Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
Question Number 47266 by MrW3 last updated on 07/Nov/18

Commented by MrW3 last updated on 07/Nov/18

$F i n d t h e m o m e n t o f i n i t i a o f t h e$ $s o l i d b o d y a b o u t i t s d i a g o n a l a x i s .$
	Answered by ajfour last updated on 08/Nov/18
	$L_G ^� = ∫r^� ×(ω^� ×r^� )dm G being the centroid (mass center) and origin of x, y, z axes is G. L_G = ∫[(r^� .r^� )ω^� −(ω^� .r^� )r^� ]dm l^� = ai^� +bj^� +ck^� ; r^� =xi^� +yj^� +zk^� ω^� = ω((l^� /l)) L^� =(ω/l)∫[(ai^� +bj^� +ck^� )(x^2 +y^2 +z^2 ) −(xi^� +yj^� +zk^� )(ax+by+cz)]dm =(ω/l)∫ Σ{[a(y^2 +z^2 )−x(by+cz)]i^� }dm = ((𝛒 ω)/l)∫_(−c/2) ^( c/2) ∫_(−b/2) ^( b/2) ∫_(−a/2) ^( a/2) ( Σ{[a(y^2 +z^2 )−x(by+cz)]i^� )dxdydz =((ρ ω)/l)∫∫_(−b/2) ^( b/2) (Σ{a^2 (y^2 +z^2 )}i^� )dydz =((ρω)/l)∫_(−c/2) ^( c/2) (Σ{a^2 ((b^3 /(12))+bz^2 )}i^� )dz =((ρω)/l)Σ{a^2 (((b^3 c)/(12))+((bc^3 )/(12)))}i^� = ((ρ ω)/l)Σ{a[((abc)/(12))(b^2 +c^2 )]}i^� =((ρω)/l)×((abc)/(12)){a(b^2 +c^2 )i^� +b(c^2 +a^2 )j^� +c(a^2 +b^2 )k^� } L_G ^� = ((ρabcω)/(12(√(a^2 +b^2 +c^2 ))))Σ{a(b^2 +c^2 )i^� } and since ρabc = M L_G ^� = ((Mω)/(12(√(a^2 +b^2 +c^2 ))))Σ{a(b^2 +c^2 )i^� } Generally L_G ^� = ∫r^� ×(ω^� ×r^� )dm = ∫(xi^� +yj^� +zk^� )× determinant ((i^� ,j^� ,k^� ),(ω_x ,ω_y ,ω_z ),(x,y,z))dm =∫ determinant ((i^� ,j^� ,k^� ),(x,y,z),((ω_y z−ω_z y),(ω_z x−ω_x z),(ω_x y−ω_y x)))dm =Σ[∫{y(ω_x y−ω_y x)−z(ω_z x−ω_x z)}dm]i^� =Σ[∫{ω_x (y^2 +z^2 )−ω_y (xy)−ω_z (zx)}dm]i^� and since I_x =∫(y^2 +z^2 )dm I_(xy) = ∫xydm L_x = I_x ω_x −I_(xy) ω_y −I_(zx) ω_z L_y = −I_(xy) ω_x +I_y ω_y −I_(yz) ω_z L_z = −I_(zx) ω_x −I_(yz) ω_y +I_z ω_z Inertia tensor is ((I_x ,(−I_(xy) ),(−I_(zx) )),((−I_(xy) ),I_y ,(−I_(yz) )),((−I_(zx) ),(−I_(yz) ),I_z ) ) For this very question Inertia tensor is ((((M/(12))(b^2 +c^2 )),0,0),(0,((M/(12))(c^2 +a^2 )),0),(0,0,((M/(12))(a^2 +b^2 ))) ) If chosen coordinate system is Gxyz , the pricipal axes of inertia_(−−−−−−−−−−−−−−−−) , then the centroidal mass products of inertia of a body with respect to these prinxipal axes of inertia, are zero. I_(xy) = I_(yz) = I_(zx) = 0 . And I_x , I_y , I_z are called the principal centroidal moments of ineetia of the body. About any other point fixed point O L_O ^� = r^� ×mv_(cm) ^� +L_G ^� r^� being the position vector of G relative to O. _________________________.$
	${\bar{L}}_{G} = \int \bar{r} \times (\bar{ω} \times \bar{r}) d m$ $G b e i n g t h e c e n t r o i d (m a s s c e n t e r)$ $a n d o r i g i n o f x, y, z a x e s i s G .