(8%) Problem 8: While watching the clouds pass by, you notice a European swallow
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Question
(8%) Problem 8: While watching the clouds pass by, you notice a European swallow flying horizontally h = 23.8 m above you. When the swallow is directly overhead, it drops a m 11.5 kg coconut. From your ornithological studies, you know that the air-speed of this particular species of swallow while carrying such a load is vo = 6.17 m/s ty ©theexpertta.com 20 % Part (a) Calculate the magnitude of the angular momentum L of the coconut in kg·m2/s as observed by you at the moment that it is released directly overhead 20% Part (b) Develop an expression for the x component of the radius vector r() (in meters) of the coconut with respect to you as a function of the given information, the free-fall time t, and variables available in the palette. Let t-0 be the moment that the coconut is released Express your answer in unit vector notation time t, and variables available in the palette. Let t-0 be the moment that the coconut is released. Express your answer in unit vector notation released directly overhead (1-1 sec) hits the ground × 20% Part (c) Develop an expression for the velocity vector v(t) in m/s) of the coconut as a function of the given information, the free-fall 20% Part (d) Calculate the magnitude of the angular momentum L of the coconut in kg-m2/s as observed by you one second after it is 20% Part(e) Calculate the magnitude of the angular momentum L of the coconut in kg-m2/s as observed by you immediately before it Grade Summary Deductions Potential 0% 100%Explanation / Answer
a. at the moment the bird is directly overhead
radius about immediatiate center of rotation, r = h = 23.8 m
mass of coconut, m = 11.5 kg
velocity ( horizontal) vo = 6.17 m/s
so angular momentum = mvr = 11.5*6.17*23.8 = 1688.729 kg m^2/s
b. as there is no horizontal acceleration, the horizontal speed remains vo
so x component of position vectopr at time t x(t) = vo*t
c. the vertical velocity of the coconut would change from 0 to some value v
after time t, v = gt
so, net velocity = sqroot(v^2 + Vo^2) = sqroot(6.17^2 + 9.81^2t^2) = sqroot( 38.0689 + 96.2361 t^2)
d. one minute after it is released
horizontal distance copvered = vO*1 = 6.17 m
erticaldistance covered = 0.5g*1^2 = 4.905 m
so r =sqroot((23.8-4.905)^2 + 6.17^2) = 19.876
angular moemntum = mvr = 19.876*sqroot(38.0689 + 96.2361)*11.5 = 2648.944 kg m^2/s
e. beforte it hits the ground,
time taken to fall = t
23.8 = 0.5gt^2
t = 2.202 s
so before hitting ground
r = vo*t = 13.59 m
v = 22.46 m/s
so L = mvr = 3511.021 kg m^2/s
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