A child is outside his home playing with a metal hoop and stick. He uses the stick to keep the hoop of radius 45.0 cm rotating along the road surface. At one point the hoop coasts downhill and picks up speed.
(a) If the hoop starts from rest at the top of the hill and reaches a linear speed of 7.70 m/s in 14.5 s, what is the angular acceleration, in rad/s^2, of the hoop?
(b) If the radius of the hoop were smaller, how would this affect the angular acceleration of the hoop?
(a) To find the angular acceleration of the hoop, we can first calculate the final angular velocity using the formula:
v = ωr
Where:
v = linear speed = 7.70 m/s
r = radius of the hoop = 0.45 m
ω = v/r
ω = 7.70 m/s / 0.45 m
ω ≈ 17.11 rad/s
Now, we can use the formula for angular acceleration:
α = (ωf - ωi) / t
Where:
α = angular acceleration
ωf = final angular velocity = 17.11 rad/s
ωi = initial angular velocity (0 since it starts from rest)
t = time taken = 14.5 s
α = (17.11 rad/s - 0 rad/s) / 14.5 s
α ≈ 1.18 rad/s^2
Therefore, the angular acceleration of the hoop is approximately 1.18 rad/s^2.
(b) If the radius of the hoop were smaller, the angular acceleration would increase. This is because a smaller radius would mean a smaller distance for the hoop to travel in the same amount of time, causing it to rotate faster and thus accelerate more angularly. This relationship is depicted by the formula for angular acceleration:
α = a / r
Where:
α = angular acceleration
a = linear acceleration
r = radius
As the radius decreases, the angular acceleration increases.