Question

A merry-go-round starts from rest and accelerates uniformly over 10.0 s to a final angular velocity of 4.55 rev/min.
(a) Find the maximum linear speed of a person sitting on the merry-go-round 6.00 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

Sol:

Given
Time(t)= 10 s
final angular velocity = 4.55 rev/min

(a)
The maximum linear speed of a person sitting on the merry-go-round 6 m from the center will be obtained as

Angular velocity (w) =4.55rev/min
=0.4764 rad/s

V=w*r
=0.4764 x 6
= 2.858 m/s

(b)
the person's maximum radial acceleration.

ar=rw^2
= 6 x 0.4764^2
= 1.36 m/s^2

(c)
the angular acceleration of the merry-go-round
a=(w-0)/t
=(0.4764 - 0)/10
= 0.0476 rad/s^2

(d) Find the person's tangential acceleration.
a=0.04764x 6
= 0.285 m/s^2

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