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23 January, 12:13

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.

+1
Answers (1)
  1. 23 January, 13:45
    0
    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.
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