8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten seconds later

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8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten seconds later its velocity has decreased by 15%. What is its angular velocity at a) time t = 50 s and b) t = 100 s?

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Latifah 3 years 2021-08-31T00:37:42+00:00 1 Answers 39 views 0

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    2021-08-31T00:39:02+00:00

    Answer:

    a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

    Explanation:

    The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

    \tau = F \cdot r

    \tau = m\cdot a \cdot r

    \tau = m \cdot \alpha \cdot r^{2}

    Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

    \alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

    \alpha = -3\,\frac{rad}{s^{2}}

    Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

    a) t = 50 s.

    \omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

    \omega = 50\,\frac{rad}{s}

    b) t = 100 s.

    Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

    t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

    t = 66.667\,s

    Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

    \omega = 0\,\frac{rad}{s}

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