Question

A string of length 1m is fixed at one end and carries a mass of 1 kg at the other end. The string makes `2//pi` revolutions per second around a vertical axis passing through the fixed end. Calculate (i) angle of inclination of the string with the vertical, (ii) the tension in the strong and (iii) the linear velocity of the mass.

Answer 1

As shown in Fig. 7.67, for vertical equilibrium of mass m
`T "cos" theta = "mg" …… (i) `
While for circular motion
`T "sin" theta = (mv^(2)//r) = mr omega^(2) " " …….. (ii) `
(a) So , from Eqn . (ii) , `T = [ mr omega^(2)//"sin" theta ] = m L omega^(2)`
[as r = L sin `theta`]
`therefore " " T = 10^(-1) xx 1 xx 4^(2) = 1.6 N` [ as `omega = 2 pi f = 2 pi xx ( 2//pi) = 4]`
(b) From Eqn. (i) , cos `theta = ("mg")/(T) = ( 10^(-1) xx 10)/(1.6) = (5)/(8)`,
i.e., `" " theta = cos^(-1) (5 //8)`
(c) `" " v = r omega = L "sin" theta xx omega ` [ as r = L sin `theta`]
or `v = 1 xx 0.78 xx 4 = 3.12` m/s [ as sin `theta = 0.78`]