Question

A child sits on a merry‑go‑round that has a diameter of 6.00 m. The child uses her legs to push the merry‑go‑round, making it go from rest to an angular speed of 18.0 rpm in a time of 43.0 s. What is the average angular acceleration ?avg of the merry‑go‑round in units of radians per second squared (rad/s2)? ?avg= rad/s2 What is the angular displacement Δ? of the merry‑go‑round, in units of radians (rad), during the time the child pushes the merry‑go‑round? Δ?= rad What is the maximum tangential speed ?max of the child if she rides on the edge of the platform? ?max=

Answer 1

Given

The child is on a merry-go-round with

diameter d = 6.0 m and radius r = 3.0 m

from rest the merry-go-round started so initial velocity is zero w1 = 0 rad/s
in time interval of dT = 43.0s , it attains the velocity of w2 = 18.0 rpm

we know that 1 rpm that is one rotation per minute in terms of rad/s is

   1 rpm = 2pi/60 rad/s

so the angular velocity (final) w2 = 18*2pi/60 rad/s = 1.88 rad/s

now By definition of average angular acceleration

   ?_avg = (w2-w1)/(t2-t1)

   ?_avg = (w2-w1)/dT

   ?_avg = (1.88-0)/(43) rad/s^2
  
   ?_avg = 0.044 rad/s^2

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For angular displacement

   from equations of motions Δ? = w1t + 0.5*alpha*t^2

   Δ? = 0*t +0.5*0.044*43^2 rad

   Δ? = 40.678 rad

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the maximum tangential speed ?max of the child if she rides on the edge of the platform

   is Vmax = r*W2 (here W2 is maximum angular speed )

   v max = 3*1.88 m/s = 5.64 m/s