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A flywheel with a moment of inertia of 3.45 kg·m2is initially rotating. In order to stopits rotation, a braking torque of -9.40 N·m is applied to the flywheel. Calculate the initialangular speed of the flywheel if it makes 1 complete revolution from the time the brake isapplied until it comes to rest

User StevenWhite
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1 Answer

18 votes
18 votes

Given data

*The given moment of inertia is I = 3.45 kg.m^2

*The given braking torque is T = -9.40 N.m

*The angular distance traveled is


\theta=(1*2\pi)rad_{}

*The final angular speed is


\omega=0\text{ rad/s}

The angular acceleration of the flywheel is calculated by using the torque and moment of inertia relation as


\begin{gathered} T=I\alpha \\ \alpha=(T)/(I) \\ =(-9.4)/(3.45) \\ =-2.72rad/s^2 \end{gathered}

The formula for the initial angular speed of the flywheel is given by the rotational equation of motion as


\omega^2-\omega^2_0=2a\theta

Substitute the known values in the above expression as


\begin{gathered} (0)^2-\omega^2_0=2*(-2.72)(2\pi) \\ \omega_0=\sqrt[]{2*2.72*2\pi} \\ =5.88\text{ rad/s} \end{gathered}

Hence, the initial angular speed of the flywheel is 5.88 rad/s

User JoshGough
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