Centripetal Acceleration

[logearav]

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?

[swansont]

Because it's zero. v is a constant at all times, which means if v_x = v, then v_y has to be zero. IOW the velocity at the beginning is purely horizontal.

[logearav]

Because it's zero. v is a constant at all times, which means if v_x = v, then v_y has to be zero. IOW the velocity at the beginning is purely horizontal.

/ 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?

[swansont]

/ 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?

 

When it's at the top or bottom (0 and 180), the velocity is horizontal. At 90 or 270, the velocity is all vertical.

[logearav]

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