Question

A bird flies overhead from where you stand at an altitude of 280.0ĵ m and at a velocity horizontal to the ground of 12.0î m/s. The bird has a mass of 2.4 kg. The radius vector to the bird makes an angle θ with respect to the ground. The radius vector to the bird lies in the xy-plane. What is the bird's angular momentum (in kg · m²/s) about the point where you are standing? (Express your answer in vector form.)

Answer 1

We know that Angular momentum is given by:

L = rxP

P = linear momentum = m*V

L = rx(m*V) = m*(rxV)

Now given that:

m = mass of bird = 2.4 kg

r = displacement vector suppose you're at origin = rf - ri = 280.0 j - (0 i + 0 j) = (280.0 j) m

V = Velocity of bird = (12.0 i) m/s

So,

L = 2.4*(280.0 j)x(12.0 i)

L = 2.4*280.0*12.0*(jxi)

In cross product: jxi = -k, So

L = 8064 (-k)

L = Angular momentum = (-8064 k) kg.m^2/sec

Here -k means angular momentum is directed into the ground

Use +ve value only if you need magnitude