Looks right to me.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 $$0^o$$?
Yes.e-pie said:Thanks.I was so confused.
And the change in $$\vec{v}$$ is $$180^o$$ since the direction changes?
Circular motion is the movement of an object along a circular path, where the distance between the object and a fixed point remains constant. It can occur in both horizontal and vertical planes.
Vector angular velocity is a vector quantity that describes the rate of change of an object's angular position with respect to time. It is a measure of how fast the object is rotating and in which direction.
In circular motion, the direction of the object's velocity is constantly changing, which means the direction of the object's angular velocity is also changing. The magnitude of the vector angular velocity remains constant, but the direction changes as the object moves along the circular path.
The vector angular velocity of an object in circular motion can be calculated by dividing the change in the object's angular displacement by the change in time. It can also be calculated by multiplying the angular speed (rate of change of angular displacement) by the radius of the circular path.
In circular motion, the linear velocity (tangential velocity) of an object is directly proportional to its angular velocity and the radius of the circular path. This means that as the angular velocity increases, so does the linear velocity, and vice versa.