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Maurinko [17]
3 years ago
7

50. The potential energy of a spring is 0.3 J when it is compressed 0.04 m to the left of its equilibrium

Physics
1 answer:
lozanna [386]3 years ago
6 0

Answer:

Your answer should be A. 0.25 J and moving to the right

Explanation:

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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
timama [110]

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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