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Is this equation valid for a given mechanical system moto

The Inertia of the motor is $J_m$,there is no loss in the system.then conservation of energy can be used.

the motor is revolving with speed $\omega$.and the gears ratio are equal. $v $ is the velocity of the weight $w$

I am asking that the can the total InertIa can be expressed as $$\frac{1}{2}J'_m\omega^2=\frac{1}{2}J_m\omega^2+\frac{1}{2}mv^2$$.

equivalent moment of inertia referred to the motor shaft is $J'_m$.

This was told by my teacher I think there is something wrong here. I am skeptical about this because we can't talk about adding the both terms and referring them to as total inertia referred to motor shaft.BOOK SOLUTION

jjl


enterhere

What you might be saying is

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Your teacher is right. This is called equivalent mass moment of inertia, and it is often used in system dynamics problems to simplify things.

In your case the equivalent inertia $J_m^\prime$ is given by the equation

$$J_m^\prime = \left[\frac{1}{2}J_m \omega^2 + \frac{1}{2} m v^2\right]\left[\frac{2}{\omega^2}\right]$$

or

$$J_m^\prime = J_m + m \left(\frac{v}{\omega}\right)^2$$

If you are confused, think of it this way: inertia is really a measure of how much energy it would take to slow a moving object to a stop. Note that this only works if there is a lossless gearbox.

If you are still skeptical, try determining the torque the motor must output for a given angular acceleration $\alpha$. You should find that the $T_m = J_m \alpha + rma/n$, where $r$ is the radius of the drum and $n$ is the gear ratio. Substituting $\alpha r/n$ for $a$ and dividing both sides by $\alpha$ yields $$\frac{T_m}{\alpha} = J_m + m \left(\frac{r}{n}\right)^2$$ which is equivalent to the expression we found above.

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  • $\begingroup$ but if the moment of inertia is referred to the motor shaft we can have by the conservation of energy ;as the system is lossless $\frac{1}{2}J_m \omega^2=\frac{1}{2}mv^2$ where $v$ is the velocity of the block.then the referred moment of inertia will be $J_m =\frac{mv^2}{\omega^2}$ $\endgroup$
    – Boris
    Jan 10 '17 at 4:09
  • $\begingroup$ for example it can be said that the total energy of the system is E is rotational kinetic + translational kinetic energy $\endgroup$
    – Boris
    Jan 10 '17 at 13:50
  • $\begingroup$ Not quite. $\frac{1}{2}J_m\omega^2 \neq \frac{1}{2}mv^2$. The correct energy equation is $\frac{1}{2}J_m^\prime \omega^2 = \frac{1}{2}J_m\omega^2 + \frac{1}{2}mv^2$. All of that energy comes from the electricity supplied to the motor. $\endgroup$
    – regdoug
    Jan 12 '17 at 12:38
  • $\begingroup$ So $J_m^\prime = J_m + m\frac{v^2}{\omega^2}$ $\endgroup$
    – regdoug
    Jan 12 '17 at 12:39
  • $\begingroup$ I edited the question above and added book pages further @regdoug $\endgroup$
    – Boris
    Jan 12 '17 at 13:33
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Ignoring the inertia in the gears and drum, assuming a gear ratio of 1, and denoting the radius of the drum by $r$, the total kinetic energy of the system is

$$T=\frac{1}{2}J_m\omega_m^2+\frac{1}{2}m v^2=\frac{1}{2}J_m\omega_m^2+\frac{1}{2}m r^2\omega_m^2=\frac{1}{2}(J_m+mr^2)\omega_m^2$$

Thus the equivalent moment of inertia is $J_m+mr^2$.

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