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Bumek [7]
4 years ago
15

A heavy flywheel rotating on its central axis is slowing down because of friction in its bearings. At the end of the first minut

e of slowing, its angular speed is 0.68 of its initial angular speed of 260 rev/min. Assuming a constant angular acceleration, find its angular speed (rev/min) at the end of the second minute.
Physics
1 answer:
Soloha48 [4]4 years ago
6 0

Answer:

ω₂ = 93.6 rev / min

Explanation:

ω₀ = 260 rev / min

ω₁ = 0.68*ω₀ = 0.68*(260 rev / min) = 176.8 rev / min

ω₂ = ?

t₁ = 1 min

t₂ = 2 min

We can apply the equation:

ω₁ = ω₀ + α*t₁     ⇒    α = (ω₁ - ω₀) / t₁  

⇒   α = (176.8 rev / min - 260 rev / min) / 1 min = - 83.2 rev / min²

then we can use the same formula, knowing the angular acceleration:

ω₂ = ω₀ + α*t₂   ⇒   ω₂ = (260 rev / min) + (- 83.2 rev / min²)*(2 min)

⇒   ω₂ = 93.6 rev / min

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(see the attached graphic)

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\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

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\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)

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\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}

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\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)

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