OldMidterm2

# OldMidterm2 - oldmidterm 02 PAPAGEORGE MATT Due Mar 3 2008...

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oldmidterm 02 – PAPAGEORGE, MATT – Due: Mar 3 2008, 6:00 pm 2 The period of the Ferris wheel is T = 60 s / 3 = 20 s . The speed of the wheel is v = 2 π r T = 2 π (9 m) 20 s = 2 . 82743 m / s , so the centripetal acceleration is a = v 2 r = (2 . 82743 m / s) 2 9 m = 0 . 888264 m / s 2 . The force exerted by the seat balances the gravity and provides the centripetal force, so F l = m [ g + a ] = (37 kg) (9 . 8 m / s 2 + 0 . 888264 m / s 2 ) = 395 . 466 N . Question 3, chap 6, sect 99. part 1 of 1 10 points A(n) 21 kg boy rides a roller coaster. The acceleration of gravity is 9 . 8 m / s 2 . With what force does he press against the seat when the car moving at 8 m / s goes over a crest whose radius of curvature is 11 m? Correct answer: 83 . 6182 N (tolerance ± 1 %). Explanation: The boy is executing circular motion at the top of the crest, so the net force is centripetal. Two forces act on him: (1) his weight acting down and (2) the supporting normal force from the seat which acts up vector F net = Mvectora vector F g − N = Mv 2 r ˆ k ⇒ N = vector F g M v 2 r ˆ k = M parenleftbigg g v 2 r parenrightbigg ˆ k = (21 kg) × bracketleftbigg 9 . 8 m / s 2 (8 m / s) 2 11 m bracketrightbigg ˆ k = 83 . 6182 N The force with which he presses against the seat is equal but opposite to the normal force.
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