Question

The wheels of a skateboard roll without slipping as it accelerates at 0.30 m/s2 down an 75-m-long hill.

If the skateboarder travels at 1.7 m/s at the top of the hill, what is the average angular speed of the 2.6-cm-radius wheels during the entire trip down the hill?

Express your answer to two significant figures and include appropriate units.

Answer 1

At the top of the hill, ω is:
ω = v/r
ω = (1.7 m/s) / (0.026 m)
ω = 65.384 rad/s

At the bottom of the hill, ω is:
ω² = ω₀² + 2α Δθ

Linear acceleration is constant at 0.45 m/s². So α is:
α = a/r
α = (0.30m/s²) / (0.026 m)
α = 11.538 rad/s²

We also need to find Δθ:
Δθ = L/D
Δθ = (75 m) / (0.052 m)
Δθ = 1442.31 radians

Plugging in:
ω² = ω₀² + 2α Δθ
ω² = (65.384 rad/s)² + 2(11.538 rad/s²) (1442.31 radians)=37557.813
ω = 193.798 rad/s

So the average angular speed is:
ω_avg = (ω₁ + ω₀) / 2
ω_avg = (193.798 rad/s + 65.384 rad/s) / 2
ω_avg = 129.591 rad/s