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What is the angular momentum of a 3-kg uni- form cylindrical grinding wheel of radius 0.2 m when rotating at 1500 rpm (Rotational Inertia of a cylinder is mR^2/2).

User Jongwon
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1 Answer

1 vote

Answer:


L = 9.42 kg m^2/s

Step-by-step explanation:

Angular speed of the cylinder is given as


f = 1500 rpm


f = 1500 round/60 s


\omega = 2\pi f


\omega = 2\pi(25) = 50 \pi

now moment of inertia of the cylinder is given as


I = (1)/(2)mR^2


I = (1)/(2)(3)(0.2)^2


I = 0.06 kg m^2

now we have


L = I\omega


L = (0.06)(50\pi)


L = 9.42 kg m^2/s

User Rob Johnstone
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