A merry-go-round starts from rest and accelerates uniformly over 19.5 s to a final angular velocity of 6.75 rev/min. (a)...

Question

Question

A merry-go-round starts from rest and accelerates uniformly over 19.5 s to a final angular velocity of 6.75 rev/min.

(a) Find the maximum linear speed of a person sitting on the merry-go-round 6.50 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

Answer #1

a] maximum linear speed v = w*r

= 6.75*2pi/60 rad/s *6.50 m

= 4.5946 m/s

b] maximum radial acceleration = w^2r

= [6.75*2pi/60]^2*6.50

= 3.248 m/s^2

c] angular acceleration alpha = [w2 -0]/t = [6.75*2pi/60]/19.5 = 0.03625 rad/s^2

d] tangential acceleration = alpha*r = 0.03625*6.50= 0.2356 m/s^2

Answer 2

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