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NCERT Exemplar · Q25

Q.A reference particle P moves anticlockwise at constant speed around a circle of radius RR centred at the origin O, with the xx-axis horizontal and the yy-axis vertical. At the instant considered, P lies in the first quadrant (above the xx-axis and to the right of the yy-axis), its radius vector making an angle (ωt+ϕ)(\omega t+\phi) with the positive xx-axis. Let P′P' be the foot of the perpendicular dropped from P onto the xx-axis (the xx-projection of P). What will be the sign of the velocity of the point P′P'?

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The projection P′P' has position x=Rcos⁡(ωt+ϕ)x=R\cos(\omega t+\phi) and velocity x˙=−Rωsin⁡(ωt+ϕ)\dot x=-R\omega\sin(\omega t+\phi). Since P is in the first quadrant, the angle is between 00 and 90∘90^\circ, so sin⁡\sin is positive and x˙\dot x is negative — the projection is moving in the negative xx-direction.

Concept: velocity of the projection

For P on the circle, its coordinates are

x=Rcos⁡(ωt+ϕ),y=Rsin⁡(ωt+ϕ).x=R\cos(\omega t+\phi),\qquad y=R\sin(\omega t+\phi).

The xx-projection P′P' therefore has velocity

vP′=dxdt=−Rωsin⁡(ωt+ϕ).v_{P'}=\frac{dx}{dt}=-R\omega\sin(\omega t+\phi).

Apply the given position …

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