Question

A roller coaster car loaded with passengers, has a total mass of $500$ $kg$ . The radius of curvature at the bottom point of a dip is $12m$ . The vehicle has a speed of $18m/s$ at this point. Take $g = 9.8m/{s^2}$ , calculate the force exerted on the vehicle by the track.
A) $4900N$
B) $13500N$
C) $14700N$
D) $18400N$

Answer 1

Calculate the net force acting on the vehicle and add it to weight to get the normal force exerted by the track on the vehicle. As the roller coaster is in circular motion the net force on it will contribute to the centripetal acceleration and centripetal force

Complete step by step answer:
The acceleration of the vehicle can be calculated by using the formula
$a = \dfrac{{{v^2}}}{R}$
Where $a$ = acceleration of the vehicle
$v$ = velocity of the vehicle at that point of dip
$R$ = radius of curvature of the dip
Thus,
$a = \dfrac{{{{(18)}^2}}}{{(12)}}$
$a = 27m/{s^2}$
Now the net force acting on the vehicle at the point of dip can be calculated as
${F_{net}} = m \times a$ ,
Where $m$ = mass of the vehicle = $500kg$
${F_{net}} = (500 \times 27)N$
${F_{net}} = 13500N$

${F_{net}} = {F_n} - {F_w}$
At the point of dip, the forces acting on the vehicle are the normal reaction force and the weight of the vehicle exerted by it on the track.
Thus
Weight force of vehicle
${F_w} = mg$
( $g$ is the acceleration due to the gravity )
$= 500 \times 9.8$
$= 4900N$
Thus,
Force on the vehicle by the track = ${F_n}$ = $(13500 + 4900)N$
$= 18400N$
Therefore, the correct answer is option D.

Note: The normal reaction force acts opposite to the direction of exertion of weight force ( towards the ground ). Hence the net force is equal to the difference between the normal reaction force and the weight force. The weight force can also be written as the force of gravity. The normal reaction force is the force exerted by the track on the vehicle whereas the force of gravity is exerted by the vehicle on the track. The net force acts in the upward direction because the movement is circular in nature in which the force acts towards the centre of the circle.