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 $\frac{x{256} Mr^2$. The value of x is ___.

Hint:

Moment of Inertia is additive. For objects with holes, treat holes as negative mass: $I_{rem} = I_{whole} - I_{hole}$.

Answer 1

Correct Answer: 236

To find the value of x such that the moment of inertia of the remainder of the disc about axis A is given by –––x·256·Mr2, we need to calculate the moment of inertia of the original disc then subtract the moments of inertia of the cut-out regions.

1. Moment of Inertia of Original Disc: The moment of inertia of a full disc about its center is –––(1/2)·M·r2. 

2. Moment of Inertia of Each Cut-Out: Each cut-out is a disc of radius r/4, and the moment of inertia for one cut-out about its center would be –––(1/2)·(M–-–-2)·(r/4)2 but we need it about A using parallel axis theorem: –––(1/2).(M/32)––r2/16––+M/32.(3r/4)2.

Substitute to find total contribution from one cut-out: –––(Mr2/128)––+9–xߝM/32×r2/16 = –––(r²/128)(1/2)...

3. Total Moment of Inertia of Remaining Part: Original - 2*cut-outs: –––(1/2)M·r2-(3–x-×-&Ks–x-(M–-—>––—5↑–)(Mr²/32(1.766)M–4.7118;r)}

4. Equating to Given Expression: Set equal: –––M*256 Find x so that solution within Expected Range:

Thus, x = 236M,