Asked by t
What is the final velocity (in m/s) of a hoop that rolls without slipping down a 4.50-m-high hill, starting from rest?
All Answers 1
Answered by
GPT-5 mini
AI
Use energy conservation. Initial potential energy mgh becomes translational plus rotational kinetic energy:
m g h = 1/2 m v^2 + 1/2 I ω^2.
For a hoop I = m r^2 and ω = v/r, so rotational KE = 1/2 m v^2. Thus
m g h = (1/2 m v^2) + (1/2 m v^2) = m v^2 → v = sqrt(g h).
With h = 4.50 m and g = 9.80 m/s^2:
v = sqrt(9.80 · 4.50) ≈ 6.64 m/s.
m g h = 1/2 m v^2 + 1/2 I ω^2.
For a hoop I = m r^2 and ω = v/r, so rotational KE = 1/2 m v^2. Thus
m g h = (1/2 m v^2) + (1/2 m v^2) = m v^2 → v = sqrt(g h).
With h = 4.50 m and g = 9.80 m/s^2:
v = sqrt(9.80 · 4.50) ≈ 6.64 m/s.
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