A piece of mass 1 kg is tied with a thread of length 10 m and is rotating along a horizontal circular path making an angle `theta` with the vertical. The thread can withstand a maximum tension of 12 N. What is the maximum velocity with which the stone can be rotated without snapping the thread?

Explanation

Let the length of the thread be l, mass and maximum linear velocity of the stone be m and v respectively and the maximum tension in the thread be T while rotating along a horizontal along a horizontal plane .The required centripetal force for rotation,
`(mv^(2))/(l sin theta)` = T sin`theta" "` ....(1)
and mg = T cos`theta" "`...(2)
From (1) and (2), we get, `sin^(2) theta + cos^(2) theta = (mv^(2))/(l T) + ((mg)/(T))^(2)`
or,`" " (mv^(2))/(l T) = (1 - (m^(2) g^(2))/(T^(2))) or, v = { (lT)/(m) ( 1 - (m^(2)g^(2))/(T^(2))) }^(1//2)`
Given, l = 10 m, T = 12 N and m = 1 kg
`therefore " " v = { (10 xx 12)/(1) ( 1 - (1^(2) xx 9.8^(2))/(12^(2))) }^(1//2) " " [ because g = 9.8 m cdot s^(-2)]`
or, v = 6.32 m`cdot^(-1)`.