A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its

Question

A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center (the figure ). The linear speed of a passenger on the rim is constant and equal to 7.70 m/s.What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion? What is the direction of the passenger's acceleration as she passes through the lowest point in her circular motion?What is the magnitude of the passenger's acceleration as she passes through the highest point in her circular motion? What is the direction of the passenger's acceleration as she passes through the highest point in her circular motion? How much time does it take the Ferris wheel to make one revolution?

Explanation / Answer

The radius of the Ferris wheel is r = 14.0 m The linear speed of a passenger on the rim is constant and equal to v = 7.70 m/s We take the positive y-direction as upward in both cases.Let nT be the upward normal force the seat applies to the passenger at the top of the circle,and let nB be the normal force at the bottom.At the top the acceleration has magnitude (v^2/r),but its vertical component is negative because its direction is downward.Hence ay = -(v^2/r),and Newton's second law at the top tells us that Top:Fy = nT + (-mg) = -m(v^2/r) or nT = m * (g - v^2/r) At the bottom the acceleration is upward,so ay = +(v^2/r) and Newton's second law is Bottom:Fy = nB + (-mg) = +m(v^2/r) or nB = m * (g + v^2/r) where m is the mass of the passenger and g = 9.8 m/s^2 Let the time taken by the Ferris wheel to make one revolution be t We know that v = r * w or w = (v/r) We know from the relation w^2 - wo^2 = 2zA where wo = 0,z is the angular acceleration and A = 2pi radians = 2 * 3.14 radians = 6.28 radians or z = (w^2 - wo^2/2A) We know that w = wo + zt or t = (w - wo/z)

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