Calculating Acceleration of Point P Relative to O

[Melawrghk]

Homework Statement

Point P rotates around the center O at a constant rate.The magnitude of the velocity of point P relative to O is 3.7 m/s and the distance between the two points is 90 mm. What are the magnitudes of (a) normal and (b) tangential components of the acceleration of P relative to O?

Homework Equations

a_n = v²/r

The Attempt at a Solution

So I was able to find the normal component of acceleration using the formula a_n = v²/r and got value of 152.11 m/s/s. But for tangential component, I'm confused, the point doesn't actually accelerate (it travels at a constant speed). So does that mean that a_t would equal to zero? Or am I misunderstanding something here?

Thank you!

 

[fluidistic]

I think you are right, the tangential acceleration of point P must be $$\frac{0m}{s^2}$$ otherwise it's SPEED (and not velocity as written) wouldn't be a constant with respect to point 0.

 

[Melawrghk]

Sweet, thanks!

This is for another question, but I don't want to open another topic. So I'm given equation of a path through which the particle travels y=0.2sin(pi*x) and it has a uniform speed in the x direction of 2 m/s. We're required to find velocity at x=0.25. So can I take the derivative of the path equation in order to find vertical velocity of the particle? And then combine the two (x&y direction) to find resultant?