Question

3. A person takes a ride on a 20-meter-diameter Ferris wheel that rotates with a constant speed of 10 radians per second. Calculate the magnitude and direction of the force that the seat exerts on a 10 kilogram person when he is (a) at the top of the wheel, (b) at the bottom of the wheel, and (c) halfway up the wheel. 4. A thin, uniform rod of length L and mass M swings from a pivot at one end, similar to a pendulum. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. When the rod is vertical, what are (a) the angular speed and (b) the tangental speed of its free end?

Answer 1

At the top FBD (Note that FBD is some for bottom) Now at the top centripetal force, W - N = m v^2/r = m omega^2 r N = W - m omega^2 r = mg - m omega^2 r = m (theta - omega^2 r) = 10 times (9.8 - 10^2 times 20/2) (Negative sign indicates that it is acting down wards) = -9902 M At the bottom, centripetal force N - W = mv^2/r = m omega^2 r N = W + m omega^2 r = m (theta + omega^2 r) = 10 times (9.8 + 10^2 times 20/2) = 10098 M (Acting upward) FBD of half way N = W = 10 times 9.8 = 98 N = 98 N This is normal force acting upward on the person But here f is the force that provides centripetal force f = mv^2/R = m omega^2 R = 10 times 10^2 times 20/2 = 10000 N theta = tan^-1 98/10000 = 0.56 degree with horizontal