A regulation basketball has a 22 cm di-

ameter and may be approximated as a thin

spherical shell.

How long will it take a basketball starting

from rest to roll without slipping 1.1 m down

an incline that makes an angle of 49.0° with

the horizontal? The acceleration of gravity is

9.81 m/s2.

Answer in units of s.

0.70 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √ (1.2gh)

v = √ (1.2 * 9.81 m/s² * 1.1 m sin 49.0°)

v = 3.13 m/s

The acceleration down the incline is constant, so given:

Δx = 1.1 m

v₀ = 0 m/s

v = 3.13 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (1.1 m) / (3.13 m/s + 0 m/s)

t = 0.704 s

Rounding to two significant figures, it takes 0.70 seconds.