Question

A bead of mass m is fitted on to a rod of
a length of 21 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 a with the
rod. Find the time when the bead will be
leave the rod. If I = 2m, a = 2m/s2
| and a = 60°

Answer 1

Method 1: Observation from ground frame. Let (ar) be the acceleration of the bead relative to the rod. Then arcosα is the leftward acceleration of the bead relative to the bead relative to the rod and arsinα is downward relative acceleration of the rod. If ay and ax be the absolute leftward horizontal and downward vertical acceleration of the bead , then

(a→bead)x=(a→bead,rod)x+(a→rod)x

or ax=arcosα+a ...(i)

and arsinα=ay−0

or ay=arsinα...(ii)

From FBD of the bead (projecting forces vertically and horizontally)

mg−Ncosα=marsinα ...(i)

and Nsinα=m(arcosα+a) ...(ii)

Eliminating N between (i) and (ii)

mgsinα=mar+macos(α)

or, ar=gsinα−asinα

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=mgcosα+masinα

Also in the direction along the rod, let acceleration of the bead w.r.t rod is ar.

Equation of motion of bead with rod,

mgsinα−macosα=mar

⇒ar=gsinα−acosα.