A long horizontal rod has a bead which can slide along its length, and initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about $A$ with constant angular acceleration $\alpha$. If the coefficient of friction between the rod and the bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is
A long horizontal rod has a bead which can slide along its length, and initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about $A$ with constant angular acceleration $\alpha$. If the coefficient of friction between the rod and the bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is
$\sqrt {\frac{\mu }{\alpha }} $
$\frac{\mu }{{\sqrt \alpha }}$
$\frac{1}{{\sqrt {\mu \alpha } }}$
Infinitesimal
A long horizontal rod has a bead which can slide along its length, and initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about $A$ with constant angular acceleration $\alpha$. If the coefficient of friction between the rod and the bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is
Let the bead starts slipping after time$ t$
For critical condition
Frictional force provides the centripetal force
$m{\omega ^2}L = \mu \,R = \mu \,m \times {a_t} = \mu Lm\alpha $
$⇒$ $m{(\alpha t)^2}L = \mu mL\alpha $ $⇒$ $t = \sqrt {\frac{\mu }{\alpha }} $ (As $\omega = \alpha t$)
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