Questions: To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 361 m. The car increases its speed at uniform rate of [ at = fracdvd t=3.01 mathrm~m / mathrms^2 ] until the tires start to skid. If the tires start to skid when the car reaches a speed of (31.5 mathrm~m / mathrms), what is the coefficient of static friction between the tires and the road?

Solution Steps

• The car is moving in a circular path, so it experiences centripetal force.
• The centripetal force is provided by the frictional force between the tires and the road.

• The centripetal force \( F_c \) required to keep the car moving in a circle is given by: \[ F_c = \frac{m v^2}{r} \] where \( m \) is the mass of the car, \( v \) is the speed of the car, and \( r \) is the radius of the circular track.

• The maximum static frictional force \( F_f \) that can act on the car is given by: \[ F_f = \mu_s m g \] where \( \mu_s \) is the coefficient of static friction and \( g \) is the acceleration due to gravity.

• At the point of skidding, the centripetal force equals the maximum static frictional force: \[ \frac{m v^2}{r} = \mu_s m g \]

• Cancel the mass \( m \) from both sides of the equation: \[ \frac{v^2}{r} = \mu_s g \]
• Solve for \( \mu_s \): \[ \mu_s = \frac{v^2}{r g} \]
• Substitute the given values \( v = 31.5 \, \mathrm{m/s} \), \( r = 361 \, \mathrm{m} \), and \( g = 9.81 \, \mathrm{m/s^2} \): \[ \mu_s = \frac{(31.5)^2}{361 \times 9.81} \]

Final Answer