The moment of inertia of two equal masses each of mass m at separation L connected by a rod of mass M,about an axis passing through centre and perpendicular to length of rod is

Question

The moment of inertia of two equal masses each of mass m at separation L connected by a rod of mass M,about an axis passing through centre and perpendicular to length of rod is (A)  ((M+3M)L2/12) (B)  ((M+6M)L2/12) (C)  (ML2/4) (D)  (ML2/12)

Tardigrade Admin · Accepted Answer

Moment of inertia of rod about XY I1= (ML2/12) Moment of inertia of two masses I2=m ( (L/2) )2+m ( (L/2) )2= (mL2/2) I=I1+I2= (ML2/12)+ (mL2/2) = ((M+6m)L2/12)

Q. The moment of inertia of two equal masses each of mass m at separation L connected by a rod of mass M,about an axis passing through centre and perpendicular to length of rod is

$\frac{( M + 3 M ) L ^{2}}{12}$

$\frac{( M + 6 M ) L ^{2}}{12}$

$\frac{M L ^{2}}{4}$

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

Moment of inertia of rod about XY
$I_{1} = \frac{M L ^{2}}{12}$
Moment of inertia of two masses
$I_{2} = m (\frac{L}{2})^{2} + m (\frac{L}{2})^{2} = \frac{m L ^{2}}{2}$
$I = I_{1} + I_{2} = \frac{M L ^{2}}{12} + \frac{m L ^{2}}{2}$
$= \frac{( M + 6 m ) L ^{2}}{12}$

Q. The moment of inertia of two equal masses each of mass m at separation L connected by a rod of mass M,about an axis passing through centre and perpendicular to length of rod is

12(M+3M)L2​

12(M+6M)L2​

4ML2​

12ML2​

Moment of inertia of rod about XY I1​=12ML2​ Moment of inertia of two masses I2​=m(2L​)2+m(2L​)2=2mL2​ I=I1​+I2​=12ML2​+2mL2​ =12(M+6m)L2​

$\frac{( M + 3 M ) L ^{2}}{12}$

$\frac{( M + 6 M ) L ^{2}}{12}$

$\frac{M L ^{2}}{4}$

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

Moment of inertia of rod about XY
$I_{1} = \frac{M L ^{2}}{12}$
Moment of inertia of two masses
$I_{2} = m (\frac{L}{2})^{2} + m (\frac{L}{2})^{2} = \frac{m L ^{2}}{2}$
$I = I_{1} + I_{2} = \frac{M L ^{2}}{12} + \frac{m L ^{2}}{2}$
$= \frac{( M + 6 m ) L ^{2}}{12}$