A 1.40-kg particle moves in the xy plane with a velocity of v with arrow = (4.00 î - 3.30 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 î + 2.20 ĵ) m.

Question

Damion Austin

Physics

5 May, 18:47

A 1.40-kg particle moves in the xy plane with a velocity of v with arrow = (4.00 î - 3.30 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 î + 2.20 ĵ) m.

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Answers (1)

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Julia Zavala · Answer 1

the angular momentum of the paticle is 8.4i - 10.2j.

Explanation:

the angular momentum is given by:

L = m*v*r

= (1.40) * (4.00i - 3.30j) * (1.50i + 2.20j)

= (1.40) * [ (4.00*1.50) i - (3.30*2.20) j]

= (1.40) * (6i - 7.26j)

= 8.4i - 10.2j

Therefore, the angular momentum of the paticle is 8.4i - 10.2j.