A bowling ball has a mass of 5.5 kg and a radius of 12 cm. It is released so that it rolls down alley at a rate of 12 rev/s. What's the magnitude if its angular momentum?

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A bowling ball has a mass of 5.5 kg and a radius of 12 cm. It is released so that it rolls down alley at a rate of 12 rev/s. What's the magnitude if its angular momentum?

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Answer 1 · 2016-12-03T05:09:42

Given
$M \to mass of the sphere = 5.5 k g$ $M->"mass of the sphere"=5.5kg$
$R \to radius of the sphere = 12 c m = 0.12 m$ $R->"radius of the sphere"=12cm=0.12m$
$I \to moment of inertia of the sphere$ $I->"moment of inertia of the sphere"$
$= \frac{2}{5} I R^{2} = \frac{2}{5} \times 5.5 \times {(0.12)}^{2} = 0.03168 k g m^{2}$ $=2/5IR^2=2/5xx5.5xx(0.12)^2=0.03168kgm^2$
$ω \to angular velocity = 2 π \times 12 = 24 π rad/s$ $omega-> "angular velocity"= 2pixx12=24pi" rad/s"$

$Angular momentum = I ω = 0.03168 \times 24 π \approx 2.39 k g m^{2} s^{- 1}$ $"Angular momentum"=Iomega=0.03168xx24pi~~2.39kgm^2s^-1$

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A bowling ball has a mass of 5.5 kg and a radius of 12 cm. It is released so that it rolls down alley at a rate of 12 rev/s. What's the magnitude if its angular momentum?

1 Answer

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