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vredina [299]
3 years ago
5

Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

the rotational inertia for rotation about an axis (a) passing through masses Band C? (b) passing through masses Aand C ? (c) passing through the center of the square and perpendicular to the plane of the square?
Physics
1 answer:
Zigmanuir [339]3 years ago
8 0

Answer:

Part a)

I = 1.5 kg m^2

Part b)

I = 0.75 kg m^2

Part c)

I = 1.5 kg m^2

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

I = mL^2 + mL^2 + m(0) + m(0)

I = 2mL^2

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2

I = 2m\frac{L^2}{2}

I = (3 kg)(0.50^2)

I = 0.75 kg m^2

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2

I = 4m\frac{L^2}{2}

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

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sorry - late reply...just stumbled across tis...hope u can still use it :)


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<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


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so f = 96/20 = 4.8 cm</span>
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