Question

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

Answer 1

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

Answer 2

w = 10 * 2 pi = 20 p i ( Understood)

K.E = 0.5 I w^2

K.E = P = 500t ?

Answer 3

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

Answer 4

t = 7.89 s

Answer 5

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

Answer 6

we need to make use of the friction force.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

Can't grasp it well.

Answer 14

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

Answer 15

The friction only

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Hmm, still not convinced by last part..

Wasn't moment of pully = moment caused by motor - moment of friction ?

Answer 21

The final answer must be 31.55 s

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

I think I am destroying the physics xD

Answer 26

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

Answer 27

so moment of external forces = moment of pulley

Answer 28

moment of pulley = - moment of external forces

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

I know, nevertheless I meant to the second phase

Answer 33

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

Answer 34

I am convinced by this

Answer 35

Now ?

Answer 36

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

Answer 37

Do you think your equation will be fixed ?

Answer 38

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

Answer 39

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

Answer 40

namely*

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

Nice.

Answer 46

:)

Answer 47

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

Answer 48

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

Answer 49

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.

Answer 50

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.

Answer 51

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