Centre of ss, oments, Torque Centre of ss If you support body t its center of mss (in uniform grvittionl field) it blnces perfectly. Tht s the definition of the center of mss of the body. If the body consists of m m m m 4 finite number of msses m,, m n ttched to n infinitely strong, weightless (idelized) rod with mss number i ttched t position x i, then the center of mss is t the (weighted) verge vlue of x: n x m ix i n m i The denomintor m n m i is the totl mss of the body. The numertor µ m i x i is clled the moment of the body bout the origin. This formul for the center of mss is derived in the following (optionl) section. See (6). For mny (but certinly not ll) purposes n (extended rigid) body cts like point prticle locted t its center of mss. For exmple it is very common to tret the Erth s point prticle. Here is more detiled exmple in which we think of body s being mde up of number of component prts nd compute the center of mss of the body s whole by using the center of msses of the component prts. Suppose tht we hve dumbbell which consists of left end mde up of prticles of msses m l,,, m l, locted t x l,,, x l, nd right end mde up of prticles of msses m r,,, m r,4 locted t x r,,, x r,4 nd n infinitely strong, weightless (idelized) rod joining ll of the prticles. Then the mss nd center of mss of the left end re l m l, + +m l, Xl m l,x l, + +m l, x l, l nd the mss nd center of mss of the right end re r m r, + +m r,4 Xr m r,x r, + +m r,4 x r,4 r c Joel Feldmn. 5. All rights reserved. Februry 8, 5
The mss nd center of mss of the entire dumbbell re m l, + +m l, + m r, + +m r,4 l + r x m l,x l, + +m l, x l, + m r, x r, + +m r,4 x r,4 l X l + r Xr r + l So we cn compute the center of mss of the entire dumbbell by treting it s being mde up of two point prticles, one of mss l locted t the centre of mss of the left end, nd one of mss r locted t the center of mss of the right end. Now we ll extend the bove ides to cover more generl clsses of bodies. If the body consists of mss distributed continuously long stright line, sy with mss density ρ(x)kg/m nd with x running from to b, rther thn consisting of finite number of point msses, the formul for the center of mss becomes x x ρ(x)dx ρ(x)dx Think of ρ(x)dx s the mss of the lmost point prticle between x nd x+dx. If the body is two dimensionl object, like metl plte, lying in the xy plne, its center of mss is point ( x,ȳ) with x being the (weighted) verge vlue of the x coordinte over the body nd ȳ being the (weighted) verge vlue of the y coordinte over the body. To be concrete, suppose the body fills the region { (x,y) x b, B(x) y T(x) } in the xy plne. For simplicity, we will ssume tht the density of the body is constnt, sy ρ. When the density is constnt, the center of mss is lso clled the centroid nd is thought of s the geometric center of the body. To find the centroid of the body, we use the use our stndrd slicing strtegy. We slice the body into thin verticl strips, s illustrted in the figure below. Here is detiled T(x) y y B(x) B(x) y T(x) x b x c Joel Feldmn. 5. All rights reserved. Februry 8, 5
description of generic strip. The strip hs width dx. Ech point of the strip hs essentilly the sme x coordinte. Cll it x. The top of the strip is t y T(x) nd the bottom of the strip is t y B(x). So the strip hs height T(x) B(x) re T(x) B(x)]dx mss dm ρt(x) B(x)]dx centroid, i.e. middle point, ( x, B(t)+T(x) ). In computing the centroid of the entire body, we my tret ech strip s single prticle of mss dm ρt(x) B(x)]dx locted t ( x, B(t)+T(x) ). So the mss of the entire body is ρ T(x) B(x)]dx ρa where A T(x) B(x)]dx is the re of the region. The coordintes of the centroid re x ȳ xdm B(x)+T(x) dm ρ xt(x) B(x)]dx ρ B(x)+T(x) T(x) B(x)]dx xt(x) B(x)]dx A T(x) B(x) ]dx A We cn of course lso slice up the body using horizontl slices. If the body hs constnt d y () (b) (c) L(y) R(y) x x R(y) c x L(y) density ρ nd fills the region { (x,y) L(y) x R(y), c y d } then the sme computtion s bove gives the mss of the body to be ρ d c R(y) L(y)]dy ρa () c Joel Feldmn. 5. All rights reserved. Februry 8, 5
