. Nov 17, 2013

### Pranav-Arora

I calculate potential energy from the line passing through the centre of cylinder.

$$U(\theta)=mg\left(R+\frac{L}{2}\right)(1-\cos \theta)$$
where $\theta$ is the angle rotated by the cylinder.

The minima is at $\theta=0$ and for this $\theta$, $U''(\theta)>0$ but solving gives me a negative answer which is certainly incorrect.

Last edited: Nov 17, 2013
4. Nov 17, 2013

### voko

You seem to assume that the block will slide, so that its center of mass is on the line passing through center of the drum and the point of contact. But the block will roll over the drum, and the point of contact will shift.

5. Nov 17, 2013

### Staff: Mentor

Consider the following drawing:

Does that suggest an approach? Hint: Consider the difference between the displacement of the point of contact on the face of the block to where the vertical through the center of mass passes through the same face (the line segment named Δ on the enlargement of the right).

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6. Nov 17, 2013

### Pranav-Arora

Nice drawing gneill!

Is $\Delta=R\theta-(L/2)\tan\theta$?

How do I continue with voko's approach of finding potential energy? I need the distance of CM of cube from centre of cylinder but I don't see how to apply the geometry here.

7. Nov 17, 2013

### Staff: Mentor

Yes indeed. The answer should be obvious from that...

In my drawing, extend the radius on the right by L/2 making it a total of R + L/2 in length. Then draw a perpendicular line from its end to the center of mass position. Treat the two as vectors to locate the center of mass.

8. Nov 17, 2013

### Pranav-Arora

How?

Do I have to apply the condition that $\Delta \geq 0$?

Please look at the attachment. Are you talking about the vectors shown? I still don't see how can I find the distance of CM from them. :(

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• ###### Fig1.png
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9. Nov 17, 2013

### Staff: Mentor

That's the idea. Consider the direction of the torque provided by the weight of the block about its contact point. Under what conditions is it a restoring force? When is it going to make the angle of tilt increase? When will it be exactly neutral?

You have the lengths of both, and they are perpendicular. You've even drawn in the hypotenuse!

10. Nov 17, 2013

### BruceW

I Love these kinds of problems. Problems related to something that you might just be absent-mindedly thinking about, after trying to balance some cereal on a football. Pranav, you always bring nice problems :)

11. Nov 17, 2013

### WannabeNewton

They're all from Kleppner and Kolenkow "An Introduction to Mechanics" (this particular one is from the chapter on rotational motion). Go get the book before I mail it to you myself!

12. Nov 17, 2013

### haruspex

Yes, bearing in mind that this is for arbitrarily small theta. You will need an approximation to get θ and tan θ comparable.

13. Nov 17, 2013

### BruceW

cool, I might check it out. It seems I've missed out on a real gem here :)

14. Nov 17, 2013

### haruspex

There is an ambiguity in the question. The interpretation of 'stable' so far is (I think) that it doesn't tilt at all, but it could be read as not rolling right off (assuming adequate friction).

15. Nov 17, 2013

### Staff: Mentor

I think the interpretation is that it's "stable" in the same sense that Lagrange Points are stable; A small perturbation induces a small orbit around the point rather than an escape trajectory.

So long as the cubical block is less than a certain size a small perturbation will result in oscillation rather than it tipping off of the cylinder. Larger than that critical size, any small perturbation will result in the angle increasing continuously -- i.e. falling off.

16. Nov 17, 2013

### haruspex

Yes, I agree, that's probably what's intended.
There's yet a third question one could ask: what's the largest L for which it is possible to place the block on the cylinder stably? What a rich source of puzzles.

17. Nov 18, 2013

### Pranav-Arora

Okay, I understand but looking over the sketch again, how did you find that angle between the perpendicular of length L/2 and vertical is $\theta$?

Just a guess, is it $R\theta$?

Thanks BruceW!

As WBN stated, some problems are from the book "Introduction to Mechanics by David Kleppner" and the other, which is much more interesting (sorry WBN, I prefer Irodov than Kleppner :P ) is this: https://www.amazon.com/Problems-General-Physics-I-Irodov/dp/8183552153

The above book is quite popular in India. You must check this and have a look at reviews on Amazon. :)

Last edited by a moderator: May 6, 2017
18. Nov 18, 2013

### Staff: Mentor

There are numerous right-angles joining the line segments, so anything tilted will make angle θ with respect to either the horizontal or the vertical. The trick is to determine which (w.r.t. horizontal or vertical). Fortunately the shape of the triangles gives a big hint, but even so you could follow the trail of perpendicular lines from the tilted radius vector and label the angles w.r.t. horizontal and vertical as you go. I've added a few annotations to the diagram -- see the thumbnail.

Yes, it must be because the two L/2 length line segments are parallel and the line you want is parallel to the bottom of the cube. The resulting figure is a rectangle.

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19. Nov 18, 2013

### voko

The cube is tangent to the drum at all times.

Your guess is right. Now try to explain it.

20. Nov 18, 2013

### Pranav-Arora

Nicely explained, thanks gneill!

Since $\Delta \geq 0$ and $\theta$ is very small, $L\geq 2R$.

Umm..I think I did not require it, I need the distance from the reference line to continue with the voko's approach. How do I find that?

What I get is $(L/2)/\cos(\theta)+\Delta \sin\theta+R-(x+\Delta \cos(\theta))$, is this correct?