Question

Chapter 10, Problem 043 The uniform solid block in the figure has mass 37.1 kg and edge lengths a = 0.685 m, b = 1.33 m, and c = 0.130 m. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces. Rotation axis - Number Unitsy the tolerance is +/-5% Click if you would like to Show Work for this question: Open Show Work

Answer 1

In the given problem,

Rotational Inertia about an axis passing through one corner and perpendicular to the large faces can be find using below formula

$I=\frac{M\left ( a^{2}+b^{2} \right )}{3}$

Given, Uniform solid block mass, M = 37-1 Kg Uuitorm folid block edge leng ths a= 0.685 m 6= 1133 m C = 0.130 m Roteion anis Rotitiunel inertia about an aris tuYough one Cormer and Perpeindiular to tue lerge faces, I=? Rotidionel inertia abuut an anis through Cor ner and Deapenelicular to the lerçe fecen Can be find as followes. M(a*+b") M = mass of Unitorm solid blodk where a, 5= edge lngths of Uuiform colid blodk subblidute, M= 37.1 kg a = 0.685m in eguation @. we set I 6 = l 33 m (a'ts) %3D 341 [o-6ss)+(い33) 37-1 (0.469+ i-769) 3コ1メ2. 238 %3D

83.0298 - 27.68. kg. I= 27.68 kg.m :- Rotetione Inertia about an anis through Corner and Per penliculer to the lerge oue faceA. I = 27.6 8 kg. m Devivation for rotationel Inrtia aboul an anin through oe corner and Nerpandi ular to the lave taun, I = M(ats) : Canter of Mass is at point o ut 1e d Mowint of Inertia of uuiform Lolid blodk about anis throuph is. entar of Mau o' Mla+b) 12 Moment of Inerlia ahwut the anis nas ting throuph an thown in fifure Can se al ulule ow Corner uling ACCordinig to nerallel anis theomm, Darallel Anis theorem" where, I= Mount of iestia atout anis rassing thruuth one corner I.= Moment of inert:a about the Gnis through onter of man dis tanle setwen I and I

Nuw substiluti, I, = সa) d= OA (trw tipure) in eguition we st I I = I, + Md? 12 <(v०ন + 9. m(atb)+ M(0AÍ OA an k Caluat as tollower Here According h Py thogeras theorm, (hy pote neus )´= (Bicde)t side, %3D (2 0A) = a*+sr 2 (OA) Janst DA vaz jats Susrtilute in equetion 3 OA M (a'+s) + M OA) 12 () ত = + M (Vats) M la+s) M(a+s) (ti tà) = m(atb) (t3) Mäts? ) = Ma+s) M lats)