Question

Keeping the banking angle same to increase the maximum speed with which a vehicle can travel a curved road by 10%, the radius of curvature of road has to be changed from 20m to-
A. $16m$
B. $18m$
C. $24.25m$
D. $30.5m$

Answer 1

Here, we take that the centripetal force which is acting on the vehicle is equal to the net force acting on it, and keeping the baking angle same, we obtain an equation between the velocity of the vehicle and the radius of curvature. Thus, using the equation, we put in the values of the velocities and find the radius.

Complete step by step answer:
As we know that for a curved road, the centripetal force on the object is equal to the net force applied on the vehicle. Thus, we have:
${F_{(net)}} = {F_{(cen)}}$
Now, the net force depends upon the mass, acceleration due to gravity and the tangent of the angle produced.
$mg\tan \theta = m\dfrac{{{v^2}}}{r}$
Here, m is the mass of the vehicle, g is the acceleration due to gravity, v is the velocity of the vehicle and r is the radius of the orbit/ circle in which the vehicle is moving.$\theta$ is the banking angle. Thus;
$\tan \theta = \dfrac{{{v^2}}}{{rg}}$
Now, the banking angle is kept the same, but we want to increase the velocity by 10%. Thus, if ${v_2}$ is the new velocity and ${v_1}$ is the old velocity, then${v_2}$=1.1 $\times {v_1}$. Also, if ${r_2}$ is the new radius of curvature and ${r_1}$ is the old radius of curvature, then we have;

\dfrac{{{v}_{1}}^{2}}{{{r}_{1}}}=\dfrac{{{v}_{2}}^{2}}{{{r}_{2}}} \\\ \Rightarrow {{r}_{2}}=\dfrac{{{r}_{1}}\centerdot {{v}_{2}}^{2}}{{{v}_{1}}^{2}}=20\centerdot {{(1.1)}^{2}} \\\ \therefore {{r}_{2}}=24.25m $$. **Hence, option C is the correct answer.** **Note:** Here, we ignore the resistance which is due to the friction on the curved road. Here, the car performs uniform circular motion and hence these equations are satisfied.Also, in uniform circular motion, as the motion is uniform, we assume that the speed of the object is constant and thus the radius also does not increase.