Aparticle 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

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.

its a 5.5 n

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