Question #270779

Angular Momentum



Solve the problems.




1. A small particle of mass 0.20 kg is being whirled in a horizontal circle at the end of a 20 m long string at a constant speed of 1.0 m/s. Determine ito angular momentum about its axis of rotation.



2. A bowling ball has a mans of 5.5 kg and a radius of 12.0 cm. It is released so that it rolls down the alley at a rate of 12 rev/s. Find the magnitude of its angular momentum

1
Expert's answer
2021-11-25T10:14:40-0500

1) Let's first find the angular velocity of the particle:


v=rω,v=r\omega,ω=vr=1.0 ms20 m=0.05 rads.\omega=\dfrac{v}{r}=\dfrac{1.0\ \dfrac{m}{s}}{20\ m}=0.05\ \dfrac{rad}{s}.

Finally, we can find angular momentum of the particle about its axis of rotation:


L=Iω=mr2ω,L=I\omega=mr^2\omega,L=0.20 kg×(20 m)2×0.05 rads=4.0 kg×m2s.L=0.20\ kg\times(20\ m)^2\times0.05\ \dfrac{rad}{s}=4.0\ \dfrac{kg\times m^2}{s}.

2) Let's first convert rev/s to rad/s:


ω=12 revs×2π rad1 rev=75.4 rads.\omega=12\ \dfrac{rev}{s}\times\dfrac{2\pi\ rad}{1\ rev}=75.4\ \dfrac{rad}{s}.

Finally, we can find angular momentum of the bowling ball about its axis of rotation:


L=Iω=25mr2ω,L=I\omega=\dfrac{2}{5}mr^2\omega,L=25×5.5 kg×(0.12 m)2×75.4 rads=2.39 kg×m2s.L=\dfrac{2}{5}\times5.5\ kg\times(0.12\ m)^2\times75.4\ \dfrac{rad}{s}=2.39\ \dfrac{kg\times m^2}{s}.

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