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4-1.Newton's Laws of Motion
easy
A train runs along an unbanked circular track of radius $30 \;m$ at a speed of $54\; km / h$ The mass of the train is $10^{6}\; kg$. What provides the centripetal force required for this purpose - The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Option A
Option B
Option C
Option D
Solution
Radius of the circular track, $r=3$
Speed of the train, $v=54\, km / h =15 \,m / s$
Mass of the train, $m=10^{6}\, kg$
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton's third law of motion, the wheel exerts an equal and opposite force on the rail.
This reaction force is responsible for the wear and rear of the rail
The angle of banking $\theta$, is related to the radius ( $r$ ) and speed $(v)$ by the relation:
$\tan \theta=\frac{v^{2}}{r g}$
$=\frac{(15)^{2}}{30 \times 10}=\frac{225}{300}$
$\theta=\tan ^{-1}(0.75)=36.87^{\circ}$
Therefore, the angle of banking is about $36.87^{\circ} .$
Speed of the train, $v=54\, km / h =15 \,m / s$
Mass of the train, $m=10^{6}\, kg$
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton's third law of motion, the wheel exerts an equal and opposite force on the rail.
This reaction force is responsible for the wear and rear of the rail
The angle of banking $\theta$, is related to the radius ( $r$ ) and speed $(v)$ by the relation:
$\tan \theta=\frac{v^{2}}{r g}$
$=\frac{(15)^{2}}{30 \times 10}=\frac{225}{300}$
$\theta=\tan ^{-1}(0.75)=36.87^{\circ}$
Therefore, the angle of banking is about $36.87^{\circ} .$
Standard 11
Physics
