Question

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 ${}^{\prime }{R}^{\prime }$ is:

• A. $M{R}^2/4$
• B. $39MR/4$
• C. $49MR/4$
• D. $49M{R}^2/4$

Answer 1

The correct option is D $49M{R}^2/4$

From parallel axix theorem

$I=M[\frac{L^2}{12}+\frac{L^2}{4}]+M[\frac{R}{2}{]}^2$

$I=\frac{M{L}^2}{3}+\frac{M{R}^2}{4}$

given L = 6R

$L=\frac{M}{3}(6R{)}^2+\frac{M{R}^2}{4}=\frac{49}{4}M{R}^2$

Hence option (D) is correct