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Sindrei [870]
3 years ago
9

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

that extends from the axis of the disk to the rim. At t = 0 this line lies along the x-axis and the disk is rotating with positive angular velocity ω0z. The disk has constant positive angular acceleration αz. At what time after t = 0 has the line on the disk rotated through an angle θ?
Express your answer in terms of the variables ω0z, αz, and θ.
Physics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

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