Moment of inertia of a uniform annular disc of internal radius r and external radius R…
Physics · JEE Main · NTA Exams — Rotational Motion
Moment of inertia of a uniform annular disc of internal radius r and external radius R and mass M about an axis through its centre and perpendicular to its plane is:
- \(\frac{1}{2} M\left(R^{2}-r^{2}\right)\)
- \(\frac{1}{2} M\left(R^{2}+r^{2}\right)\)
- \(\frac{M\left(R^{4}+r^{4}\right)}{2\left(R^{2}+r^{2}\right)}\)
- \(\frac{1}{2} \frac{M\left(R^{4}+r^{4}\right)}{\left(R^{2}-r^{2}\right)}\)
Answer
(B) 1 2 M (R^ 2 +r^ 2 )
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