JEE Challenger
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PhyMedAdv 2011 • Paper 1

Moment of Inertia of Four Solid Spheres at Corners of Square

Four solid spheres each of diameter 5 cm\sqrt{5}\text{ cm} and mass 0.5 kg0.5\text{ kg} are placed with their centers at the corners of a square of side 4 cm4\text{ cm}. The moment of inertia of the system about the diagonal of the square is N×10−4 kg-m2N \times 10^{-4}\text{ kg-m}^2, then N is

Official Numerical Answer9

Topics & Concepts

System of Particles and Rotational MotionMoment of Inertia

Step-by-Step Solution

To find the moment of inertia of the four-sphere system about the diagonal of the square, we analyze the contribution of each sphere individually using the parallel axis theorem.

1. Given Data:

  • Mass of each sphere, m=0.5 kgm = 0.5\text{ kg}
  • Diameter of each sphere, d=5 cm=5×10−2 md = \sqrt{5}\text{ cm} = \sqrt{5} \times 10^{-2}\text{ m}
  • Radius of each sphere, R=d2=52×10−2 mR = \frac{d}{2} = \frac{\sqrt{5}}{2} \times 10^{-2}\text{ m}
  • Side length of the square, a=4 cm=4×10−2 ma = 4\text{ cm} = 4 \times 10^{-2}\text{ m}

2. Moment of Inertia of a Single Sphere About Its Center of Mass:

The moment of inertia of a solid sphere about an axis passing through its center of mass is given by:
Icm=25mR2I_{\text{cm}} = \frac{2}{5} m R^2

3. Position of the Spheres Relative to the Diagonal Axis:

Let the four corners of the square be A,B,C,A, B, C, and DD, and let the axis of rotation be the diagonal passing through AA and CC.

  • Spheres at AA and CC:
    since their centers lie on the axis of rotation, the distance from their center of mass to the axis is h1=0h_1 = 0.
    By the parallel axis theorem:
    IA=IC=Icm=25mR2I_A = I_C = I_{\text{cm}} = \frac{2}{5} m R^2

  • Spheres at BB and DD:
    The perpendicular distance h2h_2 from the center of each of these spheres to the diagonal ACAC is equal to half the diagonal length of the square:
    h2=a22=a2h_2 = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}

    By the parallel axis theorem, the moment of inertia for each of these spheres about the diagonal is:
    IB=ID=Icm+mh22=25mR2+m(a2)2=25mR2+12ma2I_B = I_D = I_{\text{cm}} + m h_2^2 = \frac{2}{5} m R^2 + m \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{2}{5} m R^2 + \frac{1}{2} m a^2

4. Total Moment of Inertia of the System:

The total moment of inertia ItotalI_{\text{total}} is the sum of the moments of inertia of all four spheres:
Itotal=IA+IB+IC+IDI_{\text{total}} = I_A + I_B + I_C + I_D
Itotal=2Icm+2(Icm+12ma2)=4Icm+ma2I_{\text{total}} = 2 I_{\text{cm}} + 2 \left( I_{\text{cm}} + \frac{1}{2} m a^2 \right) = 4 I_{\text{cm}} + m a^2
Itotal=4(25mR2)+ma2=85mR2+ma2I_{\text{total}} = 4 \left( \frac{2}{5} m R^2 \right) + m a^2 = \frac{8}{5} m R^2 + m a^2

5. Numerical Calculation:

Substitute the given values into the equation:

  1. Calculate 85mR2\frac{8}{5} m R^2:
    R2=(52×10−2)2=54×10−4 m2R^2 = \left( \frac{\sqrt{5}}{2} \times 10^{-2} \right)^2 = \frac{5}{4} \times 10^{-4}\text{ m}^2
    85mR2=85×0.5×(54×10−4)=1×10−4 kg-m2\frac{8}{5} m R^2 = \frac{8}{5} \times 0.5 \times \left( \frac{5}{4} \times 10^{-4} \right) = 1 \times 10^{-4}\text{ kg-m}^2

  2. Calculate ma2m a^2:
    a2=(4×10−2)2=16×10−4 m2a^2 = (4 \times 10^{-2})^2 = 16 \times 10^{-4}\text{ m}^2
    ma2=0.5×16×10−4=8×10−4 kg-m2m a^2 = 0.5 \times 16 \times 10^{-4} = 8 \times 10^{-4}\text{ kg-m}^2

  3. Summing the terms:
    Itotal=1×10−4+8×10−4=9×10−4 kg-m2I_{\text{total}} = 1 \times 10^{-4} + 8 \times 10^{-4} = 9 \times 10^{-4}\text{ kg-m}^2

Given that Itotal=N×10−4 kg-m2I_{\text{total}} = N \times 10^{-4}\text{ kg-m}^2, we get:
N=9N = 9

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