A car is moving on a circular path and takes | vedclass

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Question

(B) When a car takes a turn on a circular path,the centripetal force is provided by the friction between the tires and the road. This creates a torque

Vedclass · Accepted Answer

$R_1 < R_2$

Vedclass · Answer

(B) When a car takes a turn on a circular path,the centripetal force is provided by the friction between the tires and the road. This creates a torque that tends to tilt the car outward.To maintain equilibrium,the normal reaction on the outer wheel increases while the normal reaction on the inner wheel decreases.The reaction on the inner wheel is given by $R_1 = \frac{1}{2}M \left[ g - \frac{v^2 h}{ra} \right]$.The reaction on the outer wheel is given by $R_2 = \frac{1}{2}M \left[ g + \frac{v^2 h}{ra} \right]$.Here,$r$ is the radius of the circular path,$2a$ is the distance between the two wheels,$h$ is the height of the center of gravity of the car,$M$ is the mass,and $v$ is the velocity.Comparing the two expressions,it is clear that $R_1 < R_2$.

$A$ car is moving on a circular path and takes a turn. If $R_1$ and $R_2$ are the reactions on the inner and outer wheels respectively,then:

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