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Wize AP Physics C: Mechanics Textbook > Unit 5: Rotation (14-20%)

Moment of Inertia

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Moment of Inertia of Discrete Particles


When objects are spinning, the position and distribution of the mass matters. The moment of inertia of an object is a measure of the resistance of the object to changes in its rotational motion. This quantity shows us how the mass of an object is distributed about its axis of rotation. You can think of it as the rotational analog of mass
So for a bunch of point masses mim_i at distances rir_i from the axis of rotation, the moment of inertia is


I=Σmi⋅ri2\boxed{I=\Sigma m_i\cdot r_i^2}







and it has units of SI units of kg ⋅m2kg\ \cdot m^2.

Watch Out!
Based on its definition, the value of II changes if the axis of rotation changes.



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Moment of Inertia of Objects with a Continuous Mass


When we have an extended object with a continuous mass distribution, then we use an integral to find the moment of Inertia



I=∫r2dm\boxed{I=\int_{ }^{ }r^2dm}

Example:












Watch Out!
Moment of Inertia changes if a different axis of rotation is chosen. Thus, pay attention to the axis of rotation introduced in the problem.


Here are the moments of inertia formulas for some common shapes:




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Example: Moment of Inertia of a Disk


Calculate the moment of inertia of a rotating disk with radius of R and mass M about an axis which is passing through its center and it is perpendicular to the disc.


Solution

Since we have a continuous mass distribution, we have to use integration to find its moment of inertia:


I=∫r2dmI=\int r^2dm
You can consider r to be the position of dis particles respect to the axis of rotation as shown in the above picture.

Furthermore, dmdm can be considered as the mass of all disk particles at distance rraway from the axis of rotation. These particles make a ring with radius of r and infinitesimally small thickness of drdr:



We can estimate the area of the disk covered by this ring to be: 2πrdr2\pi r drand its mass dmdmwould be its area multiplied by mass density ρ\rhowhere mass density can be found as:


ρ=MπR2dm=2πr⋅dr⋅ρdm=2πrMπR2dr\begin{aligned} \rho&=\frac{M}{\pi R^2}\\ dm&=2\pi r\cdot dr\cdot \rho\\ dm&=\frac{2\cancel{\pi}rM}{\cancel{\pi} R^2}dr \end{aligned}

Now we are ready to insert above expression for dmdm into the integral definition for moment of inertia and do the integral:


I=∫r2dmI=∫0Rr2⋅2rMR2dr=2MR2∫r=0Rr3dr=2MR44R2I=12MR2\begin{aligned} I&=\int r^2dm\\ I&=\int^R_0r^2\cdot\frac{2rM}{R^2}dr\\ &=\frac{2M}{R^2}\int^R_{r=0}r^3dr=\frac{2MR^4}{4R^2}\\ &\boxed{I=\frac12MR^2} \end{aligned}

A satellite is rotating once per minute. It has a moment of inertia of 10,000kg⋅m2. Erin, an astronaut, extends the satellite's solar panels, increasing its moment of inertia to 30,000kg⋅m2. How quickly is the satellite now rotating?

The engine of a turbine has a torque of 100 Nm. The engine rotates a 20 m long and 100 kg blade. How long does it take the blade to reach to 400 rpm? Model the blade as a thin rod.

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