For particle P revolving around the centre O | vedclass

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Question

(A) The particle $P$ is performing uniform circular motion. At $t=0$,the position vector $OP$ makes an angle of $30^{\circ}$ (or $\pi/6$ radians) with

Vedclass · Accepted Answer

$x(t)=r \cos \left(\omega t+\frac{\pi}{6}\right)$

Vedclass · Answer

(A) The particle $P$ is performing uniform circular motion. At $t=0$,the position vector $OP$ makes an angle of $30^{\circ}$ (or $\pi/6$ radians) with the positive $x$-axis.At any time $t$,the particle rotates by an angle $\omega t$ in the counter-clockwise direction.Therefore,the total angle $	heta$ made by the position vector $OP$ with the positive $x$-axis at time $t$ is $	heta = \omega t + 30^{\circ} = \omega t + \frac{\pi}{6}$.The projection of the position vector $OP$ on the $x$-axis is given by $x(t) = r \cos(	heta)$.Substituting the value of $	heta$,we get $x(t) = r \cos \left(\omega t + \frac{\pi}{6}ight)$.

For particle $P$ revolving around the centre $O$ with radius of circular path $r$ and angular velocity $\omega$,as shown in the figure,the projection of $OP$ on the $x$-axis at time $t$ is .................

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