Question

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

Answer 1

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