Circular Motion Question: Change in Vector Angular Velocity

AI Thread Summary
The discussion revolves around the change in angular velocity of a particle moving along a circular path. It clarifies that the angular velocity vector, denoted as $$\vec{\omega}$$, is always perpendicular to the plane of motion. When the particle completes half of its circular path, the change in its velocity vector is indeed $$180^o$$ due to the reversal in direction. Participants confirm the understanding that the angular velocity does not change in magnitude but changes in direction. This reinforces the concept of angular motion in circular dynamics.
e-pie
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Homework Statement
A particle is moving along a circular path of radius $$r$$ with uniform speed $$v$$ Through what angle does its angular velocity change when it completes half of circular path?
Relevant Equations
$$\vec{v}=\vec{\omega}\times \vec{r}$$
Hi i am e-pie's brother and he let me use his account.Since $$\vec{\omega}$$ is always perpendicular to the plane.Shouldn't this be $$0^o$$?
 
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e-pie said:
Problem Statement: A particle is moving along a circular path of radius $$r$$ with uniform speed $$v$$ Through what angle does its angular velocity change when it completes half of circular path?
Relevant Equations: $$\vec{v}=\vec{\omega}\times \vec{r}$$

Hi i am e-pie's brother and he let me use his account.Since $$\vec{\omega}$$ is always perpendicular to the plane.Shouldn't this be $$0^o$$?
Looks right to me.
 
Thanks.I was so confused.
And the change in $$\vec{v}$$ is $$180^o$$ since the direction changes?
 
e-pie said:
Thanks.I was so confused.
And the change in $$\vec{v}$$ is $$180^o$$ since the direction changes?
Yes.
 
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