A 146 kg motorcycle is driving around a turn with a radius of 87.8 m at a constant speed of 12.1 m/s. What is the minimum force of friction required to prevent the cycle from sliding off of the road?

Question

A 146 kg motorcycle is driving around a turn with a radius of 87.8 m at a constant speed of 12.1 m/s. What is the minimum force of friction required to prevent the cycle from sliding off of the road?

Answer 1

For this problem, the friction force is providing the centripetal force, so the friction force magnitude has to be at least as big as the centripetal force magnitude. From the radius, speed, and mass, we can find the centripetal force. The centripetal force is equal to the minimum friction force we need.

Centripetal force:

F = mv²/r = (146 kg)(12.1 m/s)²/(87.8 m) = 243 N

So you need at least 243 N of friction force.

Answer 2

Find centripetal force which is provide by a necessary frictional force. f=Fc = mv^2/r.