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    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?
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    (Original post by Nerrad)
    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].
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    (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)?
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    (Original post by Nerrad)
    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.
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    (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.
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    (Original post by Nerrad)
    Ok thanks fam.
    By conservation of energy, we have \frac{1}{2}mU^{2} - \frac{1}{2}mV^{2} =mgl\left ( 1-\cos \theta \right )\Longleftrightarrow V^{2} = \frac{2gl}{3}\left ( 3\cos \theta -2 \right ), giving us an expression for velocity when the particle has turned an arbitrary angle \theta made with the downward vertical.

    We know that \mathrm{F_{c}} = \frac{mV^{2}}{r} \Rightarrow \mathrm{F_{c}} \propto V^{2}. This means that the magnitude of the centripetal force depends on V2, so for there to be any centripetal force V^{2} > 0 must hold true.

    Thus, we have: \displaystyle \frac{2gl\left ( 3\cos \theta -2 \right )}{3} > 0 \Longleftrightarrow \cos \theta > \frac{2}{3} \Rightarrow \boxed{\left(\cos \theta \right)_{\text{min}} = \frac{2}{3}}.

    Sorry for the vague explanation last night - was on my phone and it was late!
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    (Original post by aymanzayedmannan)
    By conservation of energy, we have \frac{1}{2}mU^{2} - \frac{1}{2}mV^{2} =mgl\left ( 1-\cos \theta \right )\Longleftrightarrow V^{2} = \frac{2gl}{3}\left ( 3\cos \theta -2 \right ), giving us an expression for velocity when the particle has turned an arbitrary angle \theta made with the downward vertical.

    We know that \mathrm{F_{c}} = \frac{mV^{2}}{r} \Longleftrightarrow \mathrm{F_{c}} \propto V^{2}. This means that the magnitude of the centripetal force depends on V2, so for there to be any centripetal force V^{2} > 0 must hold true.

    Thus, we have: \displaystyle \frac{2gl\left ( 3\cos \theta -2 \right )}{3} > 0 \Longleftrightarrow \cos \theta > \frac{2}{3} \Rightarrow \boxed{\left(\cos \theta \right)_{\text{min}} = \frac{2}{3}}.

    Sorry for the vague explanation last night - was on my phone and it was late!
    Thank you this is exactly what I needed
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    (Original post by aymanzayedmannan)
    We know that \mathrm{F_{c}} = \frac{mV^{2}}{r} \Longleftrightarrow \mathrm{F_{c}} \propto V^{2}.
    Bit of pedantry here, but better nip this in the bud before it becomes a habit.

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

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

    i.e: you're saying that F = \frac{mv^2}{r} means that F \propto V^2 (this bit is true)

    But you're also saying that F \propto V^2 means that F = \frac{mv^2}{r}. (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 F= \frac{mv^2}{r} \Rightarrow F \propto V^2.
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    (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: \iff than \Longleftrightarrow \Longleftrightarrow which looks the same.

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

    i.e: you're saying that F = \frac{mv^2}{r} means that F \propto V^2 (this bit is true)

    But you're also saying that F \propto V^2 means that F = \frac{mv^2}{r}. (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 F= \frac{mv^2}{r} \Rightarrow F \propto V^2.
    Ah yes, thanks for that! Apologies for abusing notation. I typically use only right arrow in written form.
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    (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. :lol:
 
 
 
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