Question

The moment of inertia of a meter stick of mass $400g$ about an axis passing through $20cm$ mark and perpendicular to its length is

• A. $3.9\times {10}^{5}c{m}^2$
• B. $2.9\times {10}^{5}c{m}^2$
• C. $4.9\times {10}^{5}c{m}^2$
• D. $6.9\times {10}^{5}c{m}^2$

Answer 1

Answer: (D)

$6.9\times {10}^{5}c{m}^2$

Moment of inertia of metre stick about its centre is given by:

$I=\frac{M{L}^2}{12}$

$I=\frac{0.4\times 1^2}{12}$

Now moment of inertia about axis perpendicular to length of stick at 20 cm mark

$I=\frac{0.1}{3}+0.4\times {\left(\frac{3}{10}\right)}^2$

$I=\frac{0.1}{3}+\frac{3.6}{100}$

$I=\frac{10+10.8}{300}$

$I=\frac{20.8}{300}$

$I=0.069kg{m}^2$

$I=6.9\times {10}^{5}gc{m}^2$