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M3:what the hell?!

what are they asking in part d?
Reply 1
I would suggest you draw a diagram. And label the point O, A, B and C. You should be able to do the d) if you can do a), b), c) after drawing a good diagram by following the question d) sentence by sentence.
(edited 6 years ago)
Original post by Maths&physics
what are they asking in part d?


There is a small peg at point B (which is at a distance 1/2a to the left of O), which the string is going to hit, and hence the particle is going to start going in circles about B. Clearly, as far as the initial circular revolution about B is concerned, the radius of the circle centered B is 1/2a. It is asking you to find the tension in the string when the particle is vertical above B, which is the point C at a distance 1/2a above B.
(edited 6 years ago)
Original post by RDKGames
There is a small peg at point B (which is at a distance 1/2a to the left of O), which the string is going to hit, and hence the particle is going to start going in circles about B. Clearly, as far as the initial circular revolution about B is concerned, the radius of the circle centered B is 1/2a. It is asking you to find the tension in the string when the particle is vertical above B, which is the point C at a distance 1/2a above B.


does it look like that?
Original post by Maths&physics
does it look like that?


Not sure what that is.

It looks like this:
ive done:

[text] KE(C) + GPE(C) = KE(B) + GPE(B)

[text] (1/2)mv^2 + (a/2)mg = (1/2)(m)(5ga/2) + 0
Original post by Maths&physics
ive done:

[text] KE(C) + GPE(C) = KE(B) + GPE(B)


[text] (1/2)mv^2 + (a/2)mg = (1/2)(m)(5ga/2) + 0

Looks good (except that it doesnt go through B directly so you cannot say KE(B) and such)
Original post by RDKGames
Not sure what that is.

It looks like this:


perfect!
Original post by RDKGames
Looks good (except that it doesnt go through B directly so you cannot say KE(B) and such)


working put tension, ive got: T+mg(a/2)=2m(3ag/2)/a T + mg(a/2) = 2m(3ag/2) / a

the RHS is right but the is no (a/2) for the GPE height, just (1/2)? thanks
(edited 6 years ago)
Original post by Maths&physics
working put tension, ive got: Tmg(a/2)=2m(3ag/2)/a T mg(a/2) = 2m(3ag/2) / a

the RHS is right but the is no (a/2) for the GPE height, just (1/2)? thanks


I'm not sure what you're doing.. it seems to me you're mixing forces with energy which a no-no.

The velocity at CC is v2=32agv^2 = \frac{3}{2}ag

Solving forces at CC yields T+mg=m32ag12a=3mgT+mg =m \cdot \dfrac{\frac{3}{2}ag}{\frac{1}{2}a} = 3mg
Original post by RDKGames
I'm not sure what you're doing.. it seems to me you're mixing forces with energy which a no-no.

The velocity at CC is v2=32agv^2 = \frac{3}{2}ag

Solving forces at CC yields T+mg=m32ag12a=3mgT+mg =m \cdot \dfrac{\frac{3}{2}ag}{\frac{1}{2}a} = 3mg


sorry, I meant to say T + GPE(C)....

what height did you use for GPE at C?
Original post by Maths&physics
sorry, I meant to say T + GPE(C)....

what height did you use for GPE at C?


The radius of the smaller circle. But WHY are you adding energy to a force?!
Original post by RDKGames
The radius of the smaller circle. But WHY are you adding energy to a force?!


oops, I need to go over it again.
Original post by Maths&physics
oops, I need to go over it again.


Original post by RDKGames
The radius of the smaller circle. But WHY are you adding energy to a force?!


ok, yeah, I was trying to use GPE, which is not relevant to this question.

so, I got: T+mg=3mg T + mg = 3mg

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