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

Hope this helps you..

Please VOTE my answer

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