A car is turning around a corner and the radius of the circular path is 5 m. What is the maximum speed that the car can travel at while not slipping if coefficient of friction is 0.4.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

When the car is turning the corner the centripetal force is provided solely by friction. If the frictional force was not equal to the centripetal force the car would slip outwards.

The centripetal force if the car of mass m is traveling at a velocity v around a circular path of radius r is m*v^2/r. The frictional force is N*C where N is the normal force m*g and C is the coefficient of friction. Here, the coefficient of friction is 0.4

Substitute the given values in the equation m*v^2/r = m*g*C

=> v^2/5 = g*0.4

=> v^2 = 2*9.8

=> v = 4.427 m/s

The car has to be driven at a speed less than 4.427 m/s if it has to complete the turn without slipping.

Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial