{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 260 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 285 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 290 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 260 37 "Rolls and Strings and Yo- Yo Solutions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "A roll (radius " }{TEXT 258 1 "R" }{TEXT -1 7 ", mass " }{TEXT 257 1 "M" }{TEXT -1 62 ") has rope wound around it which leads via a p ulley to a mass " }{TEXT 256 1 "m" }{TEXT -1 26 " that falls under gra vity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 675 "The point of this worksheet is to explain how the motion of the cente r of mass (CM) is affected by the net force that acts on the roll: the string force, which applies where the string is unwound, and the stat ic friction force which acts at the point of contact (assuming that th ere is rolling without slipping). Both these forces contribute torques about the CM of the roll. This work shows that the direction of the f riction force is difficult to predict. There are situations where it h elps to push the CM forward, and acts against the rolling, and there a re situtations where it helps the rolling, but acts against the motion of the CM. This is a consequence of applying:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 283 3 "(i)" }{TEXT -1 156 " the \+ CM theorem: all force vectors acting on the body are applying at the l ocation of the CM, and determine the rate of change of the total linea r momentum;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 284 4 "(ii)" }{TEXT -1 543 " the angular momentum theorem: the rate of change of angular momentum is controlled by the net torque that appli es; one can use always the quantities about the CM, or about the point of contact (in which case friction does not contribute a torque), how ever, only in the case of pure rolling (no slipping) is the angular ve locity about the CM the same as the angular velocity about the point o f contact. In the case of pure rolling the part of the roll that touch es the ground at the point of contact has no motion, and static fricti on applies." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 285 5 "(iii)" }{TEXT -1 142 " the case of pure rolling is characterize d by a fixed relationship between the CM motion and the angular veloci ty that describes the rotation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 286 4 "(iv)" }{TEXT -1 148 " the mass that falls under gravity undergoes a motion that is composed of motion of the CM and the additional effect of the unwinding of the string." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "rest art;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "We define moments of ine rtia for a disk about the CM, as well as about the point of contact O; the latter is obtained from the parallel-axis theorem." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "I_CM:=M/2*R^2;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 16 "I_O:=I_CM+M*R^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Motion of the center of mass: apply the forces at the CM, namely the friction force " }{TEXT 19 1 "f" }{TEXT -1 23 ", and the s tring force " }{TEXT 19 1 "S" }{TEXT -1 115 ". We choose the same sign for both forces and let the calculationd decide which way the frictio n force will point.." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eq1 :=M*Xdd=S+f;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "For pure rolling \+ we can apply the torque about the contact point. The arm length is giv en by 2" }{TEXT 287 1 "R" }{TEXT -1 2 ". " }{TEXT 19 3 "Tdd" }{TEXT -1 59 " is the 2nd derivative of the rotation angle. The friction " } {TEXT 19 1 "f" }{TEXT -1 58 " contributes no torque, given that the ar m length is zero." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eq2:=I _O*Tdd=2*R*S;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "The height of th e mass falling under gravity is given by " }{TEXT 19 1 "Y" }{TEXT -1 29 ". We choose a different mass " }{TEXT 261 2 "m " }{TEXT -1 33 "for the bob attached to the rope." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq3:=m*Ydd=m*g-S;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The \+ CM motion " }{TEXT 19 1 "X" }{TEXT -1 67 " is linked to the angular ro tation in perfect rolling without slip:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eq4:=R*Tdd=Xdd;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "We need to connect the fall with the unravelling of wire (" } {TEXT 19 5 "R*Tdd" }{TEXT -1 33 "), and with the motion of the CM." