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Linear Acceleration of Rolling Objects Rotational Motion (Cont.) R Θ

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Linear Acceleration of Rolling Objects Rotational Motion (Cont.) R Θ Rotational Motion (cont.) Linear acceleration of rolling objects R Consider a round object (this could be a cylinder, hoop, sphere aCM or spherical shell) having mass M, radius R and rotational inertia I about its center of mass, rolling without θ slipping down an inclined plane. What is the linear acceleration of the object’s center of mass, aCM , down the incline? We analyze this as follows: The force of gravity, Mg, acting straight down is resolved into components parallel and perpendicular to the incline. aCM Since the object rolls without f R slipping there is a force of θ friction, f, acting on the object, Mgcosθ at it’s point of contact with the Mg incline, in the direction up Mgsin θ θ the incline. Newton’s 2nd law gives then for acceleration down the incline, = ∑F MaCM Mgsinθ− f = MaCM The force of friction also causes a torque around the center of mass having lever arm R so we can also write, τ=Rf = I α Solving for the friction, aCM f θ I f = α Mgcosθ R Mg Mgsin θ θ This is used in the expression derived from the 2nd law: Mgsinθ− f = MaCM I Mgsinθ− α= Ma R CM The objects angular acceleration is related to the linear acceleration of the edge that contacts the incline by, aR= α Since the object rolls without aCM slipping this has the same f θ magnitude as aCM so we have that, Mgcosθ a α= CM Mg R Mgsin θ θ Using this in, I Mgsinθ− α= Ma R CM IIa Mgsin θ−CM =Ma → − Ma − a =− Mgsin θ RR CM CM R2 CM Multiplying through by –1, I Ma+=θ a Mgsin CM R 2 CM I aCM MaCM 1+=θ Mgsin MR 2 f θ Mgcosθ I aCM 1+=θ gsin MR 2 Mg Mgsin θ θ So that ,finally, gsin θ a = CM I 1+ MR 2 gsin θ a = CM I a 1+ CM MR 2 f θ If the object is a solid Mgcosθ cylinder , 1 I= MR 2 Mg 2 Mgsin θ θ gsinθ gsin θθ gsin 2 a = = = =gsin θ (solid cylinder) CM 12 13 MR + 3 1+ 2 1 MR 2 22 If instead the object is cylindrical shell, with all its mass at the rim I = MR 2 and, gsinθθ gsin 1 a = = =gsin θ (cylindrical shell) CM MR 2 11+ 2 1+ MR 2 Angular Momentum The rotational analog to linear momentum (p = mv) is angular momentum, LI= ω Recall that linear momentum is important because given a system of objects, in the absence of external forces, no matter how the objects of the system interact with each other, their total linear momentum is conserved. I.e. with, n Pp= ∑ i i1= For any two times, PPfi= Thus if there are two objects with linear momenta p 1 and p 2 so that their total momentum is at one time, Ppi= 1i + p 2i Then no matter how they interact, collide, attract or repel each other, in the absence of external forces, this total momentum will not change so we have that, pp1f+=+ 2f pp 1i 2i Similarly, for rotational motion, in the absence of external torques (i.e.∑ τ= 0 ), the angular momentum is conserved meaning that for any two times, LLfi= Top view Example m v A boy (mass m) runs with speed v and jumps onto the edge of an initially R stationary merry-go-round (also called a carousel). M What is the angular velocity of the carousel (and the boy) after he has jumped on? final Conservation of angular momentum R requires, ω Lf = Li M Top view We must first consider the boy’s m angular momentum about the rotation v axis of the carousel when he is running r in a straight line. His angular momentum is, B=I BB ω M Where IB is his rotational inertia about the axis and ωB is his angular velocity about the axis. We treat the boy as a point object of mass m, making his rotational inertia about the carousel axis 2 IB = mr Where r is his (changing) distance to the carousel axis. What is the boy’s angular