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#engineering
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#physics
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A flywheel originally at rest is to reach an angular velocity of 36rads-1 in 60.0s
a.what constant angular acceleration must it have
b. What total angular does it turn through in 6.0s?

Apr 16

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Apr 16

a. To find the constant angular acceleration, we can use the formula:

ωf = ωi + αt

Where:

ωf = final angular velocity = 36 rad/s

ωi = initial angular velocity = 0 rad/s

α = angular acceleration (unknown)

t = time = 60.0 s

Plugging in the values, we get:

36 = 0 + α(60.0)

36 = 60α

α = 36/60

α = 0.6 rad/s^2

Therefore, the constant angular acceleration must be 0.6 rad/s^2.

b. To find the total angular distance the flywheel turns through in 6.0s, we can use the formula:

θ = ωi*t + 0.5*α*t^2

Where:

θ = total angular distance

ωi = initial angular velocity = 0 rad/s

t = time = 6.0 s

α = angular acceleration = 0.6 rad/s^2

Plugging in the values, we get:

θ = 0*6.0 + 0.5*0.6*(6.0)^2

θ = 0 + 0.5*0.6*36

θ = 0 + 0.5*21.6

θ = 0 + 10.8

θ = 10.8 radians

Therefore, the total angular distance the flywheel turns through in 6.0s is 10.8 radians.

ωf = ωi + αt

Where:

ωf = final angular velocity = 36 rad/s

ωi = initial angular velocity = 0 rad/s

α = angular acceleration (unknown)

t = time = 60.0 s

Plugging in the values, we get:

36 = 0 + α(60.0)

36 = 60α

α = 36/60

α = 0.6 rad/s^2

Therefore, the constant angular acceleration must be 0.6 rad/s^2.

b. To find the total angular distance the flywheel turns through in 6.0s, we can use the formula:

θ = ωi*t + 0.5*α*t^2

Where:

θ = total angular distance

ωi = initial angular velocity = 0 rad/s

t = time = 6.0 s

α = angular acceleration = 0.6 rad/s^2

Plugging in the values, we get:

θ = 0*6.0 + 0.5*0.6*(6.0)^2

θ = 0 + 0.5*0.6*36

θ = 0 + 0.5*21.6

θ = 0 + 10.8

θ = 10.8 radians

Therefore, the total angular distance the flywheel turns through in 6.0s is 10.8 radians.