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Two uniform rings, each having 'r' is cut and joined together as shown in the figure. The mass of the first half ring is 'm' while of the other half it is $2m$ . Find the moment of inertia of the system, passing through its centre of mass and perpendicular to its plane.

Option: 1
$m r^{2}\left[4+\frac{3}{4 \pi^{2}}\right]$

Option: 2
$m r^{2}\left[3+\frac{4}{3 \pi^{2}}\right]$

Option: 3
$m r^{2}\left[3-\frac{4}{3 \pi^{2}}\right]$

Option: 4
$m r^{2}\left[4-\frac{3}{4 \pi^{2}}\right]$

Answers (1)

best_answer

for the first half ring $I_{0}=m r^{2}$

for the second half ring $I_{0}^{\prime}=(2 m) r^{2}$

so, moment of inertia of the system about its centre

$I=I_{0}+I_{0}{ }^{\prime} \\$

$I=3 m r^{2}$

Now, locating the centre of mass of the system

$x_{c \cdot m}=\frac{m\left(-x_{1}\right)+(2 m)\left(x_{2}\right)}{3 m} \\$

$X_{c \cdot m}=\frac{\not m\left(\frac{2 r}{\pi}\right)}{3 \not m}=\frac{2 r}{3 \pi} \\$

$I_{c \cdot m}=I-(3 m) x^2_{c \cdot m} \\$

$I_{c \cdot m}=3 m r^{2}-\not 3m \times \frac{4 r^{2}}{3\not 9 \pi^{2}} \\$

$I_{c \cdot m}=m r^{2}\left[3-\frac{4}{3 \pi^{2}}\right]$

Posted by

Pankaj Sanodiya

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