Question

A merry-go-round starts from rest and accelerates uniformly over 26.5 s to a final angular velocity of 6.90 rev/min.

(a) Find the maximum linear speed of a person sitting on the merry-go-round 7.25 m from the center.


(b) Find the person's maximum radial acceleration.


(c) Find the angular acceleration of the merry-go-round.


(d) Find the person's tangential acceleration.

Answer 1

Given

   merry-go-round

accelerating from rest uniformly in 26.5 s to 6.90 rev/min

   angular speed W = 6.90*2pi/60 rad/s = 0.72257 rad/s

radius of the merry-go-round is r = 7.25m

a) linear speed is v = r*w = 7.25*0.72257 m/s = 5.239 m/s

b) radial acceleration a = v^2/r = w^2*r = 0.72257^2*7.25 m/s2 = 3.7853 m/s2

c) angular acceleration of merry-go-round is = alpha = W/t = 0.72257/26.5 rad/s2 = 0.0273 rad/s2

d) Tangential acceleration is a = r*alpha = 7.25*0.72257 m/s2 = 5.239 m/s2