Question

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration?
What is the force responsible for the centripetal acceleration in this case?
friction, normal, gravity, or weight?

Answer 1

Answer:

$a_c = 13.26 m/s^2$

Friction Force

Explanation:

As we know that centripetal force is the product of mass and centripetal acceleration

so we know that

$a_c = \frac{v^2}{R}$

so here we have

$v = 27.5 m/s$

$R = 57 m$

so we have

$a_c = \frac{27.5^2}{57}$

$a_c = 13.26 m/s^2$

This acceleration is given by the force which may be towards the center of the circular path

Here in the above case it is possible due to friction force.