Question

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

Answer 1

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 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

${\left({\overrightarrow{a}}_{bead}\right)}_x={\left({\overrightarrow{a}}_{bead,rod}\right)}_x={\left({\overrightarrow{a}}_{rod}\right)}_x$

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)

$mg-N\cos =m{a}_r\sin \alpha$.......(i)

and $N\sin \alpha =m(a_r\cos \alpha +a$........(ii)

Eliminating N between (i) and (ii)

$mg\sin \alpha =m{a}_r+ma\cos \alpha$

or $a_r=g\sin \alpha -a\cos \alpha$

Method 2 : Observation from an observer moving with rod. Considering bead w.r.t. rod, i.e., from non-inertial. A pseudo force of magnitude ma will act on the bead in the magnitude ma will act on the bead in the direction opposite to acceleator 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,

$mg\sin \alpha -ma\cos \alpha =m{a}_r$

$\Rightarrow$ $a_r=g\sin \alpha -a\cos \alpha$