A rod of length L is composed of a uniform length (L/2) of wood whose mass in mw and a uniform length (L/2) of brass whose mass is mb. The moment of inertia I of the rod about an axis perpendicular to the rod and through its centre is equal to

Question

A rod of length L is composed of a uniform length (L/2) of wood whose mass in mw and a uniform length (L/2) of brass whose mass is mb. The moment of inertia I of the rod about an axis perpendicular to the rod and through its centre is equal to (A) (mw+mb) (L2/6) (B) (mw+mb) (L2/2) (C) (mw+mb) (L2/12) (D) (mw+mb) (L2/3)

Tardigrade Admin · Accepted Answer

For a thin uniform rod, moment of inertia about an axis through its centre perpendicular to length of rod, I=(1/12) M L2 Here, M=(mw+mb) ∴ I=(1/12)(mw+mb) L2

Q. A rod of length $L$ is composed of a uniform length $\frac{L}{2}$ of wood whose mass in $m_{w}$ and a uniform length $\frac{L}{2}$ of brass whose mass is $m_{b}$ . The moment of inertia $I$ of the rod about an axis perpendicular to the rod and through its centre is equal to

$(m_{w} + m_{b}) \frac{L ^{2}}{6}$

$(m_{w} + m_{b}) \frac{L ^{2}}{2}$

$(m_{w} + m_{b}) \frac{L ^{2}}{12}$

$(m_{w} + m_{b}) \frac{L ^{2}}{3}$

For a thin uniform rod, moment of inertia about an axis through its centre perpendicular to length of rod, $I = \frac{1}{12} M L^{2}$
Here, $M = (m_{w} + m_{b})$
$∴ I = \frac{1}{12} (m_{w} + m_{b}) L^{2}$

Q. A rod of length L is composed of a uniform length 2L​ of wood whose mass in mw​ and a uniform length 2L​ of brass whose mass is mb​. The moment of inertia I of the rod about an axis perpendicular to the rod and through its centre is equal to

(mw​+mb​)6L2​

(mw​+mb​)2L2​

(mw​+mb​)12L2​

(mw​+mb​)3L2​