<p> I Moment of Inertia I p. 1 1. Point Mass ( I = mr2 )</p><p>Let's use the simple version of torque (where r sinq= r^ = r ) : t = rF From 'F = ma' we obtain: rF= r� ma (now using at = ra , we get...) rF r m( r a ) Finally we have a relationship between torque and angular acceleration: t= mr2 a and if we call mr2, the 'moment of inertia, I' then we have our</p><p>Fnet = ma in its rotational form: tnet = I a 2 2. Lotsa point masses ( I= mi r i ) That's a Greek sigma for 'S' (summation)! i 3. Continuous Mass Distribution ( I= r2 dm ) Just think of the integral symbol as an elongated 'S' for summation! −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−- 4. Hoop or Shell of mass, M, and radius, R ( I = MR2 ) I= r2 dm = ? Well, since all the little masses ( dm's ) are all at a distance of r = R, we can 'slide' out R2 as a constant to get: I= R2 dm = R 2 M (since the dm's add to M) HOOP VS SOLID CYLINDER ( I = ½ MR2 ) End View Side View</p><p>HOOP I = MR2</p><p>5. Solid Disk or Solid Cylinder of mass, M, and radius, R, (any length!) ( I = ½ MR2 ) The plan here is to sum a lot of hoops with radius, x. (See inside ring above.) So… I= r2 dm = x2 dm (where we will integrate from x = 0 to x = R) For each x or ring, mass or 'dm' varies with its area (circumference, 2p r ,times width, dx ) M So replace 'dm' with 's 2 p x dx ' and replace 's ' with (see box info below) to get: p R2 2 2骣M 3 I = 蝌x dm= xs2 p xdx = 2 p 琪 x dx M M 桫p R2 Area Density: s = = Area p R2 R 4 R 2M3 2 M x M 4 4 Also sdA= dm = s2 p x dx I = x dx = = 2 (R - 0 ) R2 R 2 4 2R x=0 0 1 2 Finally we get: I =2 MR whew! (I mean Q.E.D) I Moment of Inertia I p. 2 1 2 6. Rod of mass, M, and length, L ( ICM = 12 ML ) rotation axis through it's c.m. Here the plan is to integrate from x = 0 to x = L/2 (from the center out) and double our answer to get the I for the other half of the rod. 2 2 I1/2 = r dm= x dm M Doubling and substituting for 'dm' we get: Linear Density: l = L L L 2 2 3 2骣M 2 2 M x Also ldℓ= dm = l dx 2 I1/2 = I = 2蝌xl dx= 2琪 x dx = 桫L L 3 x=0 0 2M骣 L3 0 3 - 1 2 Finally, we get: 琪 = Icm = 12 ML L 桫3 8 3 x = L/2 x = 0</p><p>7. Parallel Axis Theorem – When rotating any mass about a line parallel to a line through 2 its center of mass: I = Icm + MD (where D is the distance between the two lines)</p><p> x = L c.m. D = L/2 (distance from c.m.) x = 0</p><p>8. Rod of mass, M, and length, L, rotated about a line at one of its ends (parallel lines...) 1 2 (I= 3 ML ) 1 2 2 1 2 2 Method 1/ Using the Parallel Axis Theorem, I = 12 ML + MD = 12 ML +M(L/2) 12 1 2 1 2 So we get: I = 12ML+ 4 ML = I = 3 ML Method 2/ Integrating (we need the practice!) I = r2 dm = ? This time we'll integrate from x = 0 to x = L (no doubling here!) x2 dm= x 2l dx = x 2 M dx So we obtain: I = 蝌 ( L ) M L Linear Density: l = L 3 L M2 M x 1 2 = x dx = = I= ML ldℓ= dm = l dx L L 3 3 Also 0 0</p><p>I Moment of Inertia I p. 3</p><p>9. Hollow Cylinder, mass M and with an outer (R2) and an inner (R1) radius. Same as the 'Solid Cylinder'. M M Just integrate from x = R1 to x = R2 Area Density: s = = 2 2 Area pR2- p R 1 I = 蝌r2 dm= x 2s dA = s x 2 2 p xdx Also sdA= dm = s2 p x dx</p><p>R 骣 M 骣 2M R2 骣 2M x4 2 M 2p x3 dx 琪 x3 dx =1 R4 - R 4 =琪 2 2 = 2 2 = 琪 2 2 2 2 2 ( 2 1 ) 桫pR2- p R 1 桫R2- R 1 x= R 桫R2- R 14 R 2 - R 1 1 R1 1 M R2+ R 2 R 2 - R 2 1 2 2 = 2 2 ( 21 )( 2 1 ) I = M R + R (This is just an exercise!) ( 2 1 ) 2 R2- R 1 2</p><p>R R 1 2</p><p> b a</p><p>1 10. Rectangular Plate with dimensions 'a' and 'b' , I = M a2+ b 2 12 ( ) Here, we'll consider long rods with length, L = b, and integrate from x = 0 to x = a/2 and