Moments of Inertia 01 Mass is a measure of inertia. In Newton's Second Law, the net force is equal to the mass of the object times the acceleration. The larger the mass, the smaller the acceleration for a given force. When we talk about rotational motion, it not only matters how much mass there is but also how it is distributed. The moment of inertia tells us how easy or hard it is to change the rotational speed of an object. A point object of mass M traveling in a circle of radius R has a moment of inertia equal to MR². If we have an object that is not a point object, like a solid disk, the material is distributed at a variety of distances from the center. The moment of inertia is different than that for a point object and often needs to be calculated with a little bit of calculus. The textbook provides a table (Figure 10.12) which gives the moment of inertia for different commonly shaped objects. Rank the moment of inertia of the following objects. (1=largest moment of inertia) A solid disk with a mass of 1.5 kg with a radius of 0.20 m spinning around its center. A hoop with a mass of 1.5 kg with a radius of 0.20 m spinning around its center. A point object with a mass of 1.5 kg traveling in a circle of radius 0.20 m. A solid sphere with a mass of 1.5 kg with a radius of 0.20 m spinning around its center.
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5 kg and radius (r) = 0.20 m i1 = 1/2 * m * r^2 i1 = 1/2 * 1.5 * (0.20)^2 i1 = 0.03 kg m^2 Show more…
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Find the moment of inertia of the four masses shown in Eig. $10-7$ relative to an axis perpendicular to the page and extending ( $a$ ) through point- $A$ and $(b)$ through point- $B$. (a) From the definition of moment of inertia, $$I_{A}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+\cdots+m_{N} r_{N}^{2}=(2.0 \mathrm{~kg}+3.0 \mathrm{~kg}+4.0 \mathrm{~kg}+5.0 \mathrm{~kg})\left(r^{2}\right)$$ where $r$ is half the length of the diagonal: $$r=\frac{1}{2} \sqrt{(1.20 \mathrm{~m})^{2}+(2.50 \mathrm{~m})^{2}}=1.39 \mathrm{~m}$$ Thus, $I_{A}=27 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (b) We cannot use the parallel-axis theorem here because neither $A$ nor $B$ is at the center of mass. Hence, we proceed as before. Because $r=1.25 \mathrm{~m}$ for the $2.0$ - and $3.0$ -kg masses, while $r=\sqrt{(1.20)^{2}+(1.25)^{2}}=1.733$ for the other two masses, $I_{B}=(2.0 \mathrm{~kg}+3.0 \mathrm{~kg})(1.25 \mathrm{~m})^{2}+(5.0 \mathrm{~kg}+4.0 \mathrm{~kg})(1.733 \mathrm{~m})^{2}=33 \mathrm{~kg} \cdot \mathrm{m}^{2}$.
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