(8%) Problem 8: While watching the clouds pass by, you notice a European swallow

Question

(8%) Problem 8: While watching the clouds pass by, you notice a European swallow flying horizontally 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 sucha 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 m /s as observed by you at the moment that it is released directly overhead. M20% Part (b) Develop an expression for the x component of the radius vector r(t) (in meters) of the coconut with respect to you as a function of the given information, the frec-fall time 1, and variables available in the palette. Let -0 be the moment that the coconut is rclcased Express your answer in unit vector notation. 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 time t, and variables available in the palette. Let 0 be the moment that the coconut is released. Express your answer in unit vector notation. 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 released directly overhead (t = 1 sec).

Explanation / Answer

Part (a) : Calculate the magnitude of an angular momentum of the coconut as observed by us at the moment that it is released directly overhead.

using a formula, we have

Linitial = m v0 h sin 900

where, m = mass of coconut = 11.5 kg

v0 = initial speed of coconut released = 6.17 m/s

h = distance from swallow to u = 23.8 m

then, we get

Linitial = (11.5 kg) (6.17 m/s) (23.8 m) (1)

Linitial = 1688.7 kg.m2/s

Linitial = 1.68 x 103 kg.m2/s

Part (d) : Calculate the magnitude of an angular momentum of the coconut as observed by us one second after it is released directly overhead (t = 1 sec).

using equation of motion, we have

v = v0 + a t

v = [(6.17 m/s) + (9.8 m/s2) (1 s)]

v = 15.97 m/s

we know that, Lfinal = m v r sin 450

Lfinal = (11.5 kg) (15.97 m/s) [(23.8 m) / 2] (0.7071)

Lfinal = 1545.3 kg.m2/s

Lfinal = 1.54 x 103 kg.m2/s

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