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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 magnitude of its angular momentum around the origin is:

Respuesta :

zeeshanabbas777
As we know that
L=r^2mw
=(0.75)^2 ×2×12
hamzaahmeds

This question involves the concepts of angular momentum and velocity.

The magnitude of the angular momentum is "6 J.s".

Angular momentum is given by the following formula:

[tex]L=mvr[/tex]

where,

L = angular momentum = ?

m = mass of block = 2 kg

r = radius of circle = 0.5 m

ω = angular speed = 12 rad/s

v = linear velocity = rω = (0.5 m)(12 rad/s) = 6 m/s

Therefore,

L = (2 kg)(6 m/s)(0.5 m)

L = 6 J.s

Learn more about angular momentum here:

https://brainly.com/question/15104254?referrer=searchResults

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