Question

The maximum speed with which a car crosses a convex bridge over a river with radius of curvature 9 m is: (given that the centre of gravity of car is 1m above the road)
A. 50 m/s
B. 30 m/s
C. 20 m/s
D. 10 m/s

Answer 1

A centripetal force is a force that makes a body follow a curved path. The direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous centre of curvature of the path. It is basically the force that keeps the body moving around a fixed point.

Complete step-by-step answer:
In non-ideal cases, the circular motion of a body will not have constant angular velocity but it will slow down or pick up pace regularly. The total acceleration of a body moving in a circular motion is the resultant of two types of accelerations, namely Centripetal and Tangential acceleration.
$\overrightarrow a = {\overrightarrow a _c} + {\overrightarrow a _t}$
Let us consider a non-uniform circular motion as shown:

In a non-uniform circular motion, the body not only changes its angular velocity per unit time but also, the direction of the velocity vector. Since the velocity vector, which is a tangent to the direction of motion of the body, keeps changing every second, we get an
additional component of acceleration. This component of acceleration is called Tangential acceleration.
Tangential acceleration, $\left| {{a_t}} \right| = r\alpha$
$\alpha$is called the angular acceleration, which means the change in angular velocity in radians per second. $\alpha = \dfrac{\omega }{t}$
There is one more component of acceleration, for the actual change in the velocity per unit time and is directed towards the center of the circle. This is called Centripetal acceleration. This is also called Radial acceleration.
Centripetal acceleration ,$\left| {{a_c}} \right| = \dfrac{{{v^2}}}{r}$
The force associated with its centripetal acceleration is called centripetal force. It is equal to the mass times the centripetal acceleration.
${F_c} = m{a_c} = \dfrac{{m{v^2}}}{r}$
Consider a car moving across the convex bridge of radius R.
If the centripetal force acting on the cyclist taking the turn exceeds the weight of the car, he will tend to move towards the center and thus, will tend to deviate from the bridge path and will fall off.
Necessary condition,
$\dfrac{{m{v^2}}}{R} \leqslant mg$
where mg is the mass of the car.
Solving,
$\dfrac{{m{v^2}}}{R} \leqslant mg$
$\to {v^2} \leqslant gR \\\ \to v \leqslant \sqrt {gR} \\\$
Thus, the minimum speed here,
$v = \sqrt {gR}$

Given centre of gravity, $c = 1m$
Radius of curvature, $R = 9m$
However, due to the centre of gravity being 1m above the car, the effective radius of curvature will be the sum of radius and the centre of gravity.
Effective radius, ${R_c} = R + c = 9 + 1 = 10m$
Assuming the acceleration due to gravity, $g = 10m{s^{ - 2}}$and substituting, we get the minimum velocity as –

$$v = \sqrt {g{R_c}} \\\ \to v = \sqrt {10 \times 10} \\\ \to v = 10m/s \\\$$

Hence, the correct option is Option D.

Note: Whenever you travel on highways, you encounter a sign which mentions a speed limit before the start of a bend or deep curve in the road. This speed is calculated on the same basis as explained here, and the vehicle speed should not exceed the mentioned speed limit, if they do not want to get their vehicles overhauled or tumbled towards the curve.