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.

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TopNewestIn 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.

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@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.

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

Thanks (in advance).

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