A body moves on a horizontal circular road of radius of r, with a tangential acceleration `a_(T)`. Coefficient of friction between the body and road surface is `mu`. It begin to slip when it's speed is `v`, then :

Options

A. `v^(2) = mu rg `
B. `mu g = (v^(2))/(r) + a_(T)`
C. `mu^(2)g^(2) = (v^(4))/(r^(2)) + a_(T)^(2)`
D. the force of friction makes an angle `tan^(-1) ((v^(2))/(a_(T) xx r))` with the direction of motion at point of slipping

Answer

B, C

Explanation

At time of slipping f = `mu`mg
`f cos theta = ma_(T) , f_(2) = (ma_(T)) + ((mv^(2))/(r))`
`mu (mg) = (ma_(T)) + ((mv^(2))/(r))`
`implies mu^(2) g^(2) = a_(T)^(2) + ((v^(2))/(r^(2)))`
Also , `tan theta = (v^(2))/(a_(T)r)`