Question

A cat rides a merry go round turning with uniform circular motion. At time t1=2.00s, the cat's velocity is gives by Vx = 3.00 m/s and Vy = 4.00 m/s. At t2 = 5.00 s,the cat's velocity is give by Vx = -3.00 m/s and Vy = -4.00 m/s. Take the coordinate system of the merry go round to have its origin at the center.

(a) What is the magnitude of the cat's centripetal acceleration?

(b) What is the magnitude of the cat's average acceleration during the time interval t2-t1?

(c) Assuming uniform circular motion before t1, what would have been the x-y coordinate location of the cat at t0= 0.00 s?

(d) What angle relative to the positive y-axis does the centripetal acceleration vector point at t= 3.00 s?

Answer 1

The given velocities if observed carefully are in the exact opposite direction as each other. This means that the cat has covered half a revolution in time interval 2s to 5s.

So, the angular velocity of the merry go round is

(a) Centripetal acceleration = v²/R

= 25/R, where R is the radius of the ride. (The magnitude of the given velocities is 5 m/s)

(b) Effectively as explained above, velocity changes from +5 m/s to -5 m/s, so, the net change in velocity = -10 m/s

Avg. acceleration = -10/3 m/s²

(c) Point P: at t=0s that is 2s before t₁, it was radians behind its position at t₁.

As shown in the diagram,

(d) At t=3s, at Point H

Angle from the x-axis.

So, relative to y-axis,