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3 years ago

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A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is th

e moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring? Assume that each spoke extends from the center to the outer ring is of negligible thickness.

1 answer:

Genrish500 [490]3 years ago

3 0

Answer:

$L^2(\dfrac{8m}{3}+m_r)$

Explanation:

m = Mass of each rod

L = Length of rod = Radius of ring

$m_r$ = Mass of ring

Moment of inertia of a spoke

$\dfrac{mL^2}{3}$

For 8 spokes

$8\dfrac{mL^2}{3}$

Moment of inertia of ring

$m_rL^2$

Total moment of inertia

$8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)$

The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is $L^2(\dfrac{8m}{3}+m_r)$ .

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In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

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Substituting this into our expression:
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