A 1.90-kg particle moves in the xy plane with a velocity of v with arrow =...

A 1.90-kg particle moves in the xy plane with a velocity of v with arrow = (4.00 i − 3.70 j) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 i + 2.20 j) m.

Solutions

Expert Solution

Given that :

mass of the particle, m = 1.9 kg

velocity, v = (4 - 3.7 ) m/s

position, r = (1.5 + 2.2 ) m/s

The angular momentum of the particle about the origin which is given as :

using an equation, L = r x p = m (r x v)

where, (r x v) =

1.5 2.2 0

4 -3.7 0

At , we have (1.5 x -3.7 - 4 x 2.2) -14.3 m²/s

then, finally L = (1.9 kg) (-14.3 m²/s)

L = -27.1 (kg.m²/s)

genius_generous answered 10 months ago

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