The angular momentum of a flywheel

AI Thread Summary
To find the initial angular velocity for part (b), use the angular momentum equation L = Iω, where L is the angular momentum and I is the moment of inertia. Given the initial angular momentum of 3.00 kg·m²/s and a moment of inertia of 0.200 kg·m², the initial angular velocity ω can be calculated as ω = L/I. The average torque for part (a) can be determined by calculating the deceleration using angular motion equations. For part (c), work done on the wheel can be derived from the torque and angle, while part (d) requires the average power to be calculated using torque and angular velocity. Understanding these relationships is crucial for solving the problem effectively.
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The angular momentum of a flywheel having a rotational inertia of 0.200 kg·m2 about its axis decreases from 3.00 to 1.800 kg·m2/s in 1.80 s.

(a) What is the average torque acting on the flywheel about its central axis during this period?
N·m
(b) Assuming a uniform angular acceleration, through what angle will the flywheel have turned?
rad
(c) How much work was done on the wheel?
J
(d) What is the average power of the flywheel?
W

Basically, all i need is how to find the initial angular velocity for part b. The rest of the variables i have solved for.
 
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The angular momentum of a rigid object about a fixed axis is given by

L = I\omega

where I is its moment of inertia bout this axis and \omega is its angular speed about the same axis.
 
The rotational components and equations are analogous to linear ones.

Equation of motion:

linear:
(1) x=x_0+vt+\frac{1}{2}at^2
(2) v=v_0+at
(3) F=ma
(4) W=Fx
(5) P=Fv
(6) a(x-x_0)=\frac{1}{2}(v^2-v_0^2)

angular:

(1) \phi=\phi_0+\omega t+\frac{1}{2}\alpha t^2
(2) \omega=\omega_0+\alpha t
(3) \tau=I\alpha
(4) W=\tau \phi
(5) P=\tau \omega
(6) \alpha(\phi-\phi_0)=\frac{1}{2}(\omega^2-\omega_0^2)

So to find the average torque in part (a), find the deceleration using angular equation 2.

Part (b), use 6.

Part (c), use 4.

Part (d), use 5.
 
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