logearav

Thanks a lot for the reply. Let me explain, the particle P is moving along the circumference of the circle. Now the velocity of the particle can be resolved into two components vsin and vcos. Since vsin is perpendicular to the vertical diameter, it has no effect on vertical diameter. My question is why the perpendicularity of vsin to OY(vertical diameter) results in no effect on OY. Hope i clarified. Thanks again for your effort in answering my query

Why none has answered my question?

Thanks a lot swansont. You have explained very beautifully. I got it.

/ Thanks for the reply. So, whenever the particle is at 0 ,90,180,270 and 360 degree of the circle, that is at the end of previous quadrant and start of successive quadrant, the velocity wont have vertical component, since it is along the radius of the circle. Did i interpret your answer right?

Revered members, Kindly see my attachment. The linear velocity acts tangentially along AH and BT. The particle moves from A to B. Velocity BT is resolved horizontally and vertically along BC and BD Change in velocity along horizontal direction = vcos(theta) - v Change in velocity along vertical direction = vsin(theta) - 0 Can i know why it is v in first case and 0 in second case?

Revered members, Kindly see my attachment. i) vcosθ in a direction parallel to OY. (ii) vsinθ in the direction perpendicular to OY. The component vsinθ has no effect along vertical diameter YOY' since it is perpendicular to OY. Revered members, why the perpendicularity of vsinθ to OY makes no effect along vertical diameter?