Physics, 12.03.2020 06:10 iiChantxx

A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with moment of inertia I , initially not rotating, drops onto the first cylinder. Since the surfaces are rough, the two eventually reach the same angular speed ωf.

a. Calculate ωf. (Use I1 for I1, I2 for I2, and omega_i for ωi, as necessary.)

b. Show that the kinetic energy of the system decreases in this interaction by calculating the ratio of the final to initial rotational energy. Express your answer in terms of ωi.

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Answer from: pr173418

a) The final angular speed of the system is $\omega_{f} = \frac{I_{1}}{I_{1}+I_{2}}\cdot \omega_{i}$ radians per second.

b) Since r < 1 , then the kinetic energy of the system decreases due to the interaction.

a) The coupling between the two cylinders can be modelled as a entirely inelastic collision, such phenomenon is modelled after the principle of angular momentum conservation, that is:

$I_{1}\cdot \omega_{i} = (I_{1}+I_{2})\cdot \omega_{f}$ (1)

Where:

$I_{1}$ - Moment of inertia of the moving cylinder, in kilogram-square meters. $I_{2}$ - Moment of inertia of the resting cylinder, in kilogram-square meters. $\omega_{i}$ - Initial angular speed of the moving cylinder, in radians per second. $\omega_{f}$ - Final angular speed of the resulting system, in radians per second.

Then the final angular speed of the system is:

$\omega_{f} = \frac{I_{1}}{I_{1}+I_{2}}\cdot \omega_{i}$

The final angular speed of the system is $\omega_{f} = \frac{I_{1}}{I_{1}+I_{2}}\cdot \omega_{i}$ radians per second.

b) The system experiment a decrease in total energy if the final kinetic energy to intial kinetic energy ratio is:

$r = \frac{\frac{1}{2}\cdot (I_{1}+I_{2})\cdot \omega_{f}^{2} }{\frac{1}{2}\cdot I_{1}\cdot \omega_{i}^{2}}$

$r = \left[\frac{(I_{1}+I_{2})}{I_{1}}\right]\cdot \left(\frac{I_{1}}{I_{1}+I_{2}} \right)^{2}$

$r = \frac{I_{1}}{I_{1}+I_{2}}$

Since r < 1 , then the kinetic energy of the system decreases due to the interaction.

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Answer from: ellycleland16

Part(a): The final angular velocity is $\omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b): The ratio of the rotational energies is $\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$ ,showing the the energy of th system will decrease.

Explanation:

Part(a):

If '' be the moment of inertia of an object and ' $\omega$ ' be its angular velocity then the angular momentum '' of the object can be written as

$L = I \omega$

If ' $I_{1}$ ' and ' $I_{2}$ ' be the moment of inertia of the two cylinders and ' $\omega_{1}$ ' and ' $\omega_{2}$ ' be the initial angular velocity of the cylinders and ' $\omega_{1}{'}$ ' and ' $\omega_{2}^'}$ ' be their respective final angular velocity, then from conservation of angular momentum,

$I_{1} \omega_{1} + I_{2} \omega_{2} = I_{1} \omega_{1}^{'} + I_{2} \omega_{2}^{'}$

Given, $\omega_{1} = \omega_{i},~\omega_{2} = 0,~\omega_{1}^{'} = \omega_{2}^{'} = \omega_{f}$ . From the above expression

$&& I_{1} \omega_{i} = (I_{1} + I_{2}) \omega_{f}\\&or,& \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b):

Initial kinetic energy

$K_{i} = \dfrac{1}{2} I_{1} \omega_{i}^{2}$

and Final kinetic energy

$K_{f} = \dfrac{1}{2}(I_{1} + I_{2}) \omega_{f}^{2}$

Substituting the value of $\omega_{f}$ ,

$&& K_{f} = \dfrac{1}{2}(I_{1} + I_{2})\dfrac{I_{1}^{2}\omega_{i}^{2}}{(I_{1} + I_{2})^{2}} = \dfrac{1}{(I_{1} + I_{2})} \dfrac{1}{2}I_{1}\omega_{i}^{2} = \dfrac{1}{(I_{1} + I_{2})} K_{i}\\&\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$

The above expression shows that the ebergy of the system will decrease.

Answer from: Quest

answer: 0

explanation: im 14

Answer from: Quest

50 meters per 5 seconds

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A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axl...

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