Question

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 5.65 m/s in 13.0 s, what is the angular acceleration, in rad/s2, of the hoop?

(b)If the radius of the hoop were smaller, how would this affect the angular acceleration of the hoop?

Answer 1

(a) ω = v/r = 5.65/(0.450) = 12.555... rad/s.
α = Δω/Δt = ω/t = 12.555.../13.0 = 0.966 rad/s^2 (≈0.966 rad/s^2).

(You can also use α = a_t/r with a_t = Δv/Δt = 5.65/13.0 ≈ 0.4346 m/s^2, so α = 0.4346/0.450 ≈ 0.966 rad/s^2.)

(b) α = a_t/r, so for the same linear acceleration a_t a smaller radius gives a larger angular acceleration (α increases inversely with r).