A long horizontal rod has a bead which can sl | vedclass

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(A) Let the bead start slipping after time $t$.For the bead to slip,the net force acting on it must exceed the maximum static frictional force.The for

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$\sqrt {\frac{\mu }{\alpha }} $

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(A) Let the bead start slipping after time $t$.For the bead to slip,the net force acting on it must exceed the maximum static frictional force.The forces acting on the bead in the rotating frame are the centrifugal force $F_c = m\omega^2 L$ (acting radially outward) and the tangential force $F_t = m a_t = m \alpha L$ (acting perpendicular to the rod).The normal force $N$ is zero as gravity is neglected and there are no other vertical forces.Wait,the friction force $f$ must balance the resultant of these forces. The maximum static friction is $f_{max} = \mu N$. Since there is no vertical force,$N$ is effectively provided by the constraint of the rod. However,in this specific problem context,the friction is assumed to be $\mu m a_t$ or similar. Re-evaluating: The bead slips when the resultant force $F_{net} = \sqrt{(m\omega^2 L)^2 + (m\alpha L)^2} = \mu N$. Assuming $N = m a_t = m \alpha L$ is not standard. Actually,the friction must oppose the tendency to move. The bead will slip when the net force required to keep it in circular motion exceeds the maximum available friction. The force required is $F = \sqrt{(m\omega^2 L)^2 + (m\alpha L)^2}$. Given the options,the intended logic is $m\omega^2 L = \mu m \alpha L$,which implies $\omega^2 = \mu \alpha$. Since $\omega = \alpha t$,we get $(\alpha t)^2 = \mu \alpha$,so $t^2 = \mu / \alpha$,or $t = \sqrt{\mu / \alpha}$.

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