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eq5:=Ydd=Xdd+R*Tdd;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "How many equations/unknowns? The u nknowns are: " }{TEXT 19 19 "Ydd, Xdd, Tdd, S, f" }{TEXT -1 416 " - so we should use five equations. Equation one is redundant for the deter mination of the motion, because we use the rotation equation about the point of contact. If we were to consider the Newton equation (angular momentum theorem) about the CM of the roll, the frictional torque wou ld enter, and we would need to determine it as well. In the present ca se we use the first equation to determine the friction force." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "sol:=solve(\{eq1,eq2,eq3,eq4 ,eq5\},\{Tdd,Xdd,Ydd,S,f\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 " Note that the friction force has the same sign as the string force, i. e., it is helping to propel the roll's CM forward." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(sol) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "m:=1; M:=2; g:=10; R:= 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Ydd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Xdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Tdd;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "We want a graph of a circle with a radius vector indicating the location of a m arker (" }{TEXT 19 3 "Tdd" }{TEXT -1 21 " is the acceleration)" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "theta:=unapply(1/2*Tdd*t^2,t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "X:=unapply(1/2*Xdd*t^ 2,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Y:=unapply(1/2*Ydd *t^2,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "with(plottools) : with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "arrowscal e:=0.25:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 566 "c := t-> [circ le([X(t),R], R, color=red,thickness=3),line([X(t),R],[X(t)+R*cos(theta (t)+Pi/2),R-R*sin(theta(t)+Pi/2)],color=blue,thickness=3),line([X(t),2 *R],[8.2,2*R],color=grey,thickness=3),line([8.2,2*R],[8.2,-Y(t)],color =grey,thickness=3),circle([8.2,-Y(t)],0.15,color=red,thickness=3),arro w([X(t),0], [X(t)+f*arrowscale,0], .2, .4, .1, color=magenta),arrow([X (t),2*R], [X(t)+S*arrowscale,2*R], .2, .4, .1, color=cyan),arrow([8.2, -Y(t)], [8.2,-Y(t)+S*arrowscale], .2, .4, .1, color=cyan),arrow([8.2,- Y(t)], [8.2,-Y(t)-m*g*arrowscale], .2, .4, .1, color=green)]:\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "PL:=[seq(display(c(2.2*i/60) ,scaling=constrained,view=[-2..8.4,-8..2.2]),i=0..60)]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plots[display](op(PL),insequence=tr ue,scaling=constrained);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 289 "The \+ friction force points forward, i.e., it acts against the rotational to rque about the CM provided by the string force, but it helps to move t he CM to the right. The system adjusts itself such that the condition \+ of pure rolling is met, i.e., such that the winding of theta (accelera tion " }{TEXT 19 3 "Tdd" }{TEXT -1 50 ") and the movement of the CM (l inear acceleration " }{TEXT 19 3 "Xdd" }{TEXT -1 19 ") are synchronize d." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "Wha t is the change in string length after, say, 2.2 seconds?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "evalf(Y(2.2)-X(2.2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 237 "Now let us re-do the problem using rotat ion about the center of mass. In this case it is evident how the frict ion force helps to propel the CM forward, but at the same time it acts as an opposite torque to the string torque about the CM." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 14 "I_CM:=M/2*R^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "I_O:=I_CM+M*R^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Motion of the center of mass: apply the forces at the CM:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eq1:=M*Xdd=S+f;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "We use the angular momentum theorem for a ngular momentum " }{TEXT 262 1 "L" }{TEXT -1 12 " and torque " }{TEXT 263 1 "N" }{TEXT -1 237 " defined about the CM: we have two torque con tributions that oppose each other. Tdd is the 2nd derivative of the ro tation angle.For rotation with slipping the angular velocity about the CM and about the point of contact are not the same." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq2:=I_CM*Tdd=R*S-R*f;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq3:=m*Ydd=m*g-S;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "eq4:=R*Tdd=Xdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eq5:=Ydd=Xdd+R*Tdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "sol:=solve(\{eq1,eq2,eq3,eq4,eq5\},\{Tdd,Xdd,Ydd,S,f \});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 375 "Evidently the answer is \+ the same as before! This formulation is more flexible, since it allows to discuss the case when the surface cannot provide the friction torq ue/force required due to an insufficient value of static friction cons tant mu. In this case the disk is not at rest with respect to the surf ace at the point of contact. In particular, we can discuss the case of " }{TEXT 19 3 "f=0" }{TEXT -1 42 " (negligible static and kinetic fri ction)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eq1:=M*Xdd=S;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "We use the angular momentum theorem for angular momentum " }{TEXT 264 1 "L" }{TEXT -1 12 " and torque " }{TEXT 265 1 "N" }{TEXT -1 80 " \+ defined about the CM: we have two torque contributions that oppose eac h other. " }{TEXT 19 3 "Tdd" }{TEXT -1 155 " is the 2nd derivative of \+ the rotation angle. For rotation with slipping the angular velocity ab out the CM and about the point of contact are not the same." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eq2:=I_CM*Tdd=R*S;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq3:=m*Ydd=m*g-S;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The CM motion " }{TEXT 19 1 "X" } {TEXT -1 253 " is unrelated to the angular rotation in rolling with to tal slip [rolling doesn't help at all to move forward - spinning car w heels on ice; but in contrast to that problem, where torque is the onl y source of motion we are pulling the disk in this case]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq4:='eq4': #eq4:=R*Tdd=Xdd;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "We need to connect the fall with t he unravelling of wire (" }{TEXT 19 5 "R*Tdd" }{TEXT -1 33 "), and wit h the motion of the CM." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 " eq5:=Ydd=Xdd+R*Tdd;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "How many e quations/unknowns? The unknowns are: " }{TEXT 19 16 "Ydd, Xdd, Tdd, S " }{TEXT -1 35 " - so we should use four equations." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=solve(\{eq1,eq2,eq3,eq5\},\{Tdd,Xdd, Ydd,S\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "The mass " }{TEXT 259 1 "m" }{TEXT -1 182 " accelerates on account of translation and ro tation as before, the effect from the rotation (wire unravels) is twic e as important as the effect of pulling the CM by the string force " } {TEXT 266 1 "S" }{TEXT -1 3 ". (" }{TEXT 19 3 "Ydd" }{TEXT -1 28 " is \+ three times as large as " }{TEXT 19 3 "Xdd" }{TEXT -1 78 "; in the rol l without slip they are contributing equally due to the condition " } {TEXT 19 11 "Xdd = R*Tdd" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 53 "Explore the above solution for a simple case such as " } {TEXT 277 1 "m" }{TEXT -1 1 "=" }{TEXT 276 1 "M" }{TEXT -1 69 ": how d oes the loss of friction manifest iteself in the fall of mass " } {TEXT 278 1 "m" }{TEXT -1 1 "?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 279 9 "Question:" }{TEXT -1 36 " How woul d we solve a problem where " }{TEXT 267 1 "f" }{TEXT -1 124 " is not z ero, but less than the amount required to keep the rotation locked wit h the CM translation (no-slip)? We would add " }{TEXT 268 1 "f" } {TEXT -1 61 " as a degree of freedom, and we would need a replacement \+ for " }{TEXT 19 3 "eq4" }{TEXT -1 26 ", which would specify how " } {TEXT 19 3 "Tdd" }{TEXT -1 5 " and " }{TEXT 19 3 "Xdd" }{TEXT -1 13 " \+ are related." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 200 " Now we consider another, related problem: A yo-yo is \+ being pulled in the horizontal direction. Let us assume that the radiu s at which the string is wound up is one-half the radius of the disk, \+ i.e., " }{TEXT 269 1 "R" }{TEXT -1 37 "/2. We go back to no-slip condi tions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "I_CM:=M/2*R^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "I_O:=I_C M+M*R^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eq1:=M*Xdd=S+f; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "The radius at which the yo-yo unwinds the string: (we pick a value very close to " }{TEXT 290 1 "R " }{TEXT -1 53 "/2, but not exactly to avoid plotting problems below) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "R1:=19*R/40;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "The torque from the pulling string has arm length " }{TEXT 19 2 "R1" }{TEXT -1 56 ", and applies on the \+ same side as in the example before:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "eq2:=I_CM*Tdd=R1*S-R*f;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq3:=m*Ydd=m*g-S;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eq4:=R*Tdd=Xdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq5:=Ydd=Xdd+R1*Tdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "sol:=solve(\{eq1,eq2,eq3,eq4,eq5\},\{Tdd,Xdd,Ydd,S,f \});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "Interesting coincidence: there will be no static friction force! When the armlength where the \+ string force applies is larger than " }{TEXT 270 1 "R" }{TEXT -1 30 "/ 2, a positive friction force " }{TEXT 271 1 "f" }{TEXT -1 39 " results , for arm lengths smaller than " }{TEXT 272 1 "R" }{TEXT -1 30 "/2 it \+ will be negative, i.e., " }{TEXT 273 1 "f" }{TEXT -1 3 "<0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 274 11 "Exercise 2:" } }{PARA 0 "" 0 "" {TEXT -1 91 "Vary the inner radius of the yo-yo, and \+ observe how the role of the friction force changes." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(sol);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "m:=1; M:=2; g:=10; R:=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Ydd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "X dd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Tdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "theta:=unapply(1/2*Tdd*t^2,t);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "X:=unapply(1/2*Xdd*t^2,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Y:=unapply(1/2*Ydd*t^2,t) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "with(plottools): with( plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "arrowscale:=0.25 :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 615 "c := t-> [circle([X(t ),R], R, color=red,thickness=3),circle([X(t),R], R1, color=red,thickne ss=3),line([X(t),R],[X(t)+R*cos(theta(t)+Pi/2),R-R*sin(theta(t)+Pi/2)] ,color=blue,thickness=3),line([X(t),R+R1],[8.2,R+R1],color=grey,thickn ess=3),line([8.2,R+R1],[8.2,-Y(t)],color=grey,thickness=3),circle([8.2 ,-Y(t)],0.15,color=red,thickness=3),arrow([X(t),R+R1], [X(t)+S*arrowsc ale,R+R1], .2, .4, .1, color=cyan),arrow([X(t),0], [X(t)+f*arrowscale, 0], .2, .4, .1, color=magenta),arrow([8.2,-Y(t)], [8.2,-Y(t)+S*arrowsc ale], .2, .4, .1, color=cyan),arrow([8.2,-Y(t)], [8.2,-Y(t)-m*g*arrows cale], .2, .4, .1, color=green)]:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "PL:=[seq(display(c(2.2*i/60),scaling=constrained,view =[-2..8.4,-8..2]),i=0..60)]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plots[display](op(PL),insequence=true,scaling=constrained);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 381 "This calculation shows that forwa rd motion without any friction provided by the surface is possible. In fact, we can see from the equations that the case when the string for ce applies at one half the radius corresponds to the situation when no friction is required to keep the rotation and the CM motion in lockst ep. It is a consequence of the moment of inertia for the disk being " }{XPPEDIT 18 0 "M*R^2/2;" "6#*(%\"MG\"\"\"*$%\"RG\"\"#F%F(!\"\"" } {TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 40 "The string unwound the \+ following amount:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "evalf( Y(2.2)-X(2.2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 281 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 51 "Modify the above yo-yo set-up such that the radius " }{TEXT 19 1 "r" }{TEXT -1 51 " at which the string is wo und up is different from " }{TEXT 282 1 "R" }{TEXT -1 11 "/2; choose \+ " }{TEXT 19 7 "r:=R/4:" }{TEXT -1 5 " and " }{TEXT 19 9 "r:=3*R/4:" } {TEXT -1 58 " respectively, and show the friction force on the diagram ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 117 "Now let us pull the string under conditions where the yo-yo is upside-dow n. Let us pick an adjustable winding radius " }{TEXT 19 2 "R1" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "I_CM:=M/2*R^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "I_O:=I_CM+M*R^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eq1:=M*Xdd=S+f;" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 50 "The torque from the pulling string has arm length " } {TEXT 19 2 "R1" }{TEXT -1 95 ", and applies on the opposite side as in the example before, i.e., enters with a negative sign:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "R1:=1*R/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eq2:=I_CM*Tdd=-R1*S-R*f;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 17 "eq3:=m*Ydd=m*g-S;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 15 "eq4:=R*Tdd=Xdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq5:=Ydd=Xdd-R1*Tdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "sol:=solve(\{eq1,eq2,eq3,eq4,eq5\},\{Tdd,Xdd,Ydd,S,f \});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 "Interesting observation: the static friction force now opposes the string force. Its torque ex ceeds the string torque to allow forward motion of the roll!" