A bead of mass m is fitted on a rod and can move on it without friction. Initially the bead is at the middle of the rod moves transitionally in the vertical plane with an accleration `a_(0)` in direction forming angle `alpha` with the rod as shown. The acceleration of bead with respect to rod is:
`z`

Explanation

Method 1: Observation from ground frame. Let `(a_r)` be the acceleration of the bead relative to the rod. Then `a_( r) cos alpha` is the leftward acceleration of the bead relative to the bead relative to the rod and `a_( r) sin alpha` is downward relative acceleration of the rod. If `a_(y)` and `a_(x)` be the absolute leftward horizontal and downward vertical acceleration of the bead , then
`(vec(a)_(bead))_(x)=(vec(a)_(bead,rod))_(x)+(vec(a)_(rod))_(x)`
or `a_(x)=a_(r) cos alpha+a` ...(i)
and `a_( r) sin alpha =a_(y)-0`
or `a_(y)=a_(r )sin alpha`...(ii)
From FBD of the bead (projecting forces vertically and horizontally)

`mg-N cos alpha =ma_(r ) sin alpha` ...(i)
and `N sin alpha =m(a_(r ) cos alpha +a)` ...(ii)
Eliminating N between (i) and (ii)
`mg sin alpha = ma_(r)+ma cos (alpha)`
or, `a_(r ) = g sin alpha -a sin alpha`
Method 2: Observation from an observer moving with rod. Consider bead w.r.t rod i.e. from non-inertial frame. A pseudo force of magnitude ma will act on the bead in the direction opposite to accelerator of rod, i.e. in right direction

The bead is not moving perpendicular to rod. Hence,
`N=mg cos alpha + ma sin alpha`
Also in the direction along the rod, let acceleration of the bead w.r.t rod is `a_(r )`.
Equation of motion of bead with rod,
`mg sin alpha - ma cos alpha = ma_(r )`
`implies a_( r) = g sin alpha - a cos alpha`.