then double our result to get the other half of the plank or plate! For each 'dx' there will be a rod with area,bdx . M M Using #7 above, the parallel axis theorem: Area Density: s = = Area ab I = 1 ML2 + Mx2 (where each rod is D=x away...) 12 Also sdA= dm = s bdx 1 2 2 each rod will have an I(or dI) of 12 b dm + x dm . a 2 2 2 骣M 2 2 1 b + x dm 1 2 2 b1 b + x dx I1/2 = (12 ) = (12 b + x)s bdx = 琪 (12 ) x=0 桫ab a 3 3 2 骣 a 骣 2 3 骣M骣1 2 x 骣M琪1 2 a ( 2 ) 骣 M ab a I1/2 = 琪 b�琪 b x = 琪�12b( 2 ) - + (0 = 0) + 琪 琪 桫ab 12 3 桫a琪 3 桫 a 桫 24 24 桫 0 桫 2 2 骣b a 1 2 2 I = 2I1/2 = 2M 琪 + = I =M( a + b ) 桫24 24 12 I Moment of Inertia I p. 4</p><p>Time Out for Geometry! Area of a Frustum: A=p (R1 + R 2 ) L (L is the slant height) Step 1/ What's the lateral area of a cone? L.A. = p rℓ (where ℓ is the slant height) Proof: Cut along a slant height and unroll.</p><p> q ℓ</p><p>2p r Notice the sector of a circle with radius, ℓ, and arc length, s =2p r . q The area of a sector when q is measured in degrees is: A = p R2 sector 360 rad q2 q 2 When q is measured in radians is: Asector = p R� Asector R 2p rad 2 q rad When q is measured in radians the arc length is: s= 2p R � s Rq 2p rad Using the figure above with s = 2p r = Rq =ℓq (since here R =ℓ) 2p r 2p r q 2 ( ℓ) 2 we get: q =� A = R ℓ � Asector prℓ ℓ sector 2 2</p><p>Step 2/ What's the area of a frustum of a cone? Area = p (R1+ R 2 )L (say R1 < R2) The area is simply the big cone area minus the little cone area. That formula is: Area = pRℓ- p R ℓ (Different?!) 2 2 1 1 L To show the two formulae are equal we'll use:</p><p>R1 R 2 (i) L = ℓ2- ℓ 1 and by : D's (ii) = � Rℓ R ℓ ℓ1 ℓ 2 1 2 2 1 Now we'll start with the first area formula and substitute L with ℓ2- ℓ 1 to get: Area =p (R1+ R 2)(ℓ 2 - ℓ 1), now we'll 'foil' and... get: Area = p (R1ℓ 2- R 1 ℓ 1 + R 2 ℓ 2 - R 2 ℓ 1), but by (ii) above R1ℓ 2- R 2 ℓ 1 = 0 (whoa!) so our first formula becomes: Area = p (R2ℓ 2- R 1 ℓ 1) =pR2ℓ 2- p R 1 ℓ 1 (whoa!) and that's our second area formula! </p><p>I Moment of Inertia I p. 5</p><p>Now what was that frustum area all about? We'll actually use that A=p (R1+ R 2 ) L to calculate the moment of inertia of a frustum-like piece of a thin spherical shell. </p><p>2 2 11. Thin Spherical Shell with mass, M, and radius, R ( I = 3 MR )</p><p>Integrate from x = 0 to x = R and double our result. M M dy dy Area Density: s = = 2 Also use: x2 + y2 = R2 � 2x 2 =0 � - x Area 4p R dx dx y Also s dA=dm = s2 p R ds Obtain slant height 'ds' with the Pythagorean Thm. dA =p(R1+ R 2 )L p (2 r) ds 2 2 2 2 2 (ds )= ( dx ) + ( dy )� ds ( dx ) + ( dy ) or where ds = L (slant height) and </p><p>2 we replace (R1+R2) with 2r. 1+ dy dx (used for arc length) and ds = ( dx ) 2 dA=2p y 1 + dy dx where 'r' is replaced by 'y', giving us: ( dx ) R 2 2 2 sy2 2 p y 1 dy dx I1/2 = 蝌r dm= ys dA = � ( dx ) and 0 R 2 2 2 2 dy 2 y+ x M3 R -x x R 1 1+ = 1 + = 1 + = = � I 2p 2 y鬃 dx ( dx) ( y) y2 2 y 2 ( 4p R ) y y 0 R R R M2 M 2 2 M 2 x3 2 2 �I= 2I1 � 2 - 2p y dx = R - x dx R x = � I MR 2 ( 4p R) 蝌 R( ) R ( 3 ) 3 0 0 0</p><p>2 2 12. Solid Sphere with mass, M, and radius, R (I= 5 MR ) At least the hard work has all been done above! Now we'll sum up (integrate) a lot of thin spherical shells from x = 0 to x = R . (Note: We don't have to double our result this time.) 22 2 2 2 I=蝌dI =( 3 dm� x) #10 (I = 3 MR ) and since dm = r dV r 4 p x dx ... R 2 2 2M 8p 4 M M I =xr4 p x dx = x dx r = = 3 蝌3( ) 4p R3 3 Volume Density: Volume 4p R 3 x=0 3 rdV= dm 5 R Also: and with x = r x 5 I =2M = 2 M R -0 = I = 2 MR2 dV = 4pr2 dx= 4 p x 2 dx R35 R 3 ( 5) 5 0</p><p>That's All Folks! </p>