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 $α$ . If the coefficient of friction between the rod and the bead is $μ$ , and gravity is neglected, then the time after which the bead starts slipping is [2000]

Question

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(1)

When we are giving an angular acceleration $A$ to the rod, the bead has instantaneous acceleration crinst $α$ The bead has a tendency to move away from the centre. But due to the friction between the bead and the rod, this does not happen. If instantaneous angular velocity is $ω$ then

Here, necessary frictional force is provided by frictional force

$m L ω^{2} = μ (m a) \Rightarrow m L ω^{2} = μ m L α$

$\Rightarrow ω^{2} = μ α$

$Using ω = ω_{0} + α t,$

$\Rightarrow ω = α t$

$∴$ $α^{2} t^{2} = μ α \Rightarrow t = \sqrt{\frac{μ}{α}}$

Solution

1 $\sqrt{\frac{μ}{α}}$

2 $\frac{μ}{\sqrt{α}}$

3 $\frac{1}{\sqrt{μ α}}$

4 infinitesimal

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Solution

1 μα

2 μα

3 1μα

4 infinitesimal

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1 $\sqrt{\frac{μ}{α}}$

2 $\frac{μ}{\sqrt{α}}$

3 $\frac{1}{\sqrt{μ α}}$