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

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, $$N = 9$$

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