Question 754: Centre of Mass and Rotational Motion

Question

Four solid spheres each of diameter $\sqrt 5 $ cm and mass 0.5 kg are placed with their centers at the corners of
a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N $ 	imes $ 10
−4
kg-m
2
, then N is
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l2jmnfog/8094121c-b091-4804-8cba-c8d29a7506da/859e0af0-c749-11ec-a2bf-852b719f1924/file-1l2jmnfoh.png?format=png)

Accepted Answer

The moment of inertia of each sphere about an axis passing through its centre is ${2 \over 5}m{r^2}$.
The moment of inertia of sphere B and sphere D about X $-$ X' is
${I_B} = {I_D} = {2 \over 5}m{r^2}$.
Using parallel axes theorem, the moment of inertia of sphere A and sphere C about X $-$ X' is
${I_A} = {I_C} = {2 \over 5}m{r^2} + m{d^2}$
The moment of inertia of the system about the diagonal is
$I = {I_A} + {I_B} + {I_C} + {I_D} = {8 \over 5}m{r^2} + 2m{d^2}$
$m = 0.5$ kg
$d = {a \over {\sqrt 2 }} = {4 \over {\sqrt 2 }}$ cm
$r = {{\sqrt 5 } \over 2}$ cm
$	herefore$ $I = {8 \over 5} 	imes 0.5 	imes {\left( {{{\sqrt 5 } \over 2} 	imes {{10}^{ - 2}}} ight)^2} + 2 	imes 0.5 	imes {\left( {{4 \over {\sqrt 2 }} 	imes {{10}^{ - 2}}} ight)^2}$
$ = 9 	imes {10^{ - 4}}$ kg m
2
Hence, $N = 9$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l4i9um2i/2edff618-e4a6-470a-8044-f09979480e7c/1ca5d9a0-ee23-11ec-a735-e70f40715860/file-1l4i9um2j.png?format=png)

Centre of Mass and Rotational Motion

Four solid spheres each of diameter \sqrt 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N \times 10 −4 kg-m 2 , then N is

Integer / Numeric Answer Type