where A d R(y) L(y)]dy is the re of the region, nd gives the coordintes of the c centroid to be x ȳ d c R(y)+L(y) dm d c ydm ρ d c R(y)+L(y) R(y) L(y)]dy ρ d c yr(y) L(y)]dy d c R(y) L(y) ]dx A d c yr(y) L(y)]dy A (b) (c) Exmple Find the x coordinte of the centroid (centre of grvity) of the plne region R tht lies in the first qudrnt x, y nd inside the ellipse 4x +9y 6. (The re bounded by the ellipse x + y is πb squre units.) b y 4x +9y 6 x Solution. In stndrd form 4x + 9y 6 is x + y. So, on R, x runs from to 9 4 nd R hs re A π π. For ech fixed x, between nd, y runs from to 4 x. So, pplying (.b) with, b, T(x) x nd B(x), 9 9 x A xt(x)dx A Sub in u x, du 9 xdx. 9 x 9 4 π x 9 4 u / udu π / x 9 dx 4 π ] 9 x x 9 dx 4 ] 4 π π Exmple Exmple Find the centroid of the qurter circulr disk x, y, x +y r. y x +y r x c Joel Feldmn. 5. All rights reserved. 4 Februry 8, 5
Solution. By symmetry, x ȳ. The re of the qurter disk is A 4 πr. By (.b) with, b r, T(x) r x nd B(x), x A r x r x dx To evlute the integrl, sub in u r x, du xdx. r x r x dx du u / u r / ] r r So x 4 r ] 4r πr π As we observed bove, we should hve x ȳ. But, just for prctice, let s compute ȳ by the integrl formul (.c), gin with, b r, T(x) r x nd B(x), () s expected. r ȳ A r x x πr 4r π ( r x ) dx πr ] r r r πr ( r x ) dx Exmple Exmple Find the centroid of the region R in the digrm. R (,) qurter circle Solution. By symmetry, x ȳ. The region R is squre with one qurter of circle of rdius removed nd so hs re 4 π 6 π 4. The top of R is y T(x). The c Joel Feldmn. 5. All rights reserved. 5 Februry 8, 5
bottom is y B(x) with B(x) x when x nd B(x) when x. So ȳ x x ] x A ]dx+ x ]dx 4 x 6 π +x x ] x dx 4 4 ] by () with r 6 π 44 48 π Exmple Exmple 4 Prove tht the centroid of ny tringle is locted t the point of intersection of the medins. A medin of tringle is line segment joining vertex to the midpoint of the opposite side. Solution. Choose coordinte system so tht the vertices of the tringle re locted t (,), (,b) nd (c,). (In the figure below, is negtive.) The line joining (,) nd (,b) (,b) x b (b y) (,) x c b (b y) (c,) hs eqution bx+y b. (Check tht (,) nd (,b) both relly re on this line.) The line joining (c,) nd (,b) hs eqution bx+cy bc. (Check tht (c,) nd (,b) both relly re on this line.) Hence for ech fixed y between nd b, x runs from y to c c y. b b We ll use horizontl strips to compute x nd ȳ. We could just pply () with c, d b, R(y) c(b y) (which is gotten by solving bx+cy bc for x) nd L(y) (b y) (which b b is gotten by solving bx+y b for x). But rther thn memorizing or looking up those formule, we ll derive them for this exmple. So consider thin strip t height y s illustrted in the figure bove. c Joel Feldmn. 5. All rights reserved. 6 Februry 8, 5
The strip hs length c l(y) b (b y) ] b (b y) c (b y) b The strip hs width dy. On this strip, y hs verge vlue y. On this strip, x hs verge vlue (b y)+ c (b y)] +c (b y). b b b As the re of the tringle is A (c )b, ȳ A b b 6 b y l(y) dy (c )b y c b x +c A b (b y) l(y) dy (c )b +c ] b b (y b) +c b b +c (b y) dy (by y ) dy ( b b ) b b b +c b (b y)c (b y) dy +c b b (y b) dy We hve found tht the centroid of the tringle is t ( x,ȳ) ( +c, b ). We shll now show tht this point lies on ll three medins. One vertex is t (,). The opposite side runs from(,b) nd (c,)nd so hs midpoint (c,b). The line from (,) to (c,b) hs slope b/ b nd so hs eqution c/ c y b (x ). As ( b b ( x ) +c ) b (c + ) b ȳ, c c c c the centroid does indeed lie on this