TS EAMCET · Physics · Rotational Motion
Moment of inertia of a uniform horizontal solid cylinder of mass \(M\) about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius \(R\) is
- A \(\frac{39 M R^2}{4}\)
- B \(\frac{30 M R^2}{4}\)
- C \(\frac{49 M R}{4}\)
- D \(\frac{49 M R^2}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{49 M R^2}{4}\)
Step-by-step Solution
Detailed explanation
Moment of inertial of solid cylinder, about the axis passing through its centre of mass and perpendicular to its plane \( I_{X Y}=M\left(\frac{l^2}{12}+\frac{R^2}{4}\right) \) From theorem of parallel axis, moment of inertial of cylinder ahout an axis passing through its edge…
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