circular motion (1 Viewer)

[doiyoubi]

the ends of a light string, of length 18a, are attached to two fixed points L, M in the same vertical line, M being at a depth 12a below L. A small smooth ring R of mass m is threaded on the string. The ring revolves in a horziontal circle with M as centre and the string taut. Show that MR=5a and that the tension in the string is 13mg/12. Find the period of revolution

A particle P is attached by two equal strings to two points A and B in the same vertical line. The system revoled with constant angular velocity "z" about AB so that P describes a horizontal circle. If AB=1.6m and z=3.5rad/s, prove that tthe tension in one string is zero, and that if z=7ad/s, the tensions in the strings are in ratio 5:3

Can someone help me with this thanks

 

[Riviet]

Draw a right triangle [like below], label top right corner as L, bottom right corner as M, bottom left one as R, and label LM as 12a. We know LR+MR gives length of string which isequal to 18a => LR+MR=18a
Rearrange for LR=18a-MR
By Pythagoras' theorem, MR²=LR²-144a²
MR²=(18a-RM)²-144a²
Expand and solve for MR to get MR=5a.
aL
q/|
t/ |
R M

Tension is given by the force parallel to LR and pulling the string outwards along LR, so let T be the tension on LR. Let upwards direction be positive and let @ be the angle opposite MR.
Sum of vertical forces=force on LM up - gravitational force pulling ring down
=Tcos@ - mg = 0
T=mg/cos@
but cos@=12/13
.'. T=13mg/12 - QED