PHY 5246: Theoretical Dynamics, Fall Assignment # 9, Solutions. y CM (θ = 0) = 2 ρ m

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1 PHY 546: Theoretical Dnamics, Fall 5 Assignment # 9, Solutions Graded Problems Problem (.a) l l/ l/ CM θ x In order to find the equation of motion of the triangle, we need to write the Lagrangian, with generalized coordinate {θ}. The potential energ is going to be based on the location of the center of mass, and we find that via CM (θ = ) = ρ m = m = 8 l l/ m l l dx d (l/ x) l/ ( l x dx ( l x ) ) l/ = 4 l l 8 = l, where the densit is ρ. This is as expected b smmetr considerations and b the fact that the CM has to be at the geometrical center of the triangle, i.e. at one third of it s height, l = l. Note the negative sign is because the triangle lies below the x-axis. So, for generalized θ, CM = ρ cosθ, and the potential energ is To calculate the kinetic energ, U = mg CM = mg l cosθ. T = T rot (about ) = I θ, we need first to find the principal moment of inertia about the axis of rotation, which is an axis perpendicular to the plane of the triangle, through. This is the onl moment of inertia we need,

2 since Therefore and the Lagrangian is ω = θê and therefore, T = ωt Î ω = θ I = θ I. I = ρ = ρ l/ l/ (l/ x) dxx d(x + ) (l/ x) ) ( l/ = ρ dxx l x = ρ ( l l 8 l l 4 ) 46 = m l 4 6 = ml 6. l L = T V = l/ d +ρ l/ +ρ T = ( ) ml θ, 6 ( ) ml The equation of motion for our generalized coordinate is 6 dx θ +mg l cosθ. (l/ x) ) ( l x d L dt θ L θ = ml 6 θ+mg l sinθ = d θ+ g l sinθ =. Under small oscillations about θ = this is the equation for a simple harmonic oscillator with frequenc g ω = l.

3 (.b) For this situation, the θ = position of the CM is CM (θ = ) = l. Therefore CM = l cosθ U = mg l cosθ. We need to calculate the moment of inertia I with respect to the new center of rotation (and axis through it). l/ l/ x l/ I = ρ l/ x dx ( /l x dx(x + ) l/ = ρ dxx d +ρ l = ρ [ ( l/ dxx x+ l ) + l 4 l 4 dx l/ ) = ρ 46 + l l4 6 = m l4 ++6 = 5 l6 6 ml. Thus the kinetic energ and Lagrangian are x d l dx ( x + l 8 )] CM θ l and the equation of motion is T = L = T V = ( 5 ml ) θ ( 5 ml ) θ +mg l cosθ, d L dt θ L θ = 5 ml θ+mg l sinθ = θ+ g l Thus the frequenc of the small oscillation is ω = 5 sinθ =. 5 g l.

4 Problem (a) CM = l cosθ θ l The cube rotates about the side perpendicular to the plane of the figure, through. We are therefore referring to the case in which the frame has axes parallel to the side of the cube and origin at one corner. In figure (), the cube is rotating about the back side, along the -axis. The corresponding moment of inertia is I = ml (also see p in Goldstein). We can find the angular velocit of the cube when one face strikes the plane, b imposing conservation of energ: E i = U i = mg l ( ml ) ω f E f = U f +T f = mg l + E i = E f mg l = mg l + ml ωf. g( ) = lω f ω f = ( ) g l.

5 z O x Figure : (b) If the cube can slide, there is no more fixed point. So, we have to calculate the kinetic energ as T = T CM trans +TCM rot = mv CM + ( 6 ml ) ω f, where: v CM = ẏ CM = l sinθ θ = l θ, and when θ = 45 circ, the cube hits the plane. With ω(θ = 45 ) = ω f, we have Or g T = ml 4 ω f + ml ω f = 5 4 lω f, ( ) = 5 4 lω f ω f = 5 ( ) g l. We could also have calculated the kinetic energ as T = Ttrans CM +TCM rot = mv CM + ICM ω f = = l ω f = Torigin rot. ( + 6) l ω f Now use v = R θ to find v CM = l ω f, and on p. 4 of Goldstein we can find I CM = ml 6. We thus obtain the same answer as above.

