In the figure,what will be the sign of the ve | vedclass

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(B) The position of the particle $P$ on the reference circle is given by $x = R \cos(\omega t + \phi)$.The velocity of the projection point $P_1$ on t

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Negative

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(B) The position of the particle $P$ on the reference circle is given by $x = R \cos(\omega t + \phi)$.The velocity of the projection point $P_1$ on the $x$-axis is the derivative of its position with respect to time:$v = \frac{dx}{dt} = -R\omega \sin(\omega t + \phi)$.As the particle $P$ moves in an anti-clockwise direction in the first quadrant (as shown in the figure),the angle $	heta = (\omega t + \phi)$ is between $0$ and $\pi/2$.In this range,$\sin(\omega t + \phi)$ is positive.Therefore,$v = -R\omega \sin(\omega t + \phi)$ will be negative.Geometrically,as $P$ moves anti-clockwise,its projection $P_1$ moves towards the origin $O$ (i.e.,from right to left),which corresponds to a negative velocity.

In the figure,what will be the sign of the velocity of the point $P_1$,which is the projection of the velocity of the reference particle $P$? $P$ is moving in a circle of radius $R$ in an anti-clockwise direction.

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