TrojanPoem
  • TrojanPoem
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Michele_Laino
  • 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
  • TrojanPoem
w = 10 * 2 pi = 20 p i ( Understood) K.E = 0.5 I w^2 K.E = P = 500t ?
TrojanPoem
  • TrojanPoem
0.5 * 2 * (20pi)^2 = 500 t

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More answers

TrojanPoem
  • TrojanPoem
t = 7.89 s
TrojanPoem
  • TrojanPoem
I think that doesn't work. I got the time needed to get 500 J of energy
TrojanPoem
  • TrojanPoem
we need to make use of the friction force.
Michele_Laino
  • Michele_Laino
after the cutting energy, the resistant moment is half of the electric moment, right?
Michele_Laino
  • Michele_Laino
so, the power developed by friction is half of the starting power which is \(500\;W\)
Michele_Laino
  • 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
  • TrojanPoem
Moment of rotating = Moment caused by motor - moment of friction
TrojanPoem
  • TrojanPoem
, the power developed by friction is half of the starting power which is 500W ? Why did you induce that ?
Michele_Laino
  • 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
  • TrojanPoem
Can't grasp it well.
Michele_Laino
  • Michele_Laino
If there is no electrical energy, what are the external forces which are acting on the pulley?
TrojanPoem
  • TrojanPoem
The friction only
Michele_Laino
  • 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
  • TrojanPoem
Let moment of friction before cutting the electricity : \[\tau _{2 } = 0.5 \tau _{1}\]
Michele_Laino
  • 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
  • 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
  • TrojanPoem
Hmm, still not convinced by last part.. Wasn't moment of pully = moment caused by motor - moment of friction ?
TrojanPoem
  • TrojanPoem
The final answer must be 31.55 s
TrojanPoem
  • TrojanPoem
After cutting the current the moment of pully = - moment of friction
Michele_Laino
  • Michele_Laino
the moment is referred to the external forces which are acting on the pulley
TrojanPoem
  • TrojanPoem
Pully moment = moment of caused by motor - external forces moment.
TrojanPoem
  • TrojanPoem
I think I am destroying the physics xD
Michele_Laino
  • Michele_Laino
if there is not electrical energy, then moment of the electrical motor is zero
TrojanPoem
  • TrojanPoem
so moment of external forces = moment of pulley
TrojanPoem
  • TrojanPoem
moment of pulley = - moment of external forces
TrojanPoem
  • TrojanPoem
moment of external forces after cutting = 0.5 before cutting moment of pulley = - 0.5 moment of external force before cutting
Michele_Laino
  • 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
  • TrojanPoem
That's why I said : Pully moment = moment of caused by motor - external forces moment.
Michele_Laino
  • Michele_Laino
I know, nevertheless I meant to the second phase
Michele_Laino
  • Michele_Laino
during second phase there is the moment due to friction forces only
TrojanPoem
  • TrojanPoem
I am convinced by this
TrojanPoem
  • TrojanPoem
Now ?
Michele_Laino
  • Michele_Laino
with my equation I got 31.55/2=15.79 seconds, so please wait...
TrojanPoem
  • TrojanPoem
Do you think your equation will be fixed ?
Michele_Laino
  • 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
  • Michele_Laino
oops.. \(M_f\) is the moment due to friction forces and developed by the same electric motor
Michele_Laino
  • Michele_Laino
namely*
Michele_Laino
  • 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
  • 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
  • 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
  • Michele_Laino
and finally: \[\huge \tau = \frac{{2I\omega _0^2}}{{{W_0}}}\]
TrojanPoem
  • TrojanPoem
Nice.
Michele_Laino
  • Michele_Laino
:)
TrojanPoem
  • TrojanPoem
Remember the ice cube question I asked you about ? You can T final as -60
Michele_Laino
  • Michele_Laino
I remember, even if, I think that my equation is right
TrojanPoem
  • 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
  • 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
  • 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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