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I'm in a physics class in which we calculated the moment of inertia of a ball rolling down a ramp using a conservation of energy equation: $${1 \over2}mv^2 + {1 \over2}Iω^2 = mgh$$

where $ I $ is the moment of inertia of the ball.

In a different class I'm in, I am designing a toy car. For some of our simulations, we need to determine the moment of inertia for the car. Could it be possible to determine $I$ for the car using a conservation of energy equation like the one above?

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  • $\begingroup$ Hopefully, you will never have to deal with moment of inertia in a real car! $\endgroup$ Dec 10, 2019 at 18:23
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    $\begingroup$ Yes, but It would be the moment of inertia of the rotating parts of the car, and not of the entire car. $\endgroup$ Dec 10, 2019 at 22:45

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Yes, the same approach can be used. The only difference is that for a car only part of the structure is rotating.

When a ball is rolling down a ramp the whole ball is rotating so the moment of inertia you measure is the moment of inertia of the whole ball. For your car only the wheels are rotating, so the rotational energy you will be measuring is the rotational energy just of the wheels, and the moment of inertia you calculate will be just the total moment of inertia of the wheels.

An option would be to remove a wheel from the car and let it roll down the slope on its own. Then the measured final velocity will give you the moment of inertia of the wheel. Multiply by four to get the MOI for the whole car.

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I guess you need the moment of inertia of the car itself and not the wheels. Therefore you could determine this using the formula for a compound pendulum $T=2\pi \sqrt{\frac{I}{mgR}}$ and doing the experiment. For the experiment you need to hang your car from a rope or something similar and let it swing, measure the period $T$ and the radius $R$ to the center of the car and measure the mass $m$ of your car. If you put these values into the formula and solve for $I$ you get the needed moment of inertia.

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A rough cut estimate for a passenger vehicle's moment of inertia about the axis normal to the pavement is $m \cdot a \cdot b$, where $m$ is mass, $a$ is the distance from center of mass to front axle, and $b$ is the distance from center of mass to rear axle.

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  • $\begingroup$ Yes, this is the lumped mass model, where each axle bears a certain percentage of the total mass and the MMOI is $$ I = m_a a^2 + m_b b^2 $$ by using $$ \begin{aligned} m_a & = \frac{b}{a+b} m \\ m_b & = \frac{a}{a+b} m \end{aligned}$$ you arrive at your formula. It would be nice if you included this calculations in your answer in order to enlighten the reader. $\endgroup$ May 10, 2022 at 19:25

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