Point of Maximal Tension for Vertical Circular Motion

[vijayram]

Homework Statement

A heavy mass m is attached to thin wire and whirled in a vertical circle. Then the wire is most likely to break

A only when mass is at the lowest
B somewhere between lowest point and horizontal point
C only when mass is horizontal
D only when mass is at highest point

Homework Equations

T equal to mgcos(theta) + mv²/r

The Attempt at a Solution

The wire will break when tension is maximum it will be maximum at the bottom ie theta equal to zero so I marked a but answer given is between lowest and horizontal point[/B]

 

[PeroK]

The question is ambiguous. What does "most likely" mean, precisely? And what does "only" mean?

I would have gone for A, but the word "only" would have made me a bit suspicious.

It must be most likely when the tension is greatest, as you calculated.

 

[Truman I]

So the solution is saying that the wire could break somewhere along where the red here touches the outside of the circle.

Perhaps the best way to go about this problem is process of elimination. We understand that C and D are obviously wrong. D because it just doesn't make sense, and C because it is far more likely that the wire will break at some other point.That leaves A and B. Both have in common the fact that the wire could break at the lowest point. At this point, A could only be correct in certain circumstances. It would depend on the exact problem and numbers given, but there are only certain scenarios where an object being swung in a vertical circle could only break at the bottom of its swing. It is possible that the tension in the horizontal point could be enough to break the swing as well as anywhere else in the bottom half of that circle. That is why B is correct.

 

[TomHart]

This is really a poorly-worded question. When it says "only at its lowest", or "only when horizontal", or "only at its highest", it effectively means "exactly" at those angles (0, 90, 180, or 270°). There is zero tolerance on "only". Therefore, the wire is infinitely more likely to break within any non-zero range of angle centered around the lowest point than it would be at any exact angle.

It would be like setting up a lottery where, in order to win, you have to guess the "exact" random number between 0 and 1.
Customer: "Well, how many digits is that?"
Seller: "Well it's not a number of digits, you have to get it exactly right."
Customer: "Okay, I'll give it a shot."
Seller: "Do you want to pick your own random number or would you like a quick pick."
Customer: "Just give me a quick pick."
Seller: "This is amazing! I've never seen anyone come as close to winning as you got. You matched the first 1, 579,687 digits of the random number. Unfortunately, you missed the next digit. Want to try again?"

Also, the angle that it breaks depends on the rate of the angular acceleration. If I chose an infinitesimally small acceleration, I could guarantee that the wire would always break at the lowest point. And if you were so generous as to put a tolerance on "lowest", I could even perform the experimental verification is less than infinite time. :)

Sorry to go on so long about this, but this question just kind of bugged me. :)

 

[haruspex]

At this point, A could only be correct in certain circumstances.

I disagree. It asks for "most likely". The maximum likelihood of breaking is precisely at the lowest point.

 

[CWatters]

If the required answer is B then it's a poor question.

Is it moving at constant velocity?
Is the thin wire stiff enough that it can be radial?