Emmanuel Baah
Apr 16
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?
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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.
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