A Ferris wheel with radius 9.7 m rotates at a constant rate, completing one revo

Question

A Ferris wheel with radius 9.7 m rotates at a constant rate, completing one revolution every 37.8 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.231 rad/s2 when a passenger is at the top of the wheel. Find the magnitude and direction of the passenger's acceleration at that time. I know that the magnitude is 2.25667 m/s^2, but I am not sure about the direction (degrees below direction of travel)

Explanation / Answer

Without g, the radial acceleration Ar = w^2 r = (2pif)^2 r = (2*pi/37.8)^2 * 9.7 = .268 pointing up and outward along the radius r = 9.7 m. The tangential acceleration (deceleration) is At = alpha*r = .231*9.7 = -2.24 m/sec^2, the negative sign indicates pointing back away from the direction of rotation (deceleration) and perpendicular to r. So the vectors, including g, look something like: Ar ^ | -----------> At | | | v g As you expected the vector sum to be pointing downward, that can only be if you are including gravity g. Otherwise, the sum of vectors would be pointing outward (up) and back. So with g included, Ar' = Ar - g = .268 - 9.8 = - 9.5 m/sec^2 inward along the radius So that A = sqrt(At^2 + Ar'^2) = sqrt(2.24^2 + 9.5^2) = 9.76 m/sec^2 And arctan(Ar'/At) = arctan(-9.5/-2.24) = 76.73 deg below the tangent line (direction of motion) but still pointing backward against the turn direction.

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