Shaskolskaya & Eltsin Problem 53

Physics · mechanics

Problem

If the driver applies his brakes, the car will stop when its kinetic energy has been expended in work against the force of friction. On the other hand, when the car turns the same force of friction will play the part of a centripetal force which makes the car move along the arc of a circle. In the case in which the brakes are applied

\frac{mv^2}{2} = Fx,

where F is the force of friction, and x is the distance which the car will travel after the brakes are first applied. Hence

x = mv^2/2F

. Evidently, if the car is not to be smashed up, we must have

x = \frac{mv^2}{2F} \leq a

F \geq \frac{mv^2}{2a}

In the case where the car is turned

F = \frac{mv^2}{R}

116 SOLUTIONS and if the car is not to be smashed up, we must have

R = \frac{mv^2}{F} \geq a

F \geq \frac{mv^2}{a}

To avoid crashing into the wall, braking requires half the force of friction which a turn requires. Clearly it is better to brake rather than to turn. III. Statics

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