Circular motion - determining expression for a position vector

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SUMMARY

The discussion focuses on deriving the position vector for a particle moving in circular motion with a constant angular velocity, omega. The correct expression for the position vector is given as r(t) = Rcos(omega*t)xhat + Rsin(omega*t)yhat, which accurately represents the particle's position at any time t. The initial condition at t = 0 confirms that the particle starts at (0, R) on the positive y-axis, aligning with the expected behavior of circular motion. The confusion arises from an incorrect interpretation of the position vector at t = 0 in one of the responses.

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1. Consider a particle--the small red block in the figure--that is constrained to move in a circle of radius R. We can specify its position solely by theta(t), the angle that the vector from the origin to the block makes with our chosen reference axis at time t. Following the standard conventions we measure theta(t) in the counterclockwise direction from the positive x axis.

Determine an expression for the position vector of a particle that starts on the positive y-axis at t = 0 (i.e., at t = 0, (x_0, y_0) = (0, R)) and subsequently moves with constant omega.
Express your answer in terms of R, omega, t, and unit vectors x_unit and y_unit.




2. r(t) = Rcos(omega*t)xhat + Rsin(omega*t)yhat



3. R*yhat + Rcos(omega*t)xhat
 
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I may be confused - I see what looks like one question and two answers.
I don't see the answer to #1. When t = 0 your #2 answer works out to (R,0). And when t=0 your #3 answer works out to (R,R). The answer for #1 should work out to (0,R) when t = 0.
 

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