M3: vertical circles Watch

Maths&physics
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for part b,

for T @ B: I used the triangle sin(green angle) = 2a/3a

T - mg[sin(sin^-1(2/3))] = m[(2ag)^1/2]/3a

T - mg(2/3) = mg(2/3)

where have a gone wrong?
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RDKGames
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(Original post by Maths&physics)
for part b,

for T @ B: I used the triangle sin(green angle) = 2a/3a

T - mg[sin(sin^-1(2/3))] = m[(2ag)^1/2]/3a

T - mg(2/3) = mg(2/3)

where have a gone wrong?
Why are you trying to work out the tension at a particular point in motion, when the question asks you for the "range of values" hence you need to determine the general T at an arbitrary angle.
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(Original post by RDKGames)
Why are you trying to work out the tension at a particular point in motion, when the question asks you for the "range of values" hence you need to determine the general T at an arbitrary angle.
I assumed the range would be from B to where we worked out V, which I've labelled as point C.

And the mark scheme is: x < T < y

I've assumed x and y are some values for mg at B and C respectively.
Last edited by Maths&physics; 3 weeks ago
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(Original post by RDKGames)
Why are you trying to work out the tension at a particular point in motion, when the question asks you for the "range of values" hence you need to determine the general T at an arbitrary angle.
I've looked at the mark scheme and it's for general values for T (not @ B and C)

how do I know that was what they were asking for?
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RDKGames
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(Original post by Maths&physics)
I've looked at the mark scheme and it's for general values for T (not @ B and C)

how do I know that was what they were asking for?
Range means give the lower and upper bound for T.

Obviously the range is easy to work out when we know what the function T(\theta) is explicitly. So that's what you find.
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(Original post by RDKGames)
Range means give the lower and upper bound for T.

Obviously the range is easy to work out when we know what the function T(\theta) is explicitly. So that's what you find.
so the equation I've got is: T -mgsin(some angle) = m[(V^2)/r)

why would the min value be @ V = 0, and the max value when theta = Pi/2. ????
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RDKGames
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(Original post by Maths&physics)
so the equation I've got is: T -mgsin(some angle) = m[(V^2)/r)

why would the min value be @ V = 0, and the max value when theta = Pi/2. ????
Max value is obviously going to be when the particle is directly below the point A. That's because at every other point in motion, T only has to support some component of the mass of the bead, but when the bead is directly below A, the tension supports the ENTIRE mass... which is obviously greater than any component of it.

So that's why tension is greatest at that point, and the angle in this scenario is precisely theta = pi/2.

To see why min. tension occurs at v=0, its easier to first take v^2 = 2ag(3\sin \theta - 1) and rearrange it for \sin \theta = \dfrac{v^2 + 2ag}{6ag}

Then sub it into T - mg \sin \theta = \dfrac{mv^2}{3a} to get T - \dfrac{mv^2 + 2mag}{6a} = \dfrac{mv^2}{3a}

Hence T = \dfrac{3mv^2 + 2mag}{6a} which is a quadratic in v. Obviously it is minimised when v=0.
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(Original post by RDKGames)
Max value is obviously going to be when the particle is directly below the point A. That's because at every other point in motion, T only has to support some component of the mass of the bead, but when the bead is directly below A, the tension supports the ENTIRE mass... which is obviously greater than any component of it.

So that's why tension is greatest at that point, and the angle in this scenario is precisely theta = pi/2.

To see why min. tension occurs at v=0, its easier to first take v^2 = 2ag(3\sin \theta - 1) and rearrange it for \sin \theta = \dfrac{v^2 + 2ag}{6ag}

Then sub it into T - mg \sin \theta = \dfrac{mv^2}{3a} to get T - \dfrac{mv^2 + 2mag}{6a} = \dfrac{mv^2}{3a}

Hence T = \dfrac{3mv^2 + 2mag}{6a} which is a quadratic in v. Obviously it is minimised when v=0.
ok, but in general terms, why is T a minimum when V = 0?
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(Original post by RDKGames)
Max value is obviously going to be when the particle is directly below the point A. That's because at every other point in motion, T only has to support some component of the mass of the bead, but when the bead is directly below A, the tension supports the ENTIRE mass... which is obviously greater than any component of it.

So that's why tension is greatest at that point, and the angle in this scenario is precisely theta = pi/2.

