A customer sits in an amusement park ride in which the compartment is to be pull

Question

A customer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a y axis with an acceleration magnitude of 1.24 g, with g = 9.80 m/s2. A 0.649 g coin rests on the customer's knee. Once the motion begins and in unit-vector notation, what are projections of the coin's acceleration relative to the ground on (a)x-axis, (b)y-axis, (c)z-axis and relative to the customer on (d)x-axis, (e)y-axis, (f)z-axis? (g) How long does the coin take to reach the compartment ceiling, 2.4 m above the knee? What are projections of the actual force on the coin on (h)x-axis, (i)y-axis, (j)z-axis and projections of the apparent force according to the customer's measure of the coin's acceleration on (k)x-axis, (l)y-axis, (m)z-axis?

Explanation / Answer

a) Relative to the ground, the coin moves at 9.8m/s^2 towards the ground. In unit vector notation we write -9.8 j m/s^2. This is because the compartment (and the customer who is tied to the compartment) falls faster than g, thus "leaving the coin behind" to fall at its natural acceleration of g.

b) Relative to the customer, who is travelling at -1.24g j, the coin is travelling at an acceleration of
- g j - (-1.24g) j = -g j + 1.24g j = 0.24g j. (Because a vector B relative to a vector A, is B - A)

c) From part b), relative to the compartment, the coin is accelerating upwards at a rate of 0.24g. (This is because the customer and compartment speeds are equal). The coin must travel a distance of 2.2m relative to the compartment. We wish to determine the time. We thus use the equation:
s = ut + 0.5*at^2
where s = distance, u = initial velocity (in this case 0), a = acceleration, t = time.
Therefore we have,
2.2 = 0 + 0.5*0.24*g*t^2
Therefore,
2.2 = 0.12*9.8*t^2
Therefore,
2.2 = 1.176t^2
Therefore,
1.87 = t^2
Therefore,
t = 1.367 seconds

d) Force = mass*acceleration
= 0.567*-g j
=- 0.567*9.8 j
= -5.56 j Newtons

e) Apparent force = mass*apparent acceleration relative to customer
= 0.567*0.24*g j
= 0.567*0.24*9.8 j
= 1.33 j Newtons

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