Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d. The ratio I0/IA of moment of inertia I0 of the system about an axis passing the centroid and about center of any of the spheres IA and perpendicular to the plane of the triangle is :

Question

Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d. The ratio I0/IA of moment of inertia I0 of the system about an axis passing the centroid and about center of any of the spheres IA and perpendicular to the plane of the triangle is : (A) (15/13) (B) (13/15) (C) (13/23) (D) (23/13)

Tardigrade Admin · Accepted Answer

From parallel axis theorem I0=3×[(2/5)M((d/2))2+M((d/√3))2]=(13/10)Md2 IA=I0+3M((d/√3))2 =(13/10)Md2+Md2 =(23/10)Md2 (I0/IA)=(13/23)

$\frac{15}{13}$

$\frac{13}{15}$

$\frac{13}{23}$

$\frac{23}{13}$

From parallel axis theorem
$I_{0} = 3 \times [\frac{2}{5} M (\frac{d}{2})^{2} + M (\frac{d}{3})^{2}] = \frac{13}{10} M d^{2}$
$I_{A} = I_{0} + 3 M (\frac{d}{3})^{2}$
$= \frac{13}{10} M d^{2} + M d^{2}$
$= \frac{23}{10} M d^{2}$
$\frac{I _{0}}{I _{A}} = \frac{13}{23}$

1315​

1513​

2313​

1323​

From parallel axis theorem I0​=3×[52​M(2d​)2+M(3​d​)2]=1013​Md2 IA​=I0​+3M(3​d​)2 =1013​Md2+Md2 =1023​Md2 IA​I0​​=2313​

$\frac{15}{13}$

$\frac{13}{15}$

$\frac{13}{23}$

$\frac{23}{13}$

From parallel axis theorem
$I_{0} = 3 \times [\frac{2}{5} M (\frac{d}{2})^{2} + M (\frac{d}{3})^{2}] = \frac{13}{10} M d^{2}$
$I_{A} = I_{0} + 3 M (\frac{d}{3})^{2}$
$= \frac{13}{10} M d^{2} + M d^{2}$
$= \frac{23}{10} M d^{2}$
$\frac{I _{0}}{I _{A}} = \frac{13}{23}$