Question

A cat rides a merry-go-round turning with uniform circular motion. At time t1 = 2.00 s, the cat's velocity is (3.00 m/s)i+(4.00m/s)j, measured on a horizontal xy coordinate system. At t2= 5.00s, its velocity is (-3.00m/s)i+(-4.00)j. What are (a) the magnitude of the cat's centripetal acceleration and (b) the magnitude of the cat's average acceleration during the time interval t2-t1, which is less than a period of the motion?

Answer 1

Answer:

The time difference is t₂ - t₁ = 5.00 s - 2.00 s = 3.00 s.

After three seconds the velocity (for this object in circular motion of period T) is reversed. That means, the object takes three second time to reach the opposite side of the circle. Thus T = 2(3.00 s) = 6.00s.

(a) Time period T = 2r/v, from this radius of the circle r = Tv/2, where v is the velocity and its magnitude is v = [ (3.00 m/s)² + (4.00 m/s)² ]^1/2 = 5.00 m/s. Therefore,

radius r = (6.00 s) (5.00 m/s) / 2 = 4.77 m.

The magnitde of the object's centripetal acceleration is therefore

a = v²/r = (5.00 m/s)²/(4.77 m) = 5.24 m/s².

(b) The average acceleratio is given by

a_avg = (v₂ - v₁) / (t₂ - t₁) = [(-3.00 i -4.00 j) m/s] - [(3.00 i + 4.00 j)m/s] / (5.00 s - 2.00 s)

a_avg = (-2.00 m/s²) i + (-2.67 m/s²) j.

which implies la_avgl = [ (-2.00 m/s²)² + (-2.67 m/s²)² ]^1/2 = 3.33 m/s².