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Using formula for physical pendulum but unable to solve.

Harshit Bhansali , 6 Years ago
Grade 12
anser 2 Answers
Arun

Moment of inertia for disc -

I=frac{1}{2} MR^{2}

 

=\frac{MR^{2}}{2}-[\frac{3}{32}MR^{2}]

I_{remain}=\frac{13}{32}MR^{2} 

- wherein

About an axis perpendicular to the plane of disc & passing through its centre .

 

 I=I_{remain}+I_\frac{R}{2}

I_{remain}=I-I_\frac{R}{2}

=\frac{MR^{2}}{2}- \left[ \frac{ \frac{M}{4} \left( \frac{R}{2} \right )^{2} }{2} + \frac{M}{4} \left( \frac{R}{2} \right )^{2} \right ]

=\frac{MR^{2}}{2}-[\frac{MR^{2}}{32}+\frac{MR^{2}}{16}]

= \frac{MR^2}{2}-\frac{3MR^2}{32}

I_{remaining}=\frac{13MR^2}{32}

 

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Last Activity: 6 Years ago
Vikas TU
Dear student 
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