A block of mass `m` is kept on rough horizontal turn table at a distance r from the center of table. The coefficient of friction between torn table and block is `mu` Now the turn - table starts rotating with uniform angular acceleration a
a. Find the time after which slipping accure between block and turn - table
b. Find the angle made by friction force at the point of slipping

Explanation

a. `a_(t) +alpha r`
speed after time `t`
`(dv)/(dt) = alpha r rArr v = 0 + alpha rt`
Centeripetal acceleration,` a_(t) = (v^(2))/(R ) = a^(2) rt^(2)`
Net accceleration, `a_("net") = sqrt(a_(1)^(2) +a_(2)^(2)`
`= sqrt(a^(2) r^(2) + a^(4) r^(2)t^(4))`
Block just start slipping
`mu m g = ma_("net") = m sqrt(a^(2) r^(2) + a^(4) r^(2)t^(4))`
` t = ((mu^(2) g^(2) - a^(2) r^(2))/(alpha^(2) r^(2)))^(1//4)`
`= [((mu g)/(alpha^(2) r))^(2) - ((1)/(alpha))^(2)]^(1//4)`
b. `tan theta = (a_(c))/(a_(t)) = (alpha^(2) rt^(2))/(alpha r)`
or `theta = tan^(-1) (alpha r^(2))`