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3 years ago

6

A grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 12

rad/s over the next 4.0 seconds. Assume that the angular acceleration is constant. Through how many revolutions does the grindstone turn during the 4.0-second interval?

1 answer:

PSYCHO15rus [73]3 years ago

6 0

Answer: 6.36

Explanation:

Given

Radius of grindstone, r = 4 m

Initial angular speed of grindstone, w(i) = 8 rad/s

Final angular speed of the grindstone, w(f) = 12 rad/s

Time used, t = 4 s

Angular acceleration of the grinder,

α = Δw / t

α = w(f) - w(i) / t

α = (12 - 8) / 4

α = 4/4 = 1 rad/s²

Number of complete revolution in 4s =

Δθ = w(i).t + 1/2.α.t²

Δθ = 8 * 4 + 1/2 * 1 * 4²

Δθ = 32 + 1/2 * 16

Δθ = 32 + 8

Δθ = 40 rad/s

40 rad/s = 40/2π rpm = 6.36 rpm

Therefore, the grindstone does 6.36 revolutions during the 4 s interval

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Explanation:

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$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

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$A = \pi r^2$

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3 years ago

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Answer:

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Explanation:

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At a radius of 3.6m and velocity of 16.12m/s,

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Force = Mass (m) * Acceleration (a)

36 = m * 72.182

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F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

m = m

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F = 12051.3 * 36/72.182

F = 6010.457N

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