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nata0808 [166]
3 years ago
13

Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of R = 1.22 m and a mass of 67.0 kg . To

prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3510 m/s2 .What is the maximum kinetic energy that can be stored in the fly-wheel?
Physics
1 answer:
eduard3 years ago
5 0

Answer: 71.7 KJ

Explanation:

The rotational kinetic energy of a rotating body can be written as follows:

Krot = ½ I ω2

Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:

Fc = m. ac

It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:

ac =  ω2 r

We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.

As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.

Replacing in the expression for the Krot, we have:

Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ

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Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

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To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

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304 = 1/2(13.2496)a

304 = 6.6248a

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a = 45.89m/s²

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Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radi
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Answer:

1.635×10^-3m

Explanation:

Young modulus is the ratio of the tensile stress of a material to its tensile strain.

Young modulus = Tensile stress/tensile strain

Tensile stress = Force/Area

Given force = 130N

Area = Πr² = Π×(1.55×10^-3)²

Area = 4.87×10^-6m²

Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²

Tensile strain = extension/original length

Tensile strain = e/3.9

Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²

2×10¹¹N/m² = 8.39×10^7/{e/3.9)}

2×10¹¹ = (8.39×10^7×3.9)/e

2×10¹¹e = 3.27×10^8

e = 3.27×10^8/2×10¹¹

e = 1.635×10^-3m

The stretch of the steel wire will be

1.635×10^-3m

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