$ $L_{G} = \int [(\bar{r} . \bar{r}) \bar{ω} - (\bar{ω} . \bar{r}) \bar{r}] d m$ $\bar{l} = a \hat{i} + b \hat{j} + c \hat{k}; \bar{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $\bar{ω} = ω (\frac{\bar{l}}{l})$ $\bar{L} = \frac{ω}{l} \int [(a \hat{i} + b \hat{j} + c \hat{k}) (x^{2} + y^{2} + z^{2})$ $- (x \hat{i} + y \hat{j} + z \hat{k}) (a x + b y + c z)] d m$ $= \frac{ω}{l} \int Σ {[a (y^{2} + z^{2}) - x (b y + c z)] \hat{i}} d m$ $= \frac{ρ ω}{l} \int_{- c / 2}^{c / 2} \int_{- b / 2}^{b / 2} \int_{- a / 2}^{a / 2} (Σ {[a (y^{2} + z^{2}) - x (b y + c z)] \hat{i}) d x d y d z$ $= \frac{ρ ω}{l} \int \int_{- b / 2}^{b / 2} (Σ {a^{2} (y^{2} + z^{2})} \hat{i}) d y d z$ $= \frac{ρ ω}{l} \int_{- c / 2}^{c / 2} (Σ {a^{2} (\frac{b^{3}}{12} + {b z}^{2})} \hat{i}) d z$ $= \frac{ρ ω}{l} Σ {a^{2} (\frac{b^{3} c}{12} + \frac{{b c}^{3}}{12})} \hat{i}$ $= \frac{ρ ω}{l} Σ {a [\frac{a b c}{12} (b^{2} + c^{2})]} \hat{i}$ $= \frac{ρ ω}{l} \times \frac{a b c}{12} {a (b^{2} + c^{2}) \hat{i} + b (c^{2} + a^{2}) \hat{j}$ $+ c (a^{2} + b^{2}) \hat{k}}$ ${\bar{L}}_{G} = \frac{ρ a b c ω}{12 \sqrt{a^{2} + b^{2} + c^{2}}} Σ {a (b^{2} + c^{2}) \hat{i}}$ $a n d s i n c e ρ a b c = M$ ${\bar{L}}_{G} = \frac{M ω}{12 \sqrt{a^{2} + b^{2} + c^{2}}} Σ {a (b^{2} + c^{2}) \hat{i}}$ $G e n e r a l l y$ ${\bar{L}}_{G} = \int \bar{r} \times (\bar{ω} \times \bar{r}) d m$ $= \int (x \hat{i} + y \hat{j} + z \hat{k}) \times \| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ ω_{x} & ω_{y} & ω_{z} \\ x & y & z \end{matrix} \| d m$ $= \int \| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ ω_{y} z - ω_{z} y & ω_{z} x - ω_{x} z & ω_{x} y - ω_{y} x \end{matrix} \| d m$ $= Σ [\int {y (ω_{x} y - ω_{y} x) - z (ω_{z} x - ω_{x} z)} d m] \hat{i}$ $= Σ [\int {ω_{x} (y^{2} + z^{2}) - ω_{y} (x y) - ω_{z} (z x)} d m] \hat{i}$ $a n d s i n c e$ $I_{x} = \int (y^{2} + z^{2}) d m$ $I_{x y} = \int x y d m$ $L_{x} = I_{x} ω_{x} - I_{x y} ω_{y} - I_{z x} ω_{z}$ $L_{y} = - I_{x y} ω_{x} + I_{y} ω_{y} - I_{y z} ω_{z}$ $L_{z} = - I_{z x} ω_{x} - I_{y z} ω_{y} + I_{z} ω_{z}$ $I n e r t i a t e n s o r i s$ $(\begin{matrix} I_{x} & - I_{x y} & - I_{z x} \\ - I_{x y} & I_{y} & - I_{y z} \\ - I_{z x} & - I_{y z} & I_{z} \end{matrix})$ $F o r t h i s v e r y q u e s t i o n$ $I n e r t i a t e n s o r i s$ $(\begin{matrix} \frac{M}{12} (b^{2} + c^{2}) & 0 & 0 \\ 0 & \frac{M}{12} (c^{2} + a^{2}) & 0 \\ 0 & 0 & \frac{M}{12} (a^{2} + b^{2}) \end{matrix})$ $I f c h o s e n c o o r d i n a t e s y s t e m i s$ $G x y z, t h e \underset{- - - - - - - - - - - - - - - -}{p r i c i p a l a x e s o f i n e r t i a},$ $t h e n t h e c e n t r o i d a l m a s s p r o d u c t s$ $o f i n e r t i a o f a b o d y w i t h r e s p e c t$ $t o t h e s e p r i n x i p a l a x e s o f i n e r t i a,$ $a r e z e r o . I_{x y} = I_{y z} = I_{z x} = 0 .$ $A n d I_{x}, I_{y}, I_{z} a r e c a l l e d t h e$ $p r i n c i p a l c e n t r o i d a l m o m e n t s$ $o f i n e e t i a o f t h e b o d y .$ $A b o u t a n y o t h e r p o i n t f i x e d p o i n t O$ ${\bar{L}}_{O} = \bar{r} \times m {\bar{v}}_{c m} + {\bar{L}}_{G}$ $\bar{r} b e i n g t h e p o s i t i o n v e c t o r o f G$ $r e l a t i v e t o O .$ $_________________________.$
	Commented by MrW3 last updated on 07/Nov/18