medin. In this computtion we hve implicitly ssumed tht c so tht the denomintor c. In the event tht c, the medin runs from (,) to (, ) b nd so hs eqution x. When c we lso hve x +c, so tht the centroid still lies on the medin. Another vertex is t (c,). The opposite side runs from (,) nd (,b) nd so hs midpoint (,b). The line from (c,) to (,b) hs slope b/ nd so hs ( +c b / c c b eqution y b (x c). As b ( x c) b c ) (+c c) b ȳ, c c c c the centroid does indeed lie on this medin. In this computtion we hve implicitly ssumed tht c so tht the denomintor c. In the event tht c, the medin runs from (c,) to ( c, ) b nd so hs eqution x c. When c we lso hve x +c c, so tht the centroid still lies on the medin. The third vertex is t (,b). The opposite side runs from (,) nd (c,) nd so hs midpoint ( +c,). The line from (,b) to ( +c,) b hs slope b nd so hs (+c)/ +c eqution y b b b b +c x. As b +c +c x b b ȳ, the centroid does indeed +c lie on this medin. This time, we hve implicitly ssumed tht +c. In the event tht +c, the medin runs from (,b) to (,) nd so hs eqution x. When +c we lso hve x +c, so tht the centroid still lies on the medin. Exmple 4 c Joel Feldmn. 5. All rights reserved. 7 Februry 8, 5
Torque (optionl) Newton s lw of motion sys tht the position x(t) of single prticle moving under the influence of force F obeys mx (t) F. Similrly, the positions x i (t), i n, of set of prticles moving under the influence of forces F i obey mx i(t) F i, i n. Often systems of interest consist of some smll number of rigid bodies. Suppose tht we re interested in the motion of single rigid body, sy piece of wood. The piece of wood is mde up of huge number of toms. So the system of equtions determining the motion of ll of the individul toms in the piece of wood is huge. On the other hnd, becuse the piece of wood is rigid, its configurtion is completely determined by the position of, for exmple, its centre of mss nd its orienttion. (Rther thn get into wht is precisely ment by orienttion, let s just sy tht it is certinly determined by, for exmple, the positions of few of the corners of the piece of wood). It is possible to extrct from the huge system of equtions tht determine the motion of ll of the individul toms, smll system of equtions tht determine the motion of the centre of mss nd the orienttion. We cn void some vector nlysis, tht is beyond the scope of this course, by ssuming tht our rigid body is moving in two rther thn three dimensions. So, imgine piece of wood moving in the xy plne. Furthermore, imgine tht the piece of wood consists of huge number of prticles joined by huge number of weightless but very strong steel rods. The steel rod joining prticle number one to prticle number two just represents force cting between prticles number one nd two. Suppose tht there re n prticles, with prticle number i hving mss m i t time t, prticle number i hs x coordinte x i (t) nd y coordinte y i (t) t time t, the externl force (grvity nd the like) cting on prticle number i hs x coordinte H i (t) nd y coordinte V i (t). Here H stnds for horizontl nd V stnds for verticl. t time t, the force cting on prticle number i, due to the steel rod joining prticle number i to prticle number j hs x coordinte H i,j (t) nd y coordinte V i,j (t). If there is no steel rod joining prticles number i nd j, just set H i,j (t) V i,j (t). In prticulr, H i,i (t) V i,i (t). The only ssumptions tht we shll mke bout the steel rod forces re (A) for ech i j, H i,j (t) H j,i (t) nd V i,j (t) V j,i (t). In words, the steel rod joining prticles i nd j pplies equl nd opposite forces to prticles i nd j. c Joel Feldmn. 5. All rights reserved. 8 Februry 8, 5
