Answer:
a) [tex] \alpha = 1.28 rad/s^{2} [/tex]
b) Option ii. The angular acceleration would increase
Explanation:
a) The angular acceleration is given by:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular speed = v/r
v: is the tangential speed = 6.35 m/s
r: is the radius = 45.0 cm = 0.45 m
[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)
t: is the time = 11.0 s
α: is the angular acceleration
Hence, the angular acceleration is:
[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]
b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).
Therefore, the correct option is ii. The angular acceleration would increase.
I hope it helps you!