	$t h a n k y o u s i r!$ $D o e s i t m e a n t h e M o I a b o u t t h e$ $d i a g o n a l a x i s i s I = I_{x} + I_{y} + I_{z}$ $= \frac{2 M (a^{2} + b^{2} + c^{2})}{3} ?$
	Commented by ajfour last updated on 08/Nov/18
	$No, MoI will be a tensor for asymmetric orientations (9 components) We need MoI to know angular momentum: For x- component L_x , we just need to attach ω_x , ω_y , and ω_z respectively to each element in first row of the inertia tensor found. similarly for L_y and L_z we shall rather use respectively the 2^(nd) and 3^(rd) row. L_x =Mω_x ( ((b^2 +c^2 )/(12))) L_x = ((Mω)/(√(a^2 +b^2 +c^2 ))){a(((b^2 +c^2 )/(12)))} L_x =((Mωa(b^2 +c^2 ))/(12(√(a^2 +b^2 +c^2 )))) = ((Mω)/(12))(b^2 +c^2 )cos α α being angle of diagonal with x- axes. So L_(diagonal) =ΣL_x cos α = ((Mω)/(12))[((a^2 (b^2 +c^2 )+b^2 (c^2 +a^2 )+c^2 (a^2 +b^2 ))/(a^2 +b^2 +c^2 ))] If b=c=a we get I_(diagonal) = ((Ma^2 )/6) .$
	$N o, M o I w i l l b e a t e n s o r f o r$ $a s y m m e t r i c o r i e n t a t i o n s$ $(9 c o m p o n e n t s)$ $W e n e e d M o I t o k n o w a n g u l a r$ $m o m e n t u m :$ $F o r x - c o m p o n e n t L_{x}, w e j u s t$ $n e e d t o a t t a c h ω_{x}, ω_{y}, a n d ω_{z}$ $r e s p e c t i v e l y t o e a c h e l e m e n t i n$ $f i r s t r o w o f t h e i n e r t i a t e n s o r$ $f o u n d .$ $s i m i l a r l y f o r L_{y} a n d L_{z} w e s h a l l$ $r a t h e r u s e r e s p e c t i v e l y t h e 2^{n d}$ $a n d 3^{r d} r o w .$ $L_{x} = M ω_{x} (\frac{b^{2} + c^{2}}{12})$ $L_{x} = \frac{M ω}{\sqrt{a^{2} + b^{2} + c^{2}}} {a (\frac{b^{2} + c^{2}}{12})}$ $L_{x} = \frac{M ω a (b^{2} + c^{2})}{12 \sqrt{a^{2} + b^{2} + c^{2}}} = \frac{M ω}{12} (b^{2} + c^{2}) \cos α$ $α b e i n g a n g l e o f d i a g o n a l w i t h$ $x - a x e s .$ $S o L_{d i a g o n a l} = Σ L_{x} \cos α$ $= \frac{M ω}{12} [\frac{a^{2} (b^{2} + c^{2}) + b^{2} (c^{2} + a^{2}) + c^{2} (a^{2} + b^{2})}{a^{2} + b^{2} + c^{2}}]$ $I f b = c = a w e g e t$ $I_{d i a g o n a l} = \frac{{M a}^{2}}{6} .$
	Commented by mr W last updated on 07/Nov/18

	$t h a n k s a l o t s i r!$
	Commented by mr W last updated on 07/Nov/18

	${I t}^{'} s a n e w I D s i r . I c a n n o t g e t t h e o l d e s t$ $b a c k a n y m o r e .$
	Commented by ajfour last updated on 08/Nov/18

	$P r e v i o u s l y i h a d o b t a i n e d M o I$ $a b o u t a c o r n e r p o i n t o n t h e$ $d i a g o n a l a x i s . N o w i t s a b o u t$ $G (m a s s c e n t e r) t h a t s t i l l l i e s$ $o n t h e d i a g o n a l a x i s (i n t h i s$ $q u e s t i o n) .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum