Physics - Rotation Question with Solution | TestHub
Two loops and of radii and are made from uniform wire. If moment of inertia of loops and is and respectively then is (Given: )
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Answer:
Solution:
Let the linear mass density of the uniform wire be \lambda.
The mass of loop A is M_A = \lambda (2\pi R_1) and the mass of loop B is M_B = \lambda (2\pi R_2).
The moment of inertia of a circular loop of mass M and radius R about its central axis is I = MR^2.
For loop A, I_A = M_A R_1^2 = \lambda (2\pi R_1) R_1^2 = 2\pi \lambda R_1^3.
For loop B, I_B = M_B R_2^2 = \lambda (2\pi R_2) R_2^2 = 2\pi \lambda R_2^3.
Taking the ratio of their moments of inertia:
\dfrac{I_A}{I_B} = \dfrac{2\pi \lambda R_1^3}{2\pi \lambda R_2^3} = \left(\dfrac{R_1}{R_2}\right)^3
Given that \dfrac{I_A}{I_B} = 27, we have:
\left(\dfrac{R_1}{R_2}\right)^3 = 27
\dfrac{R_1}{R_2} = 3
The ratio \dfrac{R_1}{R_2} is 3:1.
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