Centripetal Force of car on sloped track

Circular turns of radius r in a race track are often banked at an angle θ to allow the cars to achieve higher speeds around the turns. Assume friction is not present, and use the coordinate system specified.


Now assume that the car is moving at 17 m/s and the radius of the track is 410 m. What is the angle θ in degrees?

use tan(theta)= (v^2)/r*g
so solve (17m/s ^2 )/ (410m*9.8m/s^2)
tan^-1 (.0719)
the angle should be 4.11 degrees.

Centripetal Force Solutions

A wind turbine blade has a radius of 31 m and a mass of 1380 kg is rotating at 1 rev/s.

Assuming all of the mass is located at the end of the blade, what is the net force (centripetal force in N) acting on the end of the turbine blade?

Find the Centripetal Force

First you will want to change 1rev/s to velocity in m/s

To do so find the total distance traveled

Perimeter of circle=2pi*r  =194.77meters

Therefore: 1rev/s =194.77meters/sec

To find Force you need Mass* Acceleration

So find centripetal acceleration with equation: A(c) = v^2 /r  = 1223.83 m/s^2

Centripetal Force= 1223.83 m/s^2 * 1380kg =1688886.705 N