Question

Three solid spheres each of mass P and radius Q are arranged as shown in fig.The moment of inertia of the arrangement aboutYY'axis

Answer 1

Given that,

Mass = P

Radius = Q

According to figure,

We need to calculate the moment of inertia of sphere A

Using perpendicular theorem of moment of inertia

$I_{A}=\dfrac{2}{5}MR^2$

Where, M = mass

R = radius

Put the value into the formula

$I_{A}=\dfrac{2}{5}PQ^2$

The moment of inertia of B and C is

$I_{B}=I_{C}$

We need to calculate the moment of inertia of sphere B

Using parallel theorem of moment of inertia

$I_{B}=\dfrac{2}{5}MR^2+MR^2$

Put the value into the formula

$I_{B}=\dfrac{2}{5}PQ^2+PQ^2$

$I_{B}=\dfrac{7}{5}PQ^2$

Similarly,

The moment of inertia of sphere C is

$I_{C}=\dfrac{7}{5}PQ^2$

We need to calculate the moment of inertia of the arrangement about YY' axis

Using formula for total moment of inertia

$I=I_{A}+I_{B}+I_{C}$

Put the value into the formula

$I=\dfrac{2}{5}PQ^2+\dfrac{7}{5}PQ^2+\dfrac{7}{5}PQ^2$

$I=PQ^2(\dfrac{2}{5}+\dfrac{7}{5}+\dfrac{7}{5})$

$I=\dfrac{16}{5}PQ^2$

Hence, The moment of inertia of the arrangement about YY' is $\dfrac{16}{5}PQ^2$

Answer 2

Answer:

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