A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is -

Question

A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is - (A) (m1 m2/m1 + m2) l2 (B) (m1 + m2/m1 m2) l2 (C) (m1 + m2) l2 (D) √m1 m2 l2

Tardigrade Admin · Accepted Answer

m1r1 = m2 (l -r2) (m1 + m2)r1 = m2r r1 = (m2l/m1 + m2) and r2 =r - r1 = r - (m2l/m1+m2) = (m1l/m1 +m2) So I = I1 + I2 =m1r12 + m2r22 =m1 ((m2l/m1+m2))2 + m2((m1l/m1+m2))2 = (m1m2l2/m1+m2)

Q. A light rod of length $l$ has two masses $m_{1}$ and $m_{2}$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is -

$\frac{m _{1} m _{2}}{m _{1} + m _{2}} l^{2}$

$\frac{m _{1} + m _{2}}{m _{1} m _{2}} l^{2}$

$(m_{1} + m_{2}) l^{2}$

$m_{1} m_{2} l^{2}$

Q. A light rod of length l has two masses m1​ and m2​ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is -

m1​+m2​m1​m2​​l2

m1​m2​m1​+m2​​l2

(m1​+m2​)l2

m1​m2​​l2

m1​r1​=m2​(l−r2​) (m1​+m2​)r1​=m2​r r1​=m1​+m2​m2​l​ and r2​=r−r1​=r−m1​+m2​m2​l​=m1​+m2​m1​l​ So I=I1​+I2​ =m1​r12​+m2​r22​ =m1​(m1​+m2​m2​l​)2+m2​(m1​+m2​m1​l​)2=m1​+m2​m1​m2​l2​

Q. A light rod of length $l$ has two masses $m_{1}$ and $m_{2}$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is -

$\frac{m _{1} m _{2}}{m _{1} + m _{2}} l^{2}$

$\frac{m _{1} + m _{2}}{m _{1} m _{2}} l^{2}$

$(m_{1} + m_{2}) l^{2}$

$m_{1} m_{2} l^{2}$