The moment of inertia of a rod (length I mass m) about an axis perpendicular to the length of the rod and passing through a J point equidistant from its mid-point and one end, is

Question

The moment of inertia of a rod (length I mass m) about an axis perpendicular to the length of the rod and passing through a J point equidistant from its mid-point and one end, is (A)  (ml2/12)  (B)  (7/48)ml2  (C)  (13/48)ml2  (D)  (19/48)ml2

Tardigrade Admin · Accepted Answer

Moment of inertia of thin rod I=(ml2/12) Distance of axis CD from AB =(L/4) From theorem of parallel axis, we have ICD=IAB+M( (l/4) )2 =(ml2/12)+(ml2/16)

Q. The moment of inertia of a rod (length I mass m) about an axis perpendicular to the length of the rod and passing through a J point equidistant from its mid-point and one end, is

$\frac{m l ^{2}}{12}$

$\frac{7}{48} m l^{2}$

$\frac{13}{48} m l^{2}$

$\frac{19}{48} m l^{2}$

Moment of inertia of thin rod $I = \frac{m l ^{2}}{12}$
Distance of axis CD from AB $= \frac{L}{4}$ From theorem of parallel axis, we have $I_{C D} = I_{A B} + M (\frac{l}{4})^{2}$ $= \frac{m l ^{2}}{12} + \frac{m l ^{2}}{16}$

Q. The moment of inertia of a rod (length I mass m) about an axis perpendicular to the length of the rod and passing through a J point equidistant from its mid-point and one end, is

12ml2​

487​ml2

4813​ml2

4819​ml2

Moment of inertia of thin rod I=12ml2​ Distance of axis CD from AB =4L​ From theorem of parallel axis, we have ICD​=IAB​+M(4l​)2 =12ml2​+16ml2​

$\frac{m l ^{2}}{12}$

$\frac{7}{48} m l^{2}$

$\frac{13}{48} m l^{2}$

$\frac{19}{48} m l^{2}$

Moment of inertia of thin rod $I = \frac{m l ^{2}}{12}$
Distance of axis CD from AB $= \frac{L}{4}$ From theorem of parallel axis, we have $I_{C D} = I_{A B} + M (\frac{l}{4})^{2}$ $= \frac{m l ^{2}}{12} + \frac{m l ^{2}}{16}$