Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If the over-bridge is concave instead of being convex, the thrust on the road at the lowest position will be:(A) $mg - \dfrac{{m{v^2}}}{r}$ (B) $mg + \dfrac{{m{v^2}}}{r}$ (C) $\dfrac{{{v^2}g}}{r}$ (D) $\dfrac{{{m^2}{v^2}g}}{r}$

Last updated date: 14th Jun 2024
Total views: 392.4k
Views today: 3.92k
Verified
392.4k+ views
Hint
By drawing the free body diagram of the body at the lowest point, we can find the equation for the body due to its weight, the centrifugal force and the normal reaction acting on the body due to the road. We can find the thrust as it is equal to the normal reaction force on the body due to the road.
Formula Used: In this solution we will be using the following formula,
${F_c} = \dfrac{{m{v^2}}}{r}$
where ${F_c}$ is the centrifugal force,
$m$ is the mass of the body
$v$ is the velocity of the body
$r$ is the radius of the concave path.

Let us first draw the image of a body at the lowest point on a concave road with the forces acting on it.

Now as we can see that for the body, the weight will be acting in the downward direction and the normal reaction force acting on the body will be in the upward direction. The velocity of the body will be as given in the diagram as a tangent to the circular path.
Now let us consider the radius of this circular path to be $r$ . Now since the body is moving in a circular path, the force that is acting on the body is called centrifugal force. This force acts in a direction which is outward on the circular path. So the centrifugal force will be acting on the body in the upward direction.
So the sum of the centrifugal force on the body and the mass of the body will be balanced by the normal reaction force on the body due to the road.
Therefore, we can write
$N = mg + {F_c}$
The centrifugal acceleration is given as,
${F_c} = \dfrac{{m{v^2}}}{r}$
So substituting this in the equation we get,
$N = mg + \dfrac{{m{v^2}}}{r}$
Since this normal reaction force is the thrust on the road, so the answer will be, $mg + \dfrac{{m{v^2}}}{r}$
Hence the correct answer is option (B).

Note
If the body was moving on a convex road, then the centrifugal force on the body due to the curved path would be acting in the upward direction. Therefore, the thrust on the road will be the weight on the body minus the centrifugal force.