A stone tied to the end of a string is whirled in a horizontal circle. The mass of the stone is 1.0 kg and the string is 0.50 m long. If the stone revolves at a constant speed for 10 times in 15.71 s, (a) what is the tension in the string ? (b) What would happen to the tension in the string if the mass was doubled and all the other quanities remained the same ? (c ) What would happen to the tension in the string if the period was doubled and all the other quantities remain the same ?

A stone tied to the end of a string is whirled in a horizontal circle.

1.	A stone tied to the end of a string is whirled in a horizontal circle. The mass of the stone is 1.0 kg and the string is 0.50 m long. If the stone revolves at a constant speed for 10 times in 15.71 s, (a) what is the tension in the string ? (b) What would happen to the tension in the string if the mass was doubled and all the other quanities remained the same ? (c ) What would happen to the tension in the string if the period was doubled and all the other quantities remain the same ?
Answer» Solution :The angular velocity, `omega = (2pi N)/(t)=(2pi XX 10)/(15.71)=(31.42xx2)/(15.71)=4.0 rad s^(-1)` (a) The TENSION on the STRING `= mr omega^(2)` `=1.0xx0.5xx(4.0)^(2)=8.0 N`. (b)Keeping r and `omega` constant if m was doubled, the tension on the string would be doubled i.e., tension = 16.0 N. (c ) Keeping m and r constant if TIME period was doubled then the the tension in the string is : `T=mr((2pi)/(2T))^(2)=(1)/(4)[mr((2pi)/(T))^(2)]=(8)/(4)=2N`.