6 Problem (Goldstein 5.7) Take M as the mass of the cone, (x,x,x ) as the principal axes going through the CM, which is at (,,a = h) in the bod sstem. 4 x is the cone axis, x is perpendicular to the plane going through x and the segment OA, and x is orthogonal to the plane of x and x. As stated in the problem, R is the radius, h is the height and α is the half-angle of the cone. Due to the condition of no-slipping, the angular velocit of the cone, ω, is directed along the instantaneous axis of rotation which is the line of contact between the surface of the cone and the surface on which the cone is rolling (i.e. along the segment OA, see figure). x x O θ α a CM R x x x x A ω (a) The kinetic energ (T) can be calculated either using the center of mass as origin and taking as reference frame the sstem of principal axes described above, or using a sstem of axes parallel to that one but with origin at the apex of the cone O, which is a fixed point for the bod. In the first case, the kinetic energ is given b the sum of two terms: the kinetic energ of the center of mass (i.e. the kinetic energ of a pointlike object of mass M located at the CM of the cone and moving as CM moves) and the kinetic energ of the bod as rotating about the center of mass (expressed in terms of the principal moments of inertia and the component of the angular velocit in the principal axes frame or bod frame), T = T CM T CM trans = trans +Trot CM kinetic energ of the CM = Mv CM = Ma cos α θ.

7 In the above equation we have used that v CM = acosα θ as it rotates about ˆx. This implies that ω (angular velocit about OA) and θ are related. Thus we also have Combining these two relations we find The components of ω in the bod frame are This follows from our choice of x T CM rot = I ω + I ω v CM = asinαω. ω = v CM asinα = θcotα. () ω = (ωsinα,,ωcosα) = ( θcosα,, θ cos α sinα ). = { I = I = M ( R + h 4 I = MR. = ( M R + h 4 T = T CM trans +TCM and because of the relation (). Therefore: ) ) cos α θ + α MRcos4 sin α θ rot = Ma cos α θ + ( M R + h 4 = 9 Mh cos α θ + ( 4 Mh tan α+ 4 ) cos α θ + α MRcos4 sin α θ ) cos α θ + Mh cos α θ. To get this last line we used a = h, R = htanα, and the next line uses the trigonometric identit, 4 tan α =. cos α T = 9 Mh cos α θ 4 Mh θ 44 Mh cos α θ + Mh cos α θ = 4 Mh θ (+5cos α). In the second case, the kinetic energ is purel rotational, since O is a fixed point, and is given b: T = T O rot = J ω + J ω, where J and J are the principal moments of inertia with respect to a sstem of axes parallel to the ones drawn in the picture and with origin in O. The can be found using Steiner s theorem and the are simpl given b: J = I +M(a a a ) = I +Ma, J = I +M(a a a ) = I, where we have used that a = (,, a) is the position vector of O in the CM frame. It is then eas to see that, T = T O rot = I ω + Ma ω + I ω = Ma cos α θ + I ω + I ω = T CM trans +T CM rot.

8 (b) The components of the angular momentum are given b Therefore we know that L =, and L = L ê +L ê +L ê = I ω ê +I ω ê +I ω ê. L = I ω = ( ) M R + h cosα θ 4 = ( Mh cosα ) 4 cosα θ, L = I ω = MRcos α sinα θ = Mh tan α cos α sinα θ = Mh sinα θ.

9 Problem 4 x α ω In this problem the bod axes are the principal axes, and ω can move in the the bod fixed frame. It s eas to see that the plane is a smmetric top. Therefore, in absence of forces L will be constant and ω will precess around it. Let us calculate the moments of inertia explicitl: l x O x l/ l/ I = I = ρ dx x l/ l/d = ρ l l 8 = ml I = ρ l/ l/ l/ dx d (x + ) = ml l/ 6. Now at t =, ω = ( ωsinα, ωsinα ),ωcosα, and the angular momentum is L = (I ω,i ω,i ω ) = ml ( ωsinα, ωsinα ),ωsinα. The velocit with which ω precesses about L is (see discussion in class and in the text): where Ω pr = L I, L = (I ω +I ω +I ω) / = ml ω 6 = ml ω (+cos α) /. And so the frequenc of precession is Ω pr = (ml ω/)(+cos α) / ml / [ sin α 8 + sin α 8 +cos α = ω(+cos α) /. ] /

