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Simora [160]
4 years ago
8

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

constant tangential force of 290 N applied to its edge causes the wheel to have an angular acceleration of 0.816 rad/s2. (a) What is the moment of inertia of the wheel? kg
Physics
1 answer:
lakkis [162]4 years ago
3 0

Answer:

Explanation:

Torque produced = tangential force x radius

= 290 x .33

= 95.7 N.m

angular acceleration = Torque / moment of inertia

.816 = 95.7 / moment of inertia

moment of inertia = 95.7 / .816

= 117.28 kgm².

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  3. Multiply:                        \frac{dA}{dt} = 28800 \ m^2/year
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