Question

A particle of mass, m = 3.0kg movers with a uniform velocity, v = (5m/s)I +(3m/s)j. At the time, t, the particle passes through the point r = (2m)i - (3m)j from the origin. Calculate the magnitude and direction of the angular momentum about the origin at time t. (Angular momentum is given by L= mv x R

Answer 1

Answer:

The magnitude is 63 kg m²/s, and the direction is -k.

Explanation:

Plug the values into the equation:

L = m v×r

L = (3.0 kg) (<5 i + 3j> m/s × <2i − 3j> m)

Take the cross product.  The cross product of two dimensional vectors is:

<v₁ i + v₂ j> × <r₁ i + r₂ j> = <(v₁ r₂ − v₂ r₁) k>

Therefore:

L = (3.0 kg) <((5)(-3) − (3)(2)) k m²/s>

L = (3.0 kg) <-21 k m²/s>

Multiply:

L = <-63 k kg m²/s>

The magnitude is 63 kg m²/s, and the direction is -k.