Suppose there is a uniform circular disc of mass M ~kg and radius r ~m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x 256 M r² . The value of

Question

Suppose there is a uniform circular disc of mass M ~kg and radius r ~m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x 256 M r² . The value of x is _ _ _ _ .

Quantrex Academy · Accepted Answer

Correct answer: 0

The moment of inertia of the complete disc about axis A is I_ disc = 1 2 Mr^2 = 128Mr^2 256 . Each removed circular region has radius r/4 and mass m_i = M/(r^2/16r^2) = M/16 . For each removed disc at distance 3r/4 from axis A, using the parallel axis theorem: I_ removed = 1 2 m_i(r/4)^2 + m_i(3r/4)^2 = Mr^2 512 + 9Mr^2 256 = 19Mr^2 512 . For two removed discs: I_ 2removed = 38Mr^2 512 = 19Mr^2 256 . Therefore: I_ remainder = 128Mr^2 256 - 19Mr^2 256 = 109Mr^2 256 , so x = 109 .

Suppose there is a uniform circular disc of mass M ~kg and radius r ~m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x 256 M r² . The value of x is _ _ _ _ .

Correct answer

Step-by-step solution

More from Rotational Motion