Problem 16 - PHHS Physics

On Earth, the maximum speed without skidding for a car on a level circular curved track of radius 40m is 15m/s. This car and track are then transported to another planet for the Indy Galactic 500. The Maximum speed is 8.4m/s. What is the value of the acceleration due to gravity on this other planet?

40 m

What are we looking for? The key to solving this problem is to find the coefficient of friction between the car and the surface of the planet it’s on (denoted by μ). This will be the same for on Earth as well as at the Indy Galactic 500

How do we find μ? Take note that the centripetal force acting on the car is friction. Therefore:

F(centripetal) = F(friction)

Here’s where some Algebra kicks in: F(centripetal)

=

F(c) = (mass)(acceleration)

F(Friction) F(f) = μF(normal) =μF(gravity)

= ma

=μ(mass)(gravity) =μmg The masses on either side cancel out and we end up with this formula. We’ll use it later:

ma = μmg a = μg Acceleration = μ(gravity)

So Now We’re on Earth First, let’s find the acceleration of the car using this formula: a = v^2 / r r = 40 m v = 15 m/s

15^2 / 40 = 5.625 m/s/s

Now we find μ Remember this formula? a = μg It can also be rewritten to this: μ = a/g We know that: a = 5.625 and g on planet earth is approx. 9.8

μ = a/g μ = 5.625/9.8 μ = 0.57 Yaaaay, we found μ! Now we can travel to that other planet and find out it’s acceleration due to gravity!

Welcome to the Indy Galactic 500 First thing’s first, we need to find the acceleration of the car on this mystery planet using the same formula we used on earth. v = 8.4m/s a = 8.4^2/40 = 1.764m/s/s r = 40m

Remember: a = μg a = 1.746m/s/s μ = 0.57 Therefore: 1.746 = 0.57(gravity) Divide both sides by 0.57 and we get: The Answer:

Acceleration due to gravity = 3.09m/s/s