Circular motion, range of values of T?

Part (a) was nice and neat, but I'm a little confused about part (b).

So I resolved towards the centre of the circle and used circular motion acceleration through Newton's 2nd Law to get T = 3mgsinθ - (2/3)mg, but I'm not sure where to go from here.

I feel like I should instinctively solve for θ when T is zero, but I'm not sure, since that would just imply that it is zero.

Please help, thanks

@TeeEm

Part (a) was nice and neat, but I'm a little confused about part (b).

So I resolved towards the centre of the circle and used circular motion acceleration through Newton's 2nd Law to get T = 3mgsinθ - (2/3)mg, but I'm not sure where to go from here.

I feel like I should instinctively solve for θ when T is zero, but I'm not sure, since that would just imply that it is zero.

Please help, thanks

@TeeEm

By equating tension to the centripetal force $\frac{mv^2}{r}$

I have $T = \frac{2}{3}mg(3\sin(\theta)-1)$

Which gives $T = 0$ when $\sin(\theta)=\frac{1}{3}$ and $T=\frac{4}{3}mg$ when $\theta=\frac{\pi}{2}$

Hence $0 \leq T \leq \frac{4}{3}mg$

@TeeEm does this look right to you?

I am way too busy to look at this ...
that is why I asked the OP to tell me the paper so I can post my own personal solution

january 2010 M3 edexcel

january 2010 M3 edexcel

My answer is very wrong (sorry ). This paper is on my to-do list now...

january 2010 M3 edexcel

nah it wasn't that wrong, i just got 7/3 instead of 4/3 for the coefficient of mg, but apart from this question the rest of the paper was really nice

and thanks for your help everyone

I hope the link works

yeah it did, thanks a lot! very impressed with your presentation