Hi, can you please help me solve the following problem? It\'s notfrom a textbook

Question

Hi, can you please help me solve the following problem? It's notfrom a textbook, but we use the Giancoli Textbook as a resource.My teacher provided the answerbelow. Part a) is fine, however, I can't figure out how he got thatanswer. Please help me by providing a step by step processof how you approached theproblem. Thank you!

A roller coaster car begins a drop from a height of 30m from theground with initial speed of 10m/s. The bottom of the first troughis at a height of 2m from the ground, and its radius of curvatureat the lowest point is 10m. Then the car climbs to a peak at aheight of 15m from the ground, where the radius of curvature at thehighest point is 8m. Ignore friction and air resistance.

a) How many gs do the riders feel at the bottom of thetrough? Ans.: 7.62 g’s.

b) Would an unstrapped rider be flying off her seat at thetop of the 15m high peak? Ans: Getting through the top at thatspeed would require a downward force 4.03 times larger than theweight. The seat cannot provide a downward force, so withoutstraps, the rider will fly off her seat.

Explanation / Answer

A roller coaster car begins a drop from a height of 30m from theground with initial speed of 10m/s. The bottom of the first troughis at a height of 2m from the ground, and its radius of curvatureat the lowest point is 10m. Then the car climbs to a peak at aheight of 15m from the ground, where the radius of curvature at thehighest point is 8m. Ignore friction and air resistance. a) How many gs do the riders feel at the bottom of thetrough? let v = speed at the bottom of the trough. gain KE = loss of PE (conservation of energy) (1/2)m(v2 - 100) = mg(30-2) v2 = 100 + 56g centripetal acceleration = v2 /r = (100 + 56g)/10 =6.62g so the rider feels 7.62 g. b) if u = speed at the top of the 15 m peak, gain KE = loss of PE (conservation of energy) (1/2)m(u2 - 100) = mg(30-15) u2 = 100 + 30g centripetal acceleration = u2 /r = (100 + 30g)/8 = 5.03g>>> remember this is downward. normal gravitation provides 1g. The rest has to come from somewhereelse. The seat does not exert any downward force. so the rider will fly off. hope this helps!

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