A 1.30-kg particle moves in the xy plane with a velocity of v with arrow = (4.50 î − 3.30 ĵ) m/s. Determine the angular momentum of the part

Question

A 1.30-kg particle moves in the xy plane with a velocity of v with arrow = (4.50 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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Nick 4 years 2021-08-19T00:09:50+00:00 1 Answers 21 views 0

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  1. sori
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    2021-08-19T00:11:10+00:00
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    Answer:

    The angular momentum of the particle about the origin is \vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right].

    Explanation:

    Vectorially speaking, the angular momentum is given by the following cross product:

    \vec l = \vec r \times m\vec v

    This cross product can be solved with the help of determinants and its properties, that is:

    \vec l = \left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\m\cdot v_{x}&m\cdot v_{y}&0\end{array}\right|

    \vec l = m\left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\v_{x}& v_{y}&0\end{array}\right|

    The 3 x 3 determinant is solved by the Sarrus Law:

    \vec l = m \cdot (r_{x}\cdot v_{y} - r_{y}\cdot v_{x})k

    If m = 1.30\,kg, \vec r = 1.50\,i + 2.20\,j\,[m] and \vec v = 4.50\,i-3.30\,j\,\left[\frac{m}{s} \right], the angular momentum of the particle about the origin is:

    \vec l = (1.30\,kg)\cdot \left[\left(1.50\,m\right)\cdot\left(-3.30\,\frac{m}{s} \right)-\left(2.20\,m\right)\cdot\left(4.50\,\frac{m}{s} \right)\right]k

    \vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right]

    The angular momentum of the particle about the origin is \vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right].

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