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As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s.

The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The
component in the xy plane of the angular momentum around the origin has a magnitude of:
A. 0
B. 6.0 kg · m2/s
C. 9.0 kg · m2/s
D. 11 kg · m2/s
E. 14 kg · m2/s

1 Answer

1 vote

Answer: B) 6.0 kg * m^2/s

Explanation: To find the angular momentum of the block, we would have to use the formula for angular momentum, which is L = Iw, where L is the angular momentum, I is the moment of inertia, and w is the angular speed. We would need to find the moment of inertia, which can also be written as mr^2. We already know the mass, radius, and angular speed, so we can plug all these values into L = m * r^2 * w. Don't make the mistake of confusing r for the radius of the circle rather than the distance from the axis of rotation and the object. The r in this case is the distance from the axis of rotation and the block, not the radius of the circle. Simplifying the equation would get you 6.0 kg * m^2 /s. I attached an image for the work. Hope this helped!

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of-example-1
User Sudhir Tataraju
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