Physics, 26.11.2020 01:00 Riohxncho

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 the 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.

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Answer from: insomniacnana2

The time taken for the rotational kinetic energy to become half its initial value is 0.426 seconds.

Given the data in the question;

Moment of inertia of the wheel; I = 18.0kg.m^2

Initial kinetic energy; K_1 = 30.0J

Final kinetic energy; K_2 = 15.0J

Since it is slowing down, angular acceleration becomes negative

Angular acceleration; a_a = -0.200 rev/s^2 = -1.2566 rad/s^2

First we need to determine initial and final angular velocity. From the expression for Rotational kinetic energy:

$K_R = \frac{1}{2}Iw^2$

Where ω is the angular velocity and I is the moment of inertia around the axis of rotation.

We substitute our given values into the equation

$K_{R1} = \frac{1}{2}Iw_1^2\\\\ 30J = \frac{1}{2} * 18.0kg.m^2 * w_1^2\\\\30kg.m^2/s^2 = \frac{1}{2} * 18.0kg.m^2 * w_1^2\\\\30kg.m^2/s^2 = 9.0kg.m^2 * w_1^2\\\\w_1^2 = \frac{30kg.m^2/s^2}{9.0kg.m^2}\\\\ w_1^2 = \frac{30kg.m^2/s^2}{9.0kg.m^2}\\\\ w_1 = \sqrt{\frac{30kg.m^2/s^2}{9.0kg.m^2}} \\\\w_1 = 1.8257 rad/s$

$K_{R2} = \frac{1}{2}Iw_2^2\\\\15.0J = \frac{1}{2} * 18.0kg.m^2 * w_2^2\\\\15.0kg.m^2/s^2 = \frac{1}{2} * 18.0kg.m^2 * w_2^2\\\\15.0kg.m^2/s^2 = 9.0kg.m^2 * w_2^2\\\\w_2 = \sqrt{\frac{15.0kg.m^2/s^2}{9.0kg.m^2} } \\\\w_2 = 1.29099 rad/s$

Now, From the equation of rotational motion:

w_2 = w_1 + a_at

Where w_2 is the final angular velocity, w_1 is the initial angular velocity, a_a is the angular acceleration and is the time taken.

We substitute our values into the equation and solve for "t"

$1.29099rad/s = 1.8257rad/s + [ - 1.2566rad/s^2\ *\ t ]\\\\1.29099rad/s - 1.8257rad/s = - 1.2566rad/s^2\ *\ t \\\\-0.53471rad/s = - 1.2566rad/s^2\ *\ t \\\\t = \frac{-0.53471rad/s}{- 1.2566rad/s^2}\\\\t = 0.4255s\\\\t = 0.426s$

Therefore, the time taken for the rotational kinetic energy to become half its initial value is 0.426 seconds.

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Answer from: garrettcameron12

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:

- 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.

Answer from: Quest

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Answer from: Quest

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