Question

A thin rod of length $1m$ is fixed in a vertical position inside a train, which is moving horizontally with constant acceleration $4m{s}^{-2}.$ A bead can slide on the rod and friction coefficient between them is $0.5$. If the bead is released from rest at the top of the rod, it will reach the bottom in:

• A. $\sqrt{2}s$
• B. $1s$
• C. $2s$
• D. $0.5s$

Answer 1

The correct option is D $0.5s$

From FBD of bead, $N=m{a}_0=4m$
and $ma=mg-\mu N$ or $ma=10m-0.5\times 4m=8m\Rightarrow a=8m/s^2$
using $S=ut+\frac{1}{2}a{t}^2$ we get $1=0+0.5\times 8t^2$
or $t^2=1/4\Rightarrow t=0.5$sec