Asked by Anonymous
A merry-go-round starts from rest and accelerates uniformly over 29.0 s to a final angular velocity of 5.35 rev/min.
(a) Find the maximum linear speed of a person sitting on the merry-go-round 4.50 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.
(a) Find the maximum linear speed of a person sitting on the merry-go-round 4.50 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.
Answers
Answered by
Henry
C=6.28*4.5 = 28.3m/rev.=Circumference.
a. V=5.35rev/min * 28.3m/rev=151m/min.
c. a = (Vf - Vo) / t,
a = (5.35 - 0)rev/min / (29/60)min =
11.1m/min^2.
a. V=5.35rev/min * 28.3m/rev=151m/min.
c. a = (Vf - Vo) / t,
a = (5.35 - 0)rev/min / (29/60)min =
11.1m/min^2.
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