Further Mechanics circular motion(Cambridge CIE) driving me crazy

[Yskaskat]

"
A hollow cylinder of radius a is fixed with its axis horizontal. A particle P, of mass m, moves in part
of a vertical circle of radius a and centre O on the smooth inner surface of the cylinder. The speed of P
when it is at the lowest point A of its motion is ga 2
7 .
The particle P loses contact with the surface of the cylinder when OP makes an angle theta with the upward
vertical through O.
Show that theta is equal to 60 degrees

Hello everyone. Not sure if this is the right place to post this.

On my first attempt at this question I thought I was doing very well until I get a theta of like 78. Looking at the memo, which is attached, they state the following :

ReactionForce + mgcos(theta) = mv^2/a.

This is throwing me for a loop, because I was under the impression that you were supposed to minus these expressions, because at the bottom gravity and reaction force are in opposite directions, while at the top costheta changes signs so that is becomes plus(due to theta being bigger) thus compensating for the minus.

Also if it's relevant, I'm taking the position of zero potential energy to be the middle of the vertical circle. Thank you for any advice!

 

[skeeter]

with the zero for GPE at the cylinder center and using conservation of energy ...

[imath]\dfrac{1}{2}mv_0^2 - mga = \dfrac{1}{2}mv_f^2 + mga\cos{\theta}[/imath]

[imath]v_0^2 - 2ga = v_f^2 + 2ga\cos{\theta}[/imath]

at the point where the reaction force is zero, [imath]\dfrac{mv_f^2}{a} = mg\cos{\theta} \implies v_f^2 = ag\cos{\theta}[/imath]

substituting for [imath]v_f^2[/imath] in the energy equation yields ...

[imath]v_0^2 - 2ga = 3ag\cos{\theta}[/imath]

[imath]v_0^2 = ag(3\cos{\theta} + 2)[/imath]

now, check and clarify that speed you typed for P at the bottom of the cylinder ... what you typed doesn't make sense