Non Uniform Circular Motion

Question

A racecar travels around a circular track with a constant speed v. Describe how the car's velocity vector (magnitude and direction) behaves, as a function of time. Describe how the car's acceleration vector (magnitude and direction) behaves, as a function of time. A second racecar travels around the same track, beginning from rest and accelerating smoothly to higher and higher speeds. Describe how the car's velocity vector (magnitude and direction) behaves, as a function of time. Describe how the car's acceleration vector (magnitude and direction) behaves, as a function of time.

Answer 1

This is a great exercise for understanding centripetal acceleration.

For a race car with constant speed v = ω r and θ = ωt the position of the car on the race track is given by

r = < r cos (ω t), r sin (ω t) >

v = dr/dt = < - r ω sin(ω t), r ω cos(ω t)>

a = d²r/dt² = < - r ω² cos(ω t), -r ω² sin(ω t)>

Notice these are perpendicular as r • v = 0. This means the velocity is tangent to the circle as the car goes around the track. Also notice that r = -ω² a so the acceleration is anti-parallel to the radial vector. Also notice |a| = ω² r which is an expression from first year physics.

If the car accelerates smoothly from rest θ = 1/2 α t².

r = < r cos (1/2 α t²), r sin (1/2 α t²) >

v = dr/dt = < - r α t sin(1/2 α t²), r α t cos(1/2 α t²)>

a = d²r/dt² = < - r α sin(1/2 α t²) - r α² t² cos(1/2 α t²), r α cos(1/2 α t²) - r α² t² sin(1/2 α t²)>

Notice the perpendicular relationship still holds r • v = 0. This means the velocity is tangent to the circle as the car goes around the track. However it is no longer true r = -ω² a. As time increases the radial component of acceleration vector does increase much more than the angular component but never fully points toward the origin.