A 127 kg motorcycle is driving around a turn with a radius of 69.2 m at a constant speed of 19.6 m/s.

Question

A 127 kg motorcycle is driving around a turn with a radius of 69.2 m at a constant speed of 19.6 m/s. What is the minimum force of friction required to prevent the cycle from sliding off of the road? Report your answer to the nearest whole number. Do not include the units in your answer or the computer will mark the problem wrong

Answer 1

In order to do this problem you must use the concept of centripetal force. As an object travels a circular path, there is an acceleration toward the center of the circle called the centripetal acceleration. This acceleration has the value of v²/ r where v is the tangential velocity and r the radius of the circle. The force due to this acceleration is mv²/r,

The centripetal force is the force required to prevent the object from veering off on a tangent to its circular path. In this case the force keeping the object in the circular path is the force of friction between the tires and the road. The minimal force of friction is equal to the centripetal force.

Frictional force = 127 kg x 19.6² m²/s² / 69.2 m = 7 newtons.

Answer 2

This is a problem of curvilinear motion. On a level ground the force acting on the bicyclist the radial force which is a force that tends to push the bike off the course. This force is dented by F_r = m.a_r with m being the mass of the bike given as 127kg and a_r being the radial acceleration. The only force that keeps the bike on course in this case is the force of friction between the bike tires and the ground. For equilibrium to take place the sum of forces on the bike should be zero. Therefore:

∑F = 0 , and F_r - F_f = 0 , and F_r = F_f where F_f = Force of friction and a_r = V_t²/ R where V is the tangential velocity given as 19.6 m/s and R is the radius of the curve given as 69.2 m

F_r = m.a_r = 127 kg . [( 19.6 m/s)² / 69.2 m)] = 127 kg . 5.55 m/s² = 704.85 N = F_f say 704.9 N

The minimum friction force reported unitless to the nearest whole number so the computer will not mark it wrong is 705