A child pushes a merry go-round from rest to a final angular speed of 0.78 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 1.6 revolutions. What is the angular acceleration of the merry go-round?
v=d/t
.78rev/s = 1.6 rev / t
t= 1.6 rev/ .78 rev/s = 2.055 s
a= delta v/delta t
.78rev/s / 2.055s = .3804 rev/s/s
.3804 rev/s/s * 2pi = 2.39 rad/s
this is apparently wrong according to my online hw. Where did i go wrong?
Sex
You have correctly calculated the final time (t) it took for the merry-go-round to complete 1.6 revolutions. However, the formula for angular acceleration (a) is given by:
a = (ωf - ωi) / t
where ωf is the final angular speed and ωi is the initial angular speed (which is 0 in this case since the merry-go-round starts from rest). Let's plug in the values:
a = (0.78 rev/s - 0 rev/s) / 2.055 s
a = 0.78 rev/s / 2.055 s
a ≈ 0.3798 rev/s²
So, the angular acceleration of the merry-go-round is approximately 0.3798 rev/s². To convert it to radians per second squared (rad/s²), we need to multiply by 2π since there are 2π radians in one revolution:
a = 0.3798 rev/s² * 2π rad/rev
a ≈ 2.386 rad/s²
Therefore, the correct angular acceleration of the merry-go-round is approximately 2.386 rad/s².