To see why min. tension occurs at v=0, its easier to first take v^2 = 2ag(3\sin \theta - 1) and rearrange it for \sin \theta = \dfrac{v^2 + 2ag}{6ag}

Then sub it into T - mg \sin \theta = \dfrac{mv^2}{3a} to get T - \dfrac{mv^2 + 2mag}{6a} = \dfrac{mv^2}{3a}

Hence T = \dfrac{3mv^2 + 2mag}{6a} which is a quadratic in v. Obviously it is minimised when v=0.
ok, but in general terms, why is T a minimum when V = 0?

also, when T is max, doesnt the spring support the entire mass all the way around? I would have thought the tension being a max at the bottom is because that's where max V would be?
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RDKGames
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(Original post by Maths&physics)
ok, but in general terms, why is T a minimum when V = 0?
Generally not true since for a bead completing full vertical circular motion the min tension is when the bead is at the top of the circle, where the speed isn’t necessarily zero.

For a bead that doesn’t complete full revolutions and oscillates in the bottom half of the circle, at some point in that bottom half we must reach a zero speed which is when the bead starts going the other way. This also means that the entire KE is now converted into GPE hence the height of the bead is maximised.
Another thing to realise in addition to above is that tension reduces as the bead gains height. So when height is maximised (ie V=0) we have minimal tension.

also, when T is max, doesnt the spring support the entire mass all the way around? I would have thought the tension being a max at the bottom is because that's where max V would be?
When T is max thats when T=mg which says tension supports the entire weight.

Velocity is max at the bottom, and this coincides with the concept of energy being xonverted and conserved I’ve mentioned before. At the bottom, GPE is minimised (hence height is minimised, hence tension is maximised) and KE is maximised. So max tension here coincidesnwith max KE.
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(Original post by RDKGames)
Generally not true since for a bead completing full vertical circular motion the min tension is when the bead is at the top of the circle, where the speed isn’t necessarily zero.

For a bead that doesn’t complete full revolutions and oscillates in the bottom half of the circle, at some point in that bottom half we must reach a zero speed which is when the bead starts going the other way. This also means that the entire KE is now converted into GPE hence the height of the bead is maximised.
Another thing to realise in addition to above is that tension reduces as the bead gains height. So when height is maximised (ie V=0) we have minimal tension.



When T is max thats when T=mg which says tension supports the entire weight.

Velocity is max at the bottom, and this coincides with the concept of energy being xonverted and conserved I’ve mentioned before. At the bottom, GPE is minimised (hence height is minimised, hence tension is maximised) and KE is maximised. So max tension here coincidesnwith max KE.
why is the min tension at the top in a full circle and here, at its highest point too?
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DFranklin
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(Original post by Maths&physics)
why is the min tension at the top in a full circle and here, at its highest point too?
Tension is caused by the inwards acceleration needed to keep an object moving in a circle (the V^2/R term). By conservation of energy, KE is at a minimum when the object is at its highest point and therefore so is V.
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(Original post by DFranklin)
Tension is caused by the inwards acceleration needed to keep an object moving in a circle (the V^2/R term). By conservation of energy, KE is at a minimum when the object is at its highest point and therefore so is V.
Brilliant. Ok, that’s because that’s where GPE is at its max, therefore KE is at its min, due to conservation of energy - as mentioned by RDKGame.

I just doesn’t see the relationship between tension and velocity, as you’ve mentioned. Thanks
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DFranklin
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(Original post by Maths&physics)
Brilliant. Ok, that’s because that’s where GPE is at its max, therefore KE is at its min, due to conservation of energy - as mentioned by RDKGame.

I just doesn’t see the relationship between tension and velocity, as you’ve mentioned. Thanks
Tension is related to inwards acceleration, which is related to velocity as I said in my previous post.
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(Original post by DFranklin)
Tension is related to inwards acceleration, which is related to velocity as I said in my previous post.
Ok, so at all times the centripetal force and tension act in the same direction: towards the centre. The reason V is minimum at the top is because this is when GPE is at its max?
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DFranklin
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(Original post by Maths&physics)
Ok, so at all times the centripetal force and tension act in the same direction: towards the centre. The reason V is minimum at the top is because this is when GPE is at its max?
Yes.
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(Original post by DFranklin)
Yes.
If there was no change in direction (I assume that's why V = 0 when tension is at a minimum), and we go all the way to the top where V/tension will be a minimum, what would V = then? clearly not zero in the new case, because there will be no change in direction - I assume thats why V = 0 zero in this case (the question attached in the OP)?
Last edited by Maths&physics; 3 weeks ago
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