91Ó°ÊÓ

The lamp in the train is in a circular motion in the horizontal plane. Using Newton’s law, the equation of velocity can be determined in terms of radius of curvature and the angle of suspension of lamp.

The radius of the curve is \(r = \;{\rm{215}}\;{\rm{m}}\).

The angle of the suspended lamp is \(\theta = 16.5^\circ \)

The figure below represent the free body diagram for the lamp in the train.

Apply Newton’s law in vertical direction.

\(\begin{aligned}\sum {{F_y}} &= 0\\{F_T}\cos \theta - mg &= 0\\{F_T} &= \frac{{mg}}{{\cos \theta }}\end{aligned}\)

Here, \({F_T}\) is the tension force.

The lamp is in circular motion on the horizontal plane. Apply Newton’s law in horizontal direction.

\(\begin{aligned}\sum {{F_R}} &= m{a_R}\\{F_T}\sin \theta &= m\left( {\frac{{{v^2}}}{r}} \right)\\\left( {\frac{{mg}}{{\cos \theta }}} \right)\sin \theta &= m\left( {\frac{{{v^2}}}{r}} \right)\\v &= \sqrt {rg\tan \theta } \end{aligned}\)

The speed of train can be calculated as,

\(\begin{aligned}v &= \sqrt {rg\tan \theta } \\ &= \sqrt {\left( {215\;{\rm{m}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\tan \left( {16.5^\circ } \right)} \\ &= 25\;{\rm{m/s}}\end{aligned}\)

Thus, the velocity of train is \(25\;{\rm{m/s}}\).