KVPY SX SB Exam 2015 Question Paper

Question : 92 of 120

Marks: +1, -0

The moments of inertia of a non-uniform circular disc (of mass M and radius R) about four mutually perpendicular tangents AB, BC, CD, DA are

I_1, I_2, I_3 \and I_4,

respectively (the square ABCD circumscribes the circle). The distance of the center of mass of the disc from its geometrical center is given by -

Solution:

I_\cm + m (R + y)^2 = I^3

...(1)

I_\cm + m (R- y)^2 = I_1

...(2)
from (1) & (2)

I_1 - I_3 = m [(R - y)^2 - (R + y)^2 ]

I_1 -I_3 = m (2R) (- 2y)

...(3)

I_\cm' + m (R + x)^2 = I_4

...(4)

I_\cm' + m (R - x)^2 = I_2

...(5)
from (4) & (5)

(I_2 - I_4) = m [(R - x)_2 -(R + x)_2]

I_2-I_4 = m [(2R) (–2x)]

...(6)

(3)^2 + (6)^2

⇒ (I_1- I_3)^2 + (I_2 - I_4)^ 2 = (m_2 × 4R_2

× 4(x_2 + y_2 )

distance of CM from O =

√{x^2+y^2}=1/{4mR}√{(I_1- I_3 ) ^3+ (I _2-I_4)^2}

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