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vivado [14]
3 years ago
11

A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th

e flywheel has 30.0 J of kinetic energy. It is slowing down with an angular acceleration of magnitude 0.200 rev/s^2.
How long does it take for the rotational kinetic energy to become half its initial value, so it is 15.0 J?
Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
timama [110]3 years ago
6 0

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

\Delta E = K_{1}-K_{2} (1)

Where:

\Delta E - Energy losses, measured in joules.

K_{1}, K_{2} - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2}) (2)

Where:

I - Moment of inertia of the flywheel, measured in kilograms per square meter.

\omega_{1}, \omega_{2} - Initial and final angular speed, measured in radians per second.

If we know that K_{1} = 30\,J, K_{2} = 15\,J and I = 18\,kg\cdot m^{2}, then the initial angular speed is:

K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2} (3)

\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }

\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }

\omega_{1} \approx 1.825\,\frac{rad}{s}

\omega_{1}\approx 0.291\,\frac{rev}{s}

K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2} (4)

\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }

\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }

\omega_{2} \approx 1.291\,\frac{rad}{s}

\omega_{2} \approx 0.205\,\frac{rev}{s}

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

t = \frac{\omega_{2}-\omega_{1}}{\alpha} (5)

If we know that \omega_{1}\approx 0.291\,\frac{rev}{s}, \omega_{2} \approx 0.205\,\frac{rev}{s} and \alpha = -0.200\,\frac{rev}{s^{2}}, then the time needed is:

t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }

t = 0.43\,s

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

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As well as the Normal Force N:

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In addition, we know N=w, then \sum F_{y}=0

In the X-component:

\sum F_{x}=m.a

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Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

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m.g.sin(\theta)=m.a   (7)

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On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

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We already know the value of  d and calculated a, we have to find t:

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If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

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