Posted by John on .
A merry-go-round starts from rest and accelerates uniformly over 26.0 s to a final angular velocity of 5.50 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.
(d) Find the person's tangential acceleration.
An angular velocity is
ω =2πn= 5.5•2π/60= 0.58 rad/s.
v= ωR=0.58•4.25=2.45 m/s,
a = ω^2•R=(0.58)^2•4.25=1.43 m/s^2
Since ω = εt , an angular acceleration
ε = ω/t=0.58/26=2.23•10^-2 rad/s^2,
a = ε•R= 2.23•10^-2•4.25=0.095 m/s^2