10 Non-graded Problems Problem 5 Take a generic point P and the principal axis going through it (see figure). The principal moments of inertia are I = d rf( r) [ ] r x V I = d rf( r) [ ] r x V I = d rf( r) [ r x], where f( r) is the mass densit. Therefore I i +I j = d rf( r) [ ] r x i x j V = d rf( r) [ ] x i +x j +x k V d rf( R) [ ] x i +x j V d rf( r) [ r xk] = Ik. V V The equalit in the above statement occurs when x k =. Here we have assumed that i j k for i,j,k =,,. x x P V x Problem 6 Using clindrical coordinates, z { x = rcosθ = rsinθ z = z, x CM 4 h θ r z and we have that r = (R/h)z. First determine the inertia tensor with respect to : h π Rz/h I = ρ dz dθ dr r( +z ) πr h h π M Rz/h = dz dθ drr (r sin θ+z ) πr h [ M h R 4 h = π dz +π z R ] dz πr h 4 h 4z4 h z [ M = πr h R hπ R + ] 5 h = M(R +4h ) I = I b smmetr M h π Rz/h I = dz dθ dr r(x + ) πr h M R4 h 5 = π 4h 4 5 = MR.

11 B smmetr we can also sa that the non diagonal terms are zero (each of the chosen axes is a smmetr axis for the cone). These can also be shown explicitl. For instance: I = M πr h h dz π dθ Rz/h This vanishes because x = r sinθcosθ and π dθsinθcosθ =. Now in a similar wa the other two vanish where we have used I = M πr h I = M πr h h h π dz dz xz = rcosθz z = rsinθz π π π Rz/h dθ dθ Rz/h dθcosθ =, dθsinθ =. dr rx =. dr rxz =, dr rz = So, the chosen Cartesian axes are also principal axes with respect to the origin, and Î = { MR } MR MR To obtain ÎCM, we need first to determine (x CM, CM,z CM ). B smmetr, we know x CM = CM =, while: I CM = M h π Rz/h dz dr rz M πr h h = πr h π dz z R h z = πr h π R h 4 h 4 = 4 h. Finall, we can appl the generalized form of Steiner s parallel axis theorem, according to which which in our case a = (,, h), so we have 4 (I CM ) ij = (I ) ij M [ a δ ij a i a j ], ( ) (I CM ) = (I ) 4 h M = M(R +4h ) 9 6 h M = ( ) M R h 4 (I CM ) = (I CM ) b smmetr (I CM ) = (I ) = MR, and the non diagonal elements are still all zero.

12 Problem 7 Defining ρ as the mass densit, we use polar coordinates { x = rcosθ = rsinθ Given a coordinate sstem (x, ) which rotates with the disk (see figure ()), the location of the CM is x CM = and ȳ CM = ρ { } dxd + dxd M top bottom = ρ { R π R } π drr dθrsinθ+ drr dθrsinθ M π = ρ { } R R M 4 = ρr M = 4R 9π. Where in the last line we have used M = ρ πr +ρπr = ρπr. Again referencing Figures () and () we see that the relationship between the lab coordinates and the coordinates of the center of mass are { x CM = Rθ ȳ CM sinθ = Rθ 4R sinθ 9π CM = R ȳ CM cosθ = R 4R cosθ. 9π { ẋ CM = R θ 4R cosθ θ 9π 4R ẏ CM = sinθ θ. 9π The kinetic energ of the disk is made of terms: T = T trans +T rot, ρ ρ R r (x,) θ Figure : where T trans is the translational kinetic energ of the mass M with coordinates (x CM, CM ) and velocit (ẋ CM,ẏ CM ), and T rot is the rotational kinetic energ about the center of mass. Thus T = M(ẋ CM +ẏ CM)+ I θ. Here I is the moment