Help needed in circular motion

In circular motion, We know that the angular displacement is represented by $\theta$. If a particle moves on circle of radius R, then why is \[\hat{\theta}=-sin\theta\hat{i}+cos\theta\hat{j}\]

I am a beginner in Kinematics and Circular Motion. All kinds of help accepted.

#Mechanics #HelpMe! #Help #Pleasehelp #CircularMotion

Easy Math Editor

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Comments

In vector notation the point on a unit circle when at angular position $\theta$ is $cos \theta\hat{i} + sin \theta \hat{j}$ and its differential w.r.t. $\theta$ is given by the expression in consideration.

Rajen Kapur - 6 years ago

@Gautam Jha

That's right.

If you don't yet understand calculus, consider how circular motion is periodic, I.e. if we follow it halfway aaround the circle, its $x$ and $y$ coordinates are negated. If we draw a line from the origin to the location of the particle, we see that the components are given by sin and cos while the angle is changing at a constant rate $\theta=\omega t$ . Thus the positions and the velocities must be given by sine and cosine.

If we start motion on the $x$ axis, the velocity in the $x$ direction must start at zero while the velocity in the $y$ direction starts at positive $\omega$ . Moreover, when we cross the $y$ axis, the particle is moving in the negative $x$ direction at speed - $\omega$ , which shows that the $x$ velocity is given by the negative sine.

Josh Silverman Staff - 5 years, 11 months ago

@Nishant Rai @Ronak Agarwal @Rajen Kapur @Raghav Vaidyanathan

Thanks (in advance).

Gautam Jha - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$