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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 6.35 m/s in 11.0 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? i. The angular acceleration would decrease. ii. The angular acceleration would increase. iii. There would be no change to the angular acceleration.

Sagot :

Whitneytr12idn

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!  

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