(a) How many revolutions per minute must the apparatus shown in figure make about a vertical axis so that the cord makes an angle of `45^(@)` with the vertical ?

(b)What is the tension in the cord then? Given, `l=sqrt(2)m,a=20 cm` and `m=5.0 kg`?

Answer

C

Explanation

`(a) r=a+l sin 45^(@)=(0.2)+(sqrt(2))(1/(sqrt(2)))=1.2 m`
Now, `T cos 45^(@)=mg.....(i)`
and `T sin 45^(@)=mr omega^(2).....(ii)`
From eqn. (i) and (ii), we have
`omega=2n pi =sqrt(g/r)`
`:. n=1/(2pi)sqrt(g/r)=60/(pi)sqrt(9.8/1.2)rpm=27.3 r pm `
(b). From eq. (i), we have `T=sqrt(2)mg=(sqrt(2))(5.0)(9.8)`
`=69.3 N`