Question

QUESTION 6 This question is about moments of inertia. If the hollow plate shown in Fig. Q6 has a density of 8000 kg/m3 and a thickness of 10 mm, determine its moment of inertia about an axis directed perpendicular to the page and passing through point O. 125 mm a) 1.197 kg.m2 b) 11.97 kg.m2 c) 1.473 kg.m2 d) 0.276 kg.m2 e) None of above 250 mm Figure Q6

Answer 1

Answer is e) None of the above

​​​​​​We have,

Density, d = 8000 kg/m^{​​​​​​3}​​​

Thickness, t = 10 mm = 0.01 m

Outer Radius, R = 250 mm = 0.25 m

Inner Radius, r = 125 mm = 0.125 m

For a hollow disk, the Moment of Inertia, I = m( R^{​​​​​​2}​​​​​ + r^{​​​​2} ) / 2

m = mass of disk = Volume x Density = (π x (R^{​​​​​​2}​​​ - r^{​​​​​​2} ) x t) x d

Substitute values,

m = (π x (0.25² - 0.125²) x 0.01) x 8000

m = 11.78 kg

Substitute mass in equation for I,

I = 11.78 x (0.25² + 0.125²) /2

I = 0.46 kg.m²

The moment of Inertia is 0.46 kg.m²

Hope it helps.