Anthony Pitt
Mar 12
C?
A flywheel of mass 5 tonnes and having a radius of gyration of 2.5m is acted upon by a constant torque which produces an acceleration of 2 rad/s squared. Find (a) the rev/min 4 seconds after the angular velocity was 5 rad/s and (b) the number of revolutions turned in the 4 seconds and (c) the magnitude of the torque?
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Mar 12
(a) To find the final angular velocity in rev/min, we first need to find the final angular velocity in rad/s. We can use the formula:
ωf = ωi + αt
where:
ωf = final angular velocity
ωi = initial angular velocity = 5 rad/s
α = angular acceleration = 2 rad/s^2
t = time = 4 seconds
Plugging in the values, we get:
ωf = 5 + 2(4) = 13 rad/s
Now, to convert this to rev/min, we use the conversion factor:
1 rev = 2π rad
1 min = 60 s
So, ωf in rev/min = (13 * 60) / (2π) ≈ 123.8 rev/min
(b) To find the number of revolutions turned in the 4 seconds, we use the formula:
θ = ωi*t + 0.5*α*t^2
where:
θ = angle turned in radians
ωi = initial angular velocity = 5 rad/s
α = angular acceleration = 2 rad/s^2
t = time = 4 seconds
Plugging in the values, we get:
θ = 5*4 + 0.5*2*4^2 = 20 + 0.5*2*16 = 36 radians
To convert this to revolutions, we use the conversion factor:
1 rev = 2π rad
So, θ in revolutions = 36 / (2π) ≈ 5.73 revolutions
(c) To find the magnitude of the torque, we use the formula:
τ = I*α
where:
τ = torque
I = moment of inertia = mass * radius of gyration^2 = 5 * (2.5)^2 = 31.25 tonnes-m^2
α = angular acceleration = 2 rad/s^2
Plugging in the values, we get:
τ = 31.25 * 2 = 62.5 tonnes-m
Therefore, the magnitude of the torque is 62.5 tonnes-m.
ωf = ωi + αt
where:
ωf = final angular velocity
ωi = initial angular velocity = 5 rad/s
α = angular acceleration = 2 rad/s^2
t = time = 4 seconds
Plugging in the values, we get:
ωf = 5 + 2(4) = 13 rad/s
Now, to convert this to rev/min, we use the conversion factor:
1 rev = 2π rad
1 min = 60 s
So, ωf in rev/min = (13 * 60) / (2π) ≈ 123.8 rev/min
(b) To find the number of revolutions turned in the 4 seconds, we use the formula:
θ = ωi*t + 0.5*α*t^2
where:
θ = angle turned in radians
ωi = initial angular velocity = 5 rad/s
α = angular acceleration = 2 rad/s^2
t = time = 4 seconds
Plugging in the values, we get:
θ = 5*4 + 0.5*2*4^2 = 20 + 0.5*2*16 = 36 radians
To convert this to revolutions, we use the conversion factor:
1 rev = 2π rad
So, θ in revolutions = 36 / (2π) ≈ 5.73 revolutions
(c) To find the magnitude of the torque, we use the formula:
τ = I*α
where:
τ = torque
I = moment of inertia = mass * radius of gyration^2 = 5 * (2.5)^2 = 31.25 tonnes-m^2
α = angular acceleration = 2 rad/s^2
Plugging in the values, we get:
τ = 31.25 * 2 = 62.5 tonnes-m
Therefore, the magnitude of the torque is 62.5 tonnes-m.
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