An accident, luckily with no casualties, happens at an amusement park. The accid

Question

An accident, luckily with no casualties, happens at an amusement park. The accident involves a high speed carousel, composed by:

A series of seats attached to the at the free ends of each arm (negligible mass).

One of the seats, carrying a 65 kg passenger (mp), breaks during operation at nominal speed. The passenger is therefore released and after a short flight, he lands on a mk 150 kg pop-corn kiosk (of negligible height), luckily unharmed.

An investigation has been opened and you, as a member of the investigation committee, have been asked to reconstruct the details of the accident. The ground on which carousel and pop-corn kiosk are installed is flat (base of the shaft aligned with the kiosk). You are given the following information:

As a result of the accident, the damaged kiosk with the poor passenger is found at a distance s = 5 m with respect to its original position.

Determine

A. how fast the carousel was rotating

B. what was the radial tension on the arm due to the passenger when the accident happened

C. what was the speed of the passenger just before the impact

D. what was the distance between the kiosk and the carousel

Explanation / Answer

Part A)

We will start by finding the acceleration of the kiosk...

F = uFn (Fn = mg of the kiosk and person)

By Newton's Second Law... F = ma, thus

ma = umg (mass cancels)

a = ug

a = (.6)(9.8) = 5.88 m/s2

Then we can find the initial velocity...

vf2 = vo2 + 2ad

0 = vo2 + 2(5.88)(5)

vo = 7.67 m/s

Then by conservation of momentum...

(m + m)v = mv + mv

(150 + 65)(7.67) = 65(v)

v = 25.4 m/s

Part B)

T = mv2/r

T = 65(25.4)2/10

T = 4181 N

Part C)

The passenger had a horiizontal speed of 25.4 m/s

The vertical speed is found by vf2 = 0 + 2(9.8)(3)

vf = 7.67 m/s

The net speed is from the Pythagorean Theorem

v2 = (7.67)2 + (25.4)2

v = 26.5 m/s

Part D)

To fall the 3 m, we need the time that happened

d = vot + .5at2

3 = 0 + .5(9.8)(t)2

t = .782 sec

Then how far does the rider go in .782 sec...

d = vt

d = (25.4)(.782)

d = 19.9 m

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