Circular Motion #8

Classical Mechanics Level 3

A particle moves along the circle $x^{2 } + y^{2} = r^{2}$ in the $xy$ plane. It is known that the particle can move from $(0, r)$ to $(r, 0)$ without leaving the first quadrant. Which of the following vectors point in the same direction as the angular velocity of the particle, $\vec{\omega}$ ?

Note: In this problem, we are using the Right-Handed Coordinate System

This problem is part of the set - Circular Motion Practice

+ \hat{ k }

+ \hat{ i }

The direction of angular velocity depends on the position of the particle.

- \hat{ j }

- \hat{ k }

+ \hat{ j }

- \hat{ i }

The direction of angular velocity is undefined.

1 solution

Scott Ripperda
Dec 3, 2015

This problem is very simple using the right hand rule. To do this simply point your fingers of your right hand in the direction of (o, r) or $\hat{j}$ , which is directly in front of you, and then curl your fingers towards (r,0) or (\hat{i}), which is towards your right. If done correctly your right hand should be flipped upside down so your thumb will point towards the floor which is $\boxed{-\hat{k}}$ , which is the direction of the angular velocity. This motion gives the direction of the cross products in general and should visually be equivalent the the angular velocity of the particle.

Make sure to use the right hand otherwise you will get the wrong answer.

Circular Motion #8

1 solution

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