Physics Rotation?

[匿名]

2016-10-17 3:49 am

更新1:

Physics Rotation Answer:12.3210

更新2:

Word: A uniform bar has two small balls glued to its ends. The bar is 3.7 m long and has mass 1.2 kg, while the balls each have mass 0.5 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis perpendicular to the bar and through one of the balls. the answer's unit is kgm^2

回答 (1)

天同

2016-10-17 4:57 am

✔ 最佳答案

The moment of inertia of a bar about an axis perpendicular to it and through its centre of mass (i.e. at its middle) is given by ML^2/12, where M is the mass of the bar, L is its length.

Hence, moment of inertia of bar about its central axis
= (1.2 x 3.7^2)/12 kg.m^2
= 1.369 kg.m^2

Using the Parallel Axis Theorem, the moment of inertia of the bar about an axis at its end
= 1.369 + 1.2 x (3.7/2)^2 kg.m^2
= 5.476 kg.m^2

Moment of inertia of the ball at the far end of the bar
= 0.5 x 3.7^2 kg.m^2
= 6.845 kg.m^2

Hence, moment of inertia of the system
= (5.476 + 6.845) kg.m^2
= 12.321 kg.m^2

(note that the ball at the end of the bar where the axis passes through doesn't not contribute to the moment of inertia calculation, because it is regarded as a point mass and is located at zero distance from the axis).

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