An electric motor which gives electric power = 500 W , to turn a pull whose moment of inertia around the the rotation axis is 2 kg.m^2 with speed of 10 revolutions per second. what's the time required to stop the pull after cutting the electric current. Notice that the moment of friction is half after cutting the electricity

- TrojanPoem

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- schrodinger

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- Michele_Laino

the energy \(E\) transfer from electric motor to pulley, is:
\[E = \frac{1}{2}I\omega _0^2\]
where:
\[{\omega _0} = 20\pi \]

- TrojanPoem

w = 10 * 2 pi = 20 p i ( Understood)
K.E = 0.5 I w^2
K.E = P = 500t ?

- TrojanPoem

0.5 * 2 * (20pi)^2 = 500 t

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## More answers

- TrojanPoem

t = 7.89 s

- TrojanPoem

I think that doesn't work. I got the time needed to get 500 J of energy

- TrojanPoem

we need to make use of the friction force.

- Michele_Laino

after the cutting energy, the resistant moment is half of the electric moment, right?

- Michele_Laino

so, the power developed by friction is half of the starting power which is \(500\;W\)

- Michele_Laino

so we can write this differential equation:
\[\huge \frac{{dL}}{{dt}} = - \frac{{{W_0}}}{2}dt\]
where \(W_0=500\), and \(L\) is the work done by the electric motor

- TrojanPoem

Moment of rotating = Moment caused by motor - moment of friction

- TrojanPoem

, the power developed by friction is half of the starting power which is 500W ? Why did you induce that ?

- Michele_Laino

when I make a cutting of electrical energy, the moment of external forces, which is acting on the pulley, is only the moment related to the friction forces

- TrojanPoem

Can't grasp it well.

- Michele_Laino

If there is no electrical energy, what are the external forces which are acting on the pulley?

- TrojanPoem

The friction only

- Michele_Laino

and, according to the text of the exercise, the moment due to such friction forces is
"...is half after cutting the electricity..."

- TrojanPoem

Let moment of friction before cutting the electricity :
\[\tau _{2 } = 0.5 \tau _{1}\]

- Michele_Laino

I think that if the new moment is half of the first one, also the absorbed power of such friction forces is half of the power developed by the electrical motor

- Michele_Laino

If I integrate the equation above, I get the requested time \(\tau\), as below:
\[\huge \tau = \frac{{I\omega _0^2}}{{{W_0}}}\]

- TrojanPoem

Hmm, still not convinced by last part..
Wasn't moment of pully = moment caused by motor - moment of friction ?

- TrojanPoem

The final answer must be 31.55 s

- TrojanPoem

After cutting the current the moment of pully = - moment of friction

- Michele_Laino

the moment is referred to the external forces which are acting on the pulley

- TrojanPoem

Pully moment = moment of caused by motor - external forces moment.

- TrojanPoem

I think I am destroying the physics xD

- Michele_Laino

if there is not electrical energy, then moment of the electrical motor is zero

- TrojanPoem

so moment of external forces = moment of pulley

- TrojanPoem

moment of pulley = - moment of external forces

- TrojanPoem

moment of external forces after cutting = 0.5 before cutting
moment of pulley = - 0.5 moment of external force before cutting

- Michele_Laino

please wait a moment, we have to keep in mind that the friction is acting on the pulley also during the working of the electrical motor, namley not all tthe power developed by the electrical motor goes to accelerate such pulley

- TrojanPoem

That's why I said : Pully moment = moment of caused by motor - external forces moment.

- Michele_Laino

I know, nevertheless I meant to the second phase

- Michele_Laino

during second phase there is the moment due to friction forces only

- TrojanPoem

I am convinced by this

- TrojanPoem

Now ?

- Michele_Laino

with my equation I got 31.55/2=15.79 seconds, so please wait...

- TrojanPoem

Do you think your equation will be fixed ?

- Michele_Laino

I have to rewrite another equation. Here is my reasoning:
during the working of electric motor, we have the subsequent equation:
\[\huge {W_0} = {M_f} \cdot {\omega _0}\]
where \(W_0=500\), \M_f\) is the moment due to friction force, and \(\omega_0=10\)
Please keep in mind during the working of electric motor, we have a stationary or equilibrium situation, namley the moment applied by the motor is against the friction forces only

- Michele_Laino

oops.. \(M_f\) is the moment due to friction forces and developed by the same electric motor

- Michele_Laino

namely*

- Michele_Laino

After the electric energy cutting, the new value of \(M_f\), is:
\[\huge \frac{{{M_f}}}{2}\]
according to the text of your exercise

- Michele_Laino

so we can write the subsequent differential equation:
\[\huge \begin{gathered}
\frac{{{M_f}}}{2} = \frac{{{W_0}}}{{2{\omega _0}}} \hfill \\
\hfill \\
\frac{{{W_0}}}{{2{\omega _0}}} = - I\frac{{d\omega }}{{dt}} \hfill \\
\end{gathered} \]

- Michele_Laino

next I integrate such equation, so I get:
\[\huge \int_0^\tau {dt} = \frac{{ - 2I{\omega _0}}}{{{W_0}}}\int_{{\omega _0}}^0 {d\omega } \]

- Michele_Laino

and finally:
\[\huge \tau = \frac{{2I\omega _0^2}}{{{W_0}}}\]

- TrojanPoem

Nice.

- Michele_Laino

:)

- TrojanPoem

Remember the ice cube question I asked you about ?
You can T final as -60

- Michele_Laino

I remember, even if, I think that my equation is right

- TrojanPoem

I told you , you were right, the final answer is indeed -60 , But the question asks further more to explain what does -60 as final temperature means.

- Michele_Laino

I think that I have made an error, since I used as specific heat of ice the same value of the specific heat of water.

- Michele_Laino

I remember that, please check, if I'm wrong, the specific heat of ice is different from the specific heat of water

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