Question

A merry-go-round starts from rest and accelerates uniformly over 14.5 s to a final angular velocity of 7.65 rev/min. (a) Find the maximum linear speed of a person sitting on the merry-go-round 4.25 m from the center. (b) Find the person's maximum radial acceleration. (c) Find the angular acceleration of the merry-go-round.

(a) Find the maximum linear speed of a person sitting on the merry-go-round 4.25 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

here,

the time taken, t = 14.5 s

angular frequency , w = 7.65 rev/min = 0.8 rad/s

a)

r = 4.25 m

the maximum linear speed , v = r * w

v = 4.25 * 0.8 m/s = 3.4 m/s

b)

the person's maximum radial acceleration , ar = w^2 * r

ar = 0.8^2 * 4.25 m/s^2 = 2.72 m/s^2

c)

let the angular acceleration of the merry-go-round be alpha

using first equation of motion

w = w0 + alpha * t

0.8 = 0 + alpha * 14.5

solving for alpha

alpha = 5.5 * 10^-2 rad/s^2

d)

the person's tangential acceleration , at = r * alpha

at = 5.5 * 10^-2 * 4.25 m/s^2

at = 0.23 m/s^2