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jolli1 [7]
3 years ago
15

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

s that lies in the plane of the plate, passes through the center of the plate, and is parallel to the side with length b.
Physics
1 answer:
slega [8]3 years ago
4 0
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
where dm =M(b*dr)/(ab)
Now we find the moment of inertia by integrating from -a/2 to a/2
The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



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A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi
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The tension is 75.22 Newtons

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Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

v= \lambda f (2)

We can equate expression (1) and (2):

\sqrt{\frac{TL}{M}}=\lambda f

Solving for T

T= \frac{M(\lambda f)^2}{L} (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic N_{harmonic} is:

\lambda_{harmonic}=\frac{2l}{N_{harmonic}}

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

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We can now find T on (3) using all the values we have:

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False

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