# Mechanics 3 QuestionWatch

#1
Hi guys,
I'm stuck on the very last part of this M3 question, does anyone know how to show that T=T0(2+cos x) ?

Here's the question:
A particle P, of mass m, is attached to a fixed point O by a light inextensible string of length a, and is executing complete circular revolutions in a verticle plane. Show that when OP is inclined at an angle x to the downward vertical, the tension T in the string is given by: T=mg(3cos x - 2) + (mu^2)/a

where u is the speed of P when it passes through its lowest point. Show that if 3T0 and T0 are the greatest and least tensions in the string during the motion:
u^2 = 8ag and T= T0(2+cos x)

I've managed to solve the first bits of the question, but I just can't get this last result. Any ideas? Thanks!
0
7 years ago
#2
(Original post by Jugu)
Hi guys,
I'm stuck on the very last part of this M3 question, does anyone know how to show that T=T0(2+cos x) ?

Here's the question:
A particle P, of mass m, is attached to a fixed point O by a light inextensible string of length a, and is executing complete circular revolutions in a verticle plane. Show that when OP is inclined at an angle x to the downward vertical, the tension T in the string is given by: T=mg(3cos x - 2) + (mu^2)/a

where u is the speed of P when it passes through its lowest point. Show that if 3T0 and T0 are the greatest and least tensions in the string during the motion:
u^2 = 8ag and T= T0(2+cos x)

I've managed to solve the first bits of the question, but I just can't get this last result. Any ideas? Thanks!
T will be a maximum when x=0, and a minimum when x = 180.

So substitute into your equation for T to get two simultaneous equations.

Work out what mg is in terms of T0, and what (mu^2)/a is as well, and then sub back into the formula for T.
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