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



What you might be saying is


2 Answers 2


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

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


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

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.

  • $\begingroup$ 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
    Commented Jan 10, 2017 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
    Commented Jan 10, 2017 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
    Commented Jan 12, 2017 at 12:38
  • $\begingroup$ So $J_m^\prime = J_m + m\frac{v^2}{\omega^2}$ $\endgroup$
    – regdoug
    Commented Jan 12, 2017 at 12:39
  • $\begingroup$ I edited the question above and added book pages further @regdoug $\endgroup$
    – Boris
    Commented Jan 12, 2017 at 13:33

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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