Question

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_s10_qp_53.pdf
Q5ii)

Answer 1

why is the tension PB equal to zero?

Answer 2

because that's the point just before the thing collapses, ie the ball reduces the radius of its orbit.

so you repeat the calculation using that fact

Answer 3

just imagine you were doing this only with string QB, ie establishing this particular orbit first. and then adding PB in afterward on the basis it was changing nothing and would take no tension.

Answer 4

The horizontal component of both tensions QB and PB provide the force to the ball in order to move in a circle, then why cant we neglect force QB?

Answer 5

because QG is supporting the ball's weight. if this being done in outer space (no gravity), the tension in each string would be the same.

Answer 6

remember as soon as theta dips below 30 deg, there is no point in the lower string. it only has any impact once theta = 30 deg. at that point it starts to pull in and downward. so you are being asked to find the point at which it starts to have a point!

Answer 7

Tension PQB will have an angle 30 along the whole circular path,yes?

Answer 8

sorry PB**

Answer 9

yes.

Answer 10

What i got from your explanation to the question that the ball will still move in a circle by tension PB and tension QB has not effect on the path of the ball, it just puts on more weight on the ball downwards.

Answer 11

no. if the tension in PB = 0, it doesn't pull anything in any direction.

if we let V = the "minimum possible speed" the question asks for, then **IF**

1) actual v < V, the ball lowers and theta < 30, string B is no longer taut
2) actual v > V, there is tension in both strings, PB pulls down and in towards the centre and keeps theta at 30 deg
3) actual v = V, PB is just about taut but has no actual tension and no effect on the kinematics

Answer 12

Thanks @IrishBoy123 :)

Answer 13

cheers!