Question

Two spheres each of mass M and radius R/2 are connected with a massless rod of lenght 2R as shown in the figure. What will be the moment of inertia of the system about an axis passing through the centre f one of the spheres and perpendicular to the rod:

• A. 21/5 MR2
• B. 2/5 MR2
• C. 5/2 MR2
• D. 5/21 MR2.

Answer 1

Answer: (A)

21/5 MR2

Answer 2

Moment of inertia of the system about yy'

Iyy' = Moment of inertia of sphere P about yy' + moment of inertia of sphere Q about yy'

Moment of inertia of sphere P about yy'

$= \frac { 2 } { 5 } M \left( \frac { R } { 2 } \right) ^ { 2 } + M ( x ) ^ { 2 }$

[parallel axis theorem]

$= \frac { 2 } { 5 } M \left( \frac { R } { 2 } \right) ^ { 2 } + M ( 2 R ) ^ { 2 }$ $= \frac { M R ^ { 2 } } { 10 } + 4 M R ^ { 2 }$

Moment of inertia of sphere Q about yy' is

= $\frac { 2 } { 5 } M \left( \frac { R } { 2 } \right) ^ { 2 }$

Now $I _ { y y ^ { \prime } } = \frac { M R ^ { 2 } } { 10 } + 4 M R ^ { 2 } + \frac { 2 } { 5 } M \left( \frac { R } { 2 } \right) ^ { 2 }$ $= \frac { 21 } { 5 } M R ^ { 2 }$