Answer:
1) 23.45 rad/s²
2) 2.7 m/s²
3) t= 1.6 s
4) x ≈ 11 m
5) vfinal = 4.45 m/s
6) KErot = 16.2 J
KEtran = 41 J
KErot < KEtran
Explanation:
Step 1: Data given
mass bowling ball = 4.1 kg
radius = 0.117 meter
initial speed = 8.9 m/s
1) What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?
α = a / r = 2.774 m/s² / 0.117m = 23.45 rad/s²
2)What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?
a = µ*g = 0.28 * 9.8m/s² = 2.744 m/s² ≈ 2.7 m/s²
3) How long does it take the bowling ball to begin rolling without slipping?
This begins when ω = v / r
with
⇒ ω = α*t = 23.45 rad/s² * t
⇒ v = Vo - a*t = 8.9m/s - 2.744m/s²*t
This gives us:
23.45rad/s² * t = (8.9m/s - 2.744m/s²*t) / 0.11m
2.744*t = 8.9 - 2.744*t
t = 8.9 / 5.488 = 1.622 s ≈ 1.6 s
4) How far does the bowling ball slide before it begins to roll without slipping?
x = Vo*t - ½at² = (8.9*1.622 - ½*2.744*(1.622)²) m = 10.82 m ≈ 11 m
5) What is the magnitude of the final velocity?
v = Vo - at = 8.9m/s - 2.744m/s² * 1.622s = 4.45 m/s
6) After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:
trans KE = ½ * 4.1kg * (4.45m/s)² =40.595 J ≈ 41 J
I = (2/5)mr² = (2/5) * 4.1kg * (0.117m)² = 0.0224 kg·m²
ω = v/r = 4.45m/s / 0.117m = 38.03 rad/s, so
rot KE = ½Iω² = ½ * 0.0224kg·m² * (38.03rad/s)² = 16.2 J
16.2 J < 41 J
KErot < KEtran
(For a rolling solid sphere, KErot ≈ 2/5 * KEtran)
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