The rod's mass moment of inertia is 5kgm².
<h3>Moment of Inertia:</h3>
The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.
The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.
If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;
or = M / L = dm / dl
dm = (M / L) dl
Here the variable of the integration is the length (dl). The limits have changed from M to the required fraction of L.
![I = \frac{M}3L}[(\frac{L^3}{2^3} - \frac{-L^3}{2^3} )]\\\\I = \frac{1}{12}ML^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BM%7D3L%7D%5B%28%5Cfrac%7BL%5E3%7D%7B2%5E3%7D%20%20%20-%20%5Cfrac%7B-L%5E3%7D%7B2%5E3%7D%20%29%5D%5C%5C%5C%5CI%20%3D%20%5Cfrac%7B1%7D%7B12%7DML%5E2)
Mass of the rod = 15 kg
Length of the rod = 2.0 m
Moment of Inertia, I = 
= 5 kgm²
Therefore, the moment of inertia is 5kgm².
Learn more about moment of inertia here:
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Answer:
what if I do and b then someone else c I don't have enough time pls
Answer:
Explanation:
This problem is based on conservation of angular momentum.
moment of inertia of larger disc I₁ = 1/2 m r² , m is mass and r is radius of disc . I
I₁ = .5 x 20 x 5²
= 250 kgm²
moment of inertia of smaller disc I₂ = 1/2 m r² , m is mass and r is radius of disc . I
I₂ = .5 x 10 x 2.5²
= 31.25 kgm²
3500 rmp = 3500 / 60 rps
n = 58.33 rps
angular velocity of smaller disc ω₂ = 2πn
= 2π x 58.33
= 366.3124 rad /s
applying conservation of angular momentum
I₂ω₂ = ( I₁ +I₂) ω , ω is the common angular velocity
31.25 x 366.3124 = ( 250 +31.25) ω
ω = 40.7 rad / s .
When the bus starts moving forward, the man remains at rest,
causing him to lean back.
When the bus slows down, the man continues to move forward,
and appears to lean forward.
Both events are examples of the effect of inertia.
We want a sound wave with a wavelength of 0.52 meters or a natural fraction thereof. We'll work in MKS.
w = 0.52/n
That's length. We have speed 344 meters/second so w corresponds to a frequency of
f = 344 / w = n (344/.52)
f = 661.5 n Hertz
I don't really agree with how they're saying it, but all the fundamental talk is probably trying to tell us n=1,
Answer: 661.5 Hertz
Any multiple of that will also produce constructive interference; we can go to about n=30 before we're out of the audio range.