velocity m Top view ωB about the axis? v r When we previously considered motion on a circular trajectory we had that vt = rω where the subscript t reminds us that M this was velocity tangent to the circle. Tangent to the circle means the component of the velocity that is perpendicular to a line drawn to the center of the circle (all of v, when the trajectory is circular). For this more general case we must decompose the velocity vector into components parallel and perpendicular to the line to the rotation axis. Then, vsin θ vcosθ θ vsinθ= r ωB m θ v v or, ω=sin θ r B r 2 Using this and IB = mr M in, 2 v B =(mr ) sin θ gives, r vsin θ vcosθ θ θ v B =mrvsin θ= mv(rsin θ ) m R But notice that along his linear r trajectory r sin θ is also the point of closest approach to the axis which here is R so that, B =mv(rsin θ= ) mvR vsin θ vcosθ Since all of these m, v and R θ are constant what we have just m θ v shown is that despite the fact r that r and θ are both changing as the boy approaches the carousel his angular momentum about the M axis remains constant and is given by vsin θ vcosθ B = mvR θ m θ v Where R is the point of closest R approach to the axis. r (this must be true for angular momentum to be conserved. I.e. B calculated at any two times along this trajectory had better be the same). Having shown this once, we don’t need to re-derive it every m v time it comes up. r What needs to be remembered is that an object travelling linearly, with constant v, on a trajectory to M pass an arbitrary point has a constant angular momentum about that point given by, v m = mvR R B r Where R is the point of closest approach of the trajectory to the point. Top view The initial angular momentum of m the system is the sum of the initial v angular momenta of the boy Bi = mvR and the carousel Ci = 0 (since ωCi = 0), R Li=+= Bi Ci mvR M After the boy jumps on we have final LIf= ff ω Where If is the combined rotational inertia of the boy and the carousel. The carousel R can be treated as a uniform disk with ω 1 I= MR 2 M C 2 Top view For rotation about the same axis m rotational inertia are simply additive v so, R 1M222 If = MR +=+ mR m R 22 Then, M M 2 Lf= I ff ω= + m R ω=f L i = mvR 2 final Solving for ωf , mv ω=f M ω R + mR 2 M Check behavior if M → 0 . New example initial Suppose the boy is on the rim of R carousel at radius R from the center ω and the two have angular velocity i. ωi The boy then pulls himself along the hand rail until he is a distance R/3 M from the center. final What is ω f after this move? Lf = Li R/3 ωf Bf+=+ Cf Bi Ci 2 M R122 1 2 mω+ω=ω+ωfMR fi mR MR i 32 2 initial 2 112 22 1 2 m Rω+ω=ω+ωf MR fi mR MR i R 32 2 ωi mM 22 M +ω=+ω Rfi mR 92 2 M mM M final +ω=+ω fim 92 2 M m + R/3 ω ω=2 ω f fimM + M 92 M initial m + ω=2 ω fimM R + 92 ωi Suppose that M = 6m then, M 6m m + final ω=2 ω fim 6m + 92 R/3 ω (13+ ) f ω=f ω= ii1.29 ω 1 M + 3 9 So the angular velocity increases when the boy moves in. If the boy’s mass is m = 25 kg, the carousel’s initial mass is 6m = 150 kg the initial angular velocity was ωi = 2 rad/s and R = 1.5 m. R How much work did the boy do in pulling himself in to R/3? ωi Wnc =∆ KU +∆ M There is no ∆U here so, final Wnc =∆=− KKf K i 1122 WInc= f ω− f I i ω i R/3 22 ωf Above we found that, M rad ω=1.29 ω= 2.58 fis 221 initial I= mR + MR i 2 R 1 22 Ii =+= m 6m R 4mR 2 ωi I= 4(25kg)(1.5m)22= 225kg ⋅ m i M 2 R12 If = m + MR 32 final 1122 2 If = m R += 6mR 3.11mR 92 R/3 ωf I= 3.11(25kg)(1.5m)22= 175kg ⋅ m f M initial So with, rad I= 225kg ⋅ m2 ω=2.00 R i i s rad ω I= 175kg ⋅ m2 ω=2.58 i f f s M 11 WI= ω−22 I ω nc22 f f i i final Wnc = 582.4 J −= 450 J 132 J R/3 ωf M Summary of translational quantities and their rotational analogs .
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