A ferris wheel with a radius of 15 m is rotating at a rate of one revolution every 2 minutes How fast is a rider rising when the rider is 24 m above ground level?

Answer 1

My answer is Pi/50 ms-1. My solution uses geometry rather than calculus though. First, the cirumference of the circle = 30(Pi)m, and the time for one rotation is 120s. Therefore the speed of the rider perpendicular to the 15m radius is (30(pi))/120 = (pi)/40 ms-1 We need the velocity in the vertical direction, where the velocity along the tangent to the circle is (pi)/40. Drawing the triangle which involves this tangent, and the vertical and horizontal components of the velocity, the law of similar triangles says that the angle to which the vertical velocity is the opposite is the same as that formed by the 15m radius, and the vertical line from the diameter of the wheel which is parallel to the ground, which will be 24-15=9m long. This angle is therefore cos-1(9/15). Moving now to the triangle of velocity components, where x is the required vertical velocity, we have: x = sin(cos-1(9/15))*((pi)/40) = (4/5)*(pi)/40 =(pi)/50 The above answer is off by an order of magnitude, since 30(pi)/120 = (pi)/4 m/s The correct answer is (4/5)*(pi)/4 = (pi)/5 m/s.