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/s2, of the hoop?
rad/s2
(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 use the formula:
vf = vi + at
where:
vf = final velocity = 7.70 m/s
vi = initial velocity = 0 m/s
a = angular acceleration (unknown)
t = time = 14.5 s
Rearranging the formula to solve for angular acceleration gives us:
a = (vf - vi) / t
a = (7.70 m/s - 0 m/s) / 14.5 s
a = 0.531 rad/s^2
So, the angular acceleration of the hoop is 0.531 rad/s^2.
(b) If the radius of the hoop were smaller, the angular acceleration would increase. This is because the hoop would have to rotate faster to achieve the same linear speed as a hoop with a larger radius. The angular acceleration is directly proportional to the linear speed of the hoop, so a smaller radius would result in a greater angular acceleration.