M3 Edexcel Circular Motion

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].

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.

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 $\mathrm{F_{c}} = \frac{mV^{2}}{r} \Longleftrightarrow \mathrm{F_{c}} \propto V^{2}$. 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!

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 $F= \frac{mv^2}{r} \Rightarrow F \propto V^2$.

Ah yes, thanks for that! Apologies for abusing notation. I typically use only right arrow in written form.