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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 quantities 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 ? |
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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)=8.0N`. (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 tension in the string is : `T=mr((2pi)/(2T))^(2)=(1)/(4)[mr((2pi)/(T))^(2)]=(8)/(4)=2N`. |
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