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# M3 Edexcel Circular Motion

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1. Questions: https://gyazo.com/2d48bb53f461f66a81cecc79ab29f34c
Solutions: https://gyazo.com/cb86e298607f034035579e1a6ff93b41
My question: For their answer to Qc) I get what it means when theta = 0 and cos theta will have its max value. But why does T have minimum value when cos (theta) = 2/3, surely the minimum value of cos (theta) could be lower than 2/3?
Questions: https://gyazo.com/2d48bb53f461f66a81cecc79ab29f34c
Solutions: https://gyazo.com/cb86e298607f034035579e1a6ff93b41
My question: For their answer to Qc) I get what it means when theta = 0 and cos theta will have its max value. But why does T have minimum value when cos (theta) = 2/3, surely the minimum value of cos (theta) could be lower than 2/3?
I think you can find this from the velocity equation on a general point. My first thought is to check v > 0 [v in terms of theta].
3. (Original post by aymanzayedmannan)
I think you can find this from the velocity equation on a general point. My first thought is to check v > 0 [v in terms of theta].
But if you look at the velocity equation on a general point from Qa and Qb, none of them implies that theta = arccos(2/3)?
But if you look at the velocity equation on a general point from Qa and Qb, none of them implies that theta = arccos(2/3)?
That's what I'm saying - you have your velocity equation in terms of theta and gl, correct? Set that equation > 0. You'll see that you get cos(theta) > 2/3, the condition required for circular motion to occur.
5. (Original post by aymanzayedmannan)
That's what I'm saying - you have your velocity equation in terms of theta and gl, correct? Set that equation > 0. You'll see that you get cos(theta) > 2/3, the condition required for circular motion to occur.
Ok thanks fam.
Ok thanks fam.
By conservation of energy, we have , giving us an expression for velocity when the particle has turned an arbitrary angle made with the downward vertical.

We know that . This means that the magnitude of the centripetal force depends on V2, so for there to be any centripetal force must hold true.

Thus, we have: .

Sorry for the vague explanation last night - was on my phone and it was late!
7. (Original post by aymanzayedmannan)
By conservation of energy, we have , giving us an expression for velocity when the particle has turned an arbitrary angle made with the downward vertical.

We know that . This means that the magnitude of the centripetal force depends on V2, so for there to be any centripetal force must hold true.

Thus, we have: .

Sorry for the vague explanation last night - was on my phone and it was late!
Thank you this is exactly what I needed
8. (Original post by aymanzayedmannan)
We know that .
Bit of pedantry here, but better nip this in the bud before it becomes a habit.

First off, using \iff is a lot easier: than \Longleftrightarrow which looks the same.

Second off, when you're using the sign, then you're saying that the two statements are equivalent.

i.e: you're saying that means that (this bit is true)

But you're also saying that means that . (this bit isn't true, just because it's proportional doesn't mean it's proportional to the mass and radius too)

So, you want to say .
9. (Original post by Zacken)
Bit of pedantry here, but better nip this in the bud before it becomes a habit.

First off, using \iff is a lot easier: than \Longleftrightarrow which looks the same.

Second off, when you're using the sign, then you're saying that the two statements are equivalent.

i.e: you're saying that means that (this bit is true)

But you're also saying that means that . (this bit isn't true, just because it's proportional doesn't mean it's proportional to the mass and radius too)

So, you want to say .
Ah yes, thanks for that! Apologies for abusing notation. I typically use only right arrow in written form.
10. (Original post by aymanzayedmannan)
Ah yes, thanks for that! Apologies for abusing notation. I typically use only right arrow in written form.
No worries, it certainly does look neater, I must admit.

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