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 " assign(sol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "m:=1; M:=2; g:=10; R:=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Ydd;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Xdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Tdd;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "theta:=unapply(1/2*Tdd*t^2,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "X:=unapply(1/2*Xdd*t^2,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Y:=unapply(1/2*Ydd*t^2,t);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "with(plottools): with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "arrowscale:=0.25:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 615 "c := t-> [circle([X(t),R], R, color=red, thickness=3),circle([X(t),R], R1, color=red,thickness=3),line([X(t),R] ,[X(t)+R*cos(theta(t)+Pi/2),R-R*sin(theta(t)+Pi/2)],color=blue,thickne ss=3),line([X(t),R-R1],[8.2,R-R1],color=grey,thickness=3),line([8.2,R- R1],[8.2,-Y(t)],color=grey,thickness=3),circle([8.2,-Y(t)],0.15,color= red,thickness=3),arrow([X(t),0], [X(t)+f*arrowscale,0], .2, .4, .1, co lor=magenta),arrow([X(t),R-R1], [X(t)+S*arrowscale,R-R1], .2, .4, .1, \+ color=cyan),arrow([8.2,-Y(t)], [8.2,-Y(t)+S*arrowscale], .2, .4, .1, c olor=cyan),arrow([8.2,-Y(t)], [8.2,-Y(t)-m*g*arrowscale], .2, .4, .1, \+ color=green)]:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "PL:=[se q(display(c(2.2*i/60),scaling=constrained,view=[-2..8.4,-8..2]),i=0..6 0)]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plots[display](op(P L),insequence=true,scaling=constrained);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 443 "Note that the torque from the string force acts against \+ the rotation, and consequently agains the forward motion of the disk d ue to rotation, but that the force applies at the CM, and therefore th e disk is going to accelerate to the right! In this example (winding o rientation) the string winds itself around the yo-yo as the roll accel erates to the right. The friction force adjusts itself (if possible, i .e., if the static friction constant " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 25 " is large enough so that " }{TEXT 288 1 "f" }{TEXT -1 17 " does not exceed " }{XPPEDIT 18 0 "mu*M*g;" "6#*(%#muG\"\"\"%\"MGF %%\"gGF%" }{TEXT -1 333 " ) in such a way that the the condition of pu re rolling is satisfied. This raises the question why pure rolling is \+ desirable. The origin of this pure rolling (and of static friction in \+ general) are the molecular forces of electrostatic origin that act bet ween the particles which make up the cylinder and the table. The norma l force " }{TEXT 289 2 "Mg" }{TEXT -1 127 " helps the molecules from t he different materials to get closer, so that the bonds occur more oft en, and the material constant " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 70 " describes how effective they are. Anything that is within this limit " }{XPPEDIT 18 0 "mu*M*g;" "6#*(%#muG\"\"\"%\"MGF%%\"gGF%" } {TEXT -1 388 " is allowed to occur. Compare this condition to static f riction when trying to push a block resting on a surface. The block wo n't move until the pushing force exceeds the static friction force. In the same sense pure rolling is bound to occur until the friction requ ired for pure rolling motion (forward or backward, depending on the ca se as demonstrated above) exceeds the allowed limit." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "The yo-yo is winding it self up by the amount:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "e valf(Y(2.2)-X(2.2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 500 "We can a sk another question to understand the role of friction, namely how it \+ helps the rotation and resists the forward motion of the center of mas s, and vice versa: we can calculate the amount of work performed for t he rotation and the translation. We integrate the frictional torque ab out the CM times the angular velocity, and also the linear friction fo rce times the linear velocity. We use the sign for the friction force \+ and the frictional torque with which they enter the equations of motio n:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "W_CM:=int(f*diff(X(t) ,t),t=0..2.2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "W_rot:=in t(-R*f*diff(theta(t),t),t=0..2.2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 457 "The contributions have to cancel, because static friction impl ies that there is no net motion carried out [the forward translation o f the CM is perfectly balanced by the tangential velocity at the conta ct point due to the rotation]. Also we were able to show that one can \+ calculate the entire motion without considering the friction at all [t he rotation of the disk about the contact point about which friction c ontributes no torque due to zero arm length]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 280 11 "Exercise 4:" }} {PARA 0 "" 0 "" {TEXT -1 130 "Change the inner yo-yo radius and find o ut whether there are characteristic changes in the behaviour of the st atic friction force." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }}{MARK "122" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }