A 6.0-kg particle moves to the right at 4.0 m/s as shown. The magnitude of its angular momentum about the point O is: A) 0 kg∙m²/s B) 288 kg∙m²/s C) 144 kg∙m²/s D) 24 kg∙m²/s E) 249 kg∙m²/s

Question

Quizplus · Accepted Answer

The angular momentum L of a particle about point O is given by L = r x p, where r is the position vector from O to the particle, and p is the momentum vector of the particle. Since the particle is moving to the right, its momentum vector is p = mv to the right. We need to find the position vector r and take its cross product with p to find L.

The position vector r is perpendicular to the direction of motion of the particle and passes through O. Therefore, we can draw a line segment from the particle to O that is perpendicular to the direction of motion, and the magnitude of r is the length of this line segment.

We can use the Pythagorean theorem to find the length of this line segment: r = sqrt((1.0 m)^2 + (2.0 m)^2) = sqrt(5) m. (Note that this is the hypotenuse of a right triangle with legs of length 1.0 m and 2.0 m.)

The direction of r is perpendicular to the plane of the diagram and into the page, so we need to use the right-hand rule to determine its direction for the cross product. If we curl the fingers of our right hand from r to p, our thumb points into the page. Therefore, the direction of L is into the page.

The magnitude of L is given by L = rp sin(theta), where theta is the angle between r and p. Since r and p are perpendicular, sin(theta) = 1, so L = rp = (sqrt(5) m)(6.0 kg)(4.0 m/s) = 144 kg m^2/s. Therefore, the answer is choice C.

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