M3 Vertical Circular Motion

With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..

Reply 1

Zacken

Original post by MathsAstronomy12

With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..

Tension.

Reply 2

TheFarmerLad

Original post by Zacken

Tension.

Then how come :

https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202011%20QP%20-%20M3%20Edexcel.pdf (question 7b)

Mark scheme :

https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202011%20MS%20-%20M3%20Edexcel.pdf

bump

I think maybe one is a necessary condition and the other is necessary and sufficient or something like that? I'm not entirely sure myself. @Student403 will know.

Reply 5

EricPiphany

Original post by MathsAstronomy12

With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..

The speed has to be > 0.

Reply 6

TheFarmerLad

Original post by EricPiphany

The speed has to be > 0.

But then the particle could be moving freely under gravity, not necessarily remaining in circular motion..?

Reply 7

tiny hobbit

Original post by MathsAstronomy12

With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..

Original post by Zacken

Tension.

Depends on what the particle is attached to. If it is on the end of a rod or in a tube, it can't fall out of the circular path, so it is the speed that must remain positive (not to mention real). If it is on the end of a string or going round inside a hollow sphere, so that it can fall out of the circular path, it is the tension (or normal reaction) that must remain > = 0 (think the Simpsons movie).

Reply 8

Zacken

Original post by tiny hobbit

Aaaahh, that makes a ton of sense, thank you!

Reply 9

EricPiphany

Original post by MathsAstronomy12

But then the particle could be moving freely under gravity, not necessarily remaining in circular motion..?

Already answered above by tiny hobbit.
Accept that I would say that the speed > 0 is a condition that needs to be fulfilled either way.
The speed can never be < 0, if the maths says so, it'll never reach there.
And tension by definition is greater or equal to zero.

(edited 8 years ago)

Reply 10

TheFarmerLad

Original post by tiny hobbit

Oh, so if it was at the end of a string, T > 0? That makes sense ish as in the question above, the particle is attached to a rod, the sneaky bugger :wink:

EDIT : having reread your post, you've already answered that ^. Thank you!

(edited 8 years ago)

Reply 11

Student403

Original post by MathsAstronomy12

With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..