Question

A merry-go-round starts from rest and accelerates uniformly over 19.5 s to a final angular velocity of 6.25 rev/min.

(a) Find the maximum linear speed of a person sitting on the
merry-go-round 4.00 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

Given

angular velocity of merry ground w = 6.25 rev / min

= 6.25 ( 2*pi rad / 1rev ) ( 1 min / 60 sec )

= 0.654 rad / sec

a) maximum linear speed of a person sitting on the merry-go-round 4 m from the center

v = r*w

= 4 * 0.654

= 2.616 m/s

b) persons maximum radialc acceleration

a _{r} = v ^{2} / r

= ( 2.,616) ^{2} / 4

= 1.71 m/s ^{2}

c) the angular acceleration of the merry-go-round is

= w/t

= 0.654/19.5

= 0.0335 rad/s^{2}

d) persons tangential acceleration

a _{t} = v _{f} - v
_{i} / t

= 2.616-0/19.5

= 0.13415 m /s ^{2}

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