A bead of mass $^{'} m^{'}$ is fitted onto a rod and can move on it without friction. At the initial moment the bead is in the middle of the rod. The bead moves translationally in a horizontal plane with an acceleration $^{'} a^{'}$ in a direction forming an angle $α$ with the rod. Find the acceleration of the bead relative to the rod.

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Solution

Let $a_{r}$ be the acceleration of the bead relative to the rod. Then $a_{r} sin α$ is downward relative acceleration of the rod. If $a_{x}$ and $a_{y}$ be the absolute leftward horizontal and downward vertical acceleration of the bead
$a_{r} cos α = a_{x} - a$
( $∵$ relative acceleration is simply the vector difference between the absolute acceleration)
or $a_{x} = a_{r} cos α + a . . . (1)$
and $a_{r} sin α = a_{y} - 0$
or $a_{y} = a_{r} sin α . . . (2)$
Form $F B D$ of the bead (projecting forces vertically and horizontally)
$m g - N cos α = m a_{r} sin α . . . (3)$
and $N sin α = m (a_{r} cos α + a) . . . (4)$
Eliminating $N$ between $(3)$ and $(4)$
$m g sin α = {m a}_{r} + m a cos α$
or
$a_{r} = g sin α - a cos α$

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