(A) for ech i j, there is function i,j (t) such tht H i,j (t) i,j (t) x i (t) x j (t) ] nd V i,j (t) i,j (t) y i (t) y j (t) ]. In words, the force due to the rod joining prticles i nd j cts prllel to the line joining prticles i nd j. For (A) to be true, we need i,j (t) j,i (t). Newton s lw of motion, pplied to prticle number i, now tells us tht m i x i (t) H i(t)+ m i y i(t) V i (t)+ H i,j (t) (X i ) j V i,j (t) (Y i ) Adding up ll of the equtions (X i ), for i,,,, n nddding upll ofthe equtions (Y i ), for i,,,, n gives m i x i(t) m i y i(t) j H i (t)+ i,j n V i (t)+ i,j n H i,j (t) (Σ i X i ) V i,j (t) (Σ i Y i ) The sum i,j n H i,j(t) contins H, (t) exctly once nd it lso contins H, (t) exctly once nd these two terms cncel exctly, by ssumption (A). In this wy, ll terms in i,j n H i,j(t) with i j exctly cncel. All terms with i j re ssumed to be zero. So i,j n H i,j(t). Similrly, i,j n V i,j(t), so the equtions (Σ i X i ) nd (Σ i Y i ) simplify to m i x i (t) H i (t) (Σ i X i ) m i y i(t) V i (t) (Σ i Y i ) Denote by the totl mss of the system, by m i X(t) m i x i (t) nd Y(t) m i y i (t) the x nd y coordintes of the centre of mss of the system nd by H(t) H i (t) nd V(t) V i (t) c Joel Feldmn. 5. All rights reserved. 9 Februry 8, 5
the x nd y coordintes of the totl externl force cting on the system. In this nottion, the equtions (Σ i X i ) nd (Σ i Y i ) re X (t) H(t) Y (t) V(t) (4) So the centre of mss of the system moves just like single prticle of mss subject to the totl externl force. Now multiply eqution (Y i ) by x i (t), subtrct from it eqution (X i ) multiplied by y i (t), nd sum over i. This gives the eqution i xi (t)(y i ) y i (t)(x i ) ] : m i xi (t)y i(t) y i (t)x i(t) ] By the ssumption (A) xi (t)v i (t) y i (t)h i (t) ] + i,j n xi (t)v i,j (t) y i (t)h i,j (t) ] x (t)v, (t) y (t)h, (t) x (t), (t) y (t) y (t) ] y (t), (t) x (t) x (t) ], (t) y (t)x (t) x (t)y (t) ] x (t)v, (t) y (t)h, (t) x (t), (t) y (t) y (t) ] y (t), (t) x (t) x (t) ], (t) y (t)x (t)+x (t)y (t) ], (t) y (t)x (t)+x (t)y (t) ] So the i, j term in i,j n xi (t)v i,j (t) y i (t)h i,j (t) ] exctly cncels the i, j term. In this wy ll of the terms in i,j n xi (t)v i,j (t) y i (t)h i,j (t) ] with i j cncel. Ech term with i j is exctly zero. So i,j n xi (t)v i,j (t) y i (t)h i,j (t) ] nd m i xi (t)y i (t) y i(t)x i (t)] xi (t)v i (t) y i (t)h i (t) ] Define L(t) T(t) m i xi (t)y i (t) y i(t)x i (t)] xi (t)v i (t) y i (t)h i (t) ] In this nottion d L(t) T(t) (5) dt Eqution (5) plys the role of Newton s lw of motion for rottionl motion. T(t) is clled the torque nd plys the role of rottionl force. L(t) is clled the ngulr momentum (bout the origin) nd is mesure of the rte t which the piece of wood is rotting. For exmple, if prticle of mss m is trveling in circle of rdius r, centred on the origin, t ω rdins per unit time, then x(t) rcos(ωt), y(t) rsin(ωt) nd m x(t)y (t) y(t)x (t) ] m rcos(ωt) rωcos(ωt) rsin(ωt) ( rωsin(ωt) )] mr ω c Joel Feldmn. 5. All rights reserved. Februry 8, 5
is proportionl to ω, which is the rte of rottion bout the origin. In ny event, in order for the piece of wood to remin sttionry, equtions (4) nd (5) force H(t) V(t) T(t). Now suppose tht the piece of wood is seesw tht is long nd thin nd is lying on the x xis, supported on fulcrum t x p. Then every y i nd the torque simplifies to T(t) n x i(t)v i (t). The forces consist of grvity, m i g, cting downwrds on prticle number i, for ech i n nd the force F imposed by the fulcrum tht is pushing stright F m g m g m g m 4 g up on the prticle t x p. The net verticl force is V(t) F n m i g F g. If the seesw is to remin sttionry, this must be zero so tht F g. The totl torque (bout the origin) is T Fp m i gx i gp m i gx i If the seesw is to remin sttionry, this must lso be zero nd the fulcrum must be plced t.p m i x i (6) which is the centre of mss of the piece of wood. c Joel Feldmn. 5. All rights reserved. Februry 8, 5