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Physics-

General

Easy

Question

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 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

$\sqrt{μ / α}$
$μ / \sqrt{α}$
$1 / \sqrt{μ α}$
Infinitesimal

The correct answer is: $square root of mu divided by alpha end root$

Tangential acceleration $a equals L alpha$
$therefore blank$ Normal relation $N equals M a equals M L alpha$
$therefore$ Frictional force $F equals m N equals mu M L alpha$
For no sliding along the length frictional force $greater or equal than$ centripetal force
$i e comma mu M L alpha greater or equal than M L omega to the power of 2 end exponent$
As $omega equals omega subscript 0 end subscript plus alpha t equals t$
$therefore blank mu M L alpha greater or equal than M L open parentheses alpha t close parentheses to the power of 2 end exponent rightwards double arrow t equals square root of fraction numerator mu over denominator alpha end fraction end root$

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