Question
The uniform disc shown in the figure has a moment of inertia of $0.6 \mathrm{~kg}-\mathrm{m}^{2}$ around the axis that passes through $O$ and is perpendicular to the planc. II a scgment is cut out from the dise as shown, what is the moment of incrtia of the remaining disc ?
Step 1
6 \mathrm{~kg}-\mathrm{m}^{2}$. The mass density of the disc, denoted as $\sigma$, is given by the total mass $M$ divided by the area of the disc, which is $\pi r^2$. So, we have $\sigma = \frac{M}{\pi r^2}$. Show more…
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Consider a scmicircular disc (of mass $m$ and radius $r$ ) in $x-y$ planc, as shown in the figure. Choose the incorrect alternative (a) Moment of inertia of this disc about $x$ -axis is $\frac{m r^{2}}{4}$ (b) Moment of inertia of this disc about z-axis is $\frac{m r^{2}}{2}$ (c) Consider a point in $x-y$ plane with an axis perpendicular to the $x-y$ planc passing through the point, such that moment of incria about this axis is a constant $I_{a}$. Locus ol all sueh points is a cirele (d) Moment of inertia of this disc about axis parallel to $z$ -axis, passing through $A$ is $\frac{5 m r^{2}}{4}$
Rotational and Rolling Motion
Section B
[II] The uniform circular disk in Fig. $10-8$ has a mass of $6.5$ kg and a diameter of $80 \mathrm{~cm}$. Compute its moment of inertia about an axis perpendicular to the page $(a)$ through $G$ and $(b)$ through $A$. (a) $I_{G}=\frac{1}{2} M r^{2}=\frac{1}{2}(6.5 \mathrm{~kg})(0.40 \mathrm{~m})^{2}=0.52 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (b) By the result of $(a)$ and the parallel-axis theorem, $I_{A}=I_{G}+M h^{2}=0.52 \mathrm{~kg} \cdot \mathrm{m}^{2}+(6.5 \mathrm{~kg})(0.22 \mathrm{~m})^{2}=0.83 \mathrm{~kg} \cdot \mathrm{m}^{2}$
The uniform circular disk in Fig. $10-8$ has a mass of $6.5 \mathrm{~kg}$ and a diameter of $80 \mathrm{~cm}$. Compute its moment of inertia about an axis perpendicular to the page $(a)$ through $G$ and $(b)$ through $A$. (a) $I_{G}=\frac{1}{2} M r^{2}=\frac{1}{2}(6.5 \mathrm{~kg})(0.40 \mathrm{~m})^{2}=0.52 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (b) By the result of $(a)$ and the parallel-axis theorem, $$ I_{A}=I_{G}+M h^{2}=0.52 \mathrm{~kg} \cdot \mathrm{m}^{2}+(6.5 \mathrm{~kg})(0.22 \mathrm{~m})^{2}=0.83 \mathrm{~kg} \cdot \mathrm{m}^{2} $$
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