Find the moments of inertia for a wire that lies along $\overrightarrow { \mathbf { r } } ( t ) = 7 \cos t \tilde { \mathbf { i } } + 7 \sin t \tilde { \mathbf { j } } , 0 \leq t \leq 2 \pi$, with density $\rho ( x , y ) = 1$
A) $I _ { x } = 343 \pi , I _ { y } = 343 \pi$
B) $I _ { x } = 7 \pi , I _ { y } = 49 \pi$
C) $I _ { x } = 49 \pi , I _ { y } = 49 \pi$
D) $L _ { x } = 343 \pi , I _ { y } = 7 \pi$
E) $I _ { x } = 7 \pi , I _ { y } = 7 \pi$

Question

Quizplus · Accepted Answer

The moments of inertia for a wire with uniform density lying along a curve in the plane are given by $I_x = \int (y^2 \, dm)$ and $I_y = \int (x^2 \, dm)$, where $dm = \rho \, ds$, with $\rho$ being the density and $ds$ the differential arc length. For the given curve, which represents a circle of radius 7 (since $x = 7\cos(t)$ and $y = 7\sin(t)$), and with $\rho = 1$, the moments of inertia are computed using polar coordinates. The differential arc length for a circle is $ds = r \, d\theta$, and since the radius $r = 7$, $ds = 7 \, dt$. Thus, $I_x = \int_0^{2\pi} (7\sin(t))^2 \cdot 7 \, dt = 343\pi$ and $I_y = \int_0^{2\pi} (7\cos(t))^2 \cdot 7 \, dt = 343\pi$, making the correct answer $I_x = 343\pi$ and $I_y = 343\pi$.

Find the Moments of Inertia for a Wire That Lies r→(t)=7cos⁡ti~+7sin⁡tj~,0≤t≤2π\overrightarrow { \mathbf { r } } ( t ) = 7 \cos t \tilde { \mathbf { i } } + 7 \sin t \tilde { \mathbf { j } } , 0 \leq t \leq 2 \pir(t)=7costi~+7sintj~​,0≤t≤2π

Find the Moments of Inertia for a Wire That Lies $\overrightarrow { \mathbf { r } } ( t ) = 7 \cos t \tilde { \mathbf { i } } + 7 \sin t \tilde { \mathbf { j } } , 0 \leq t \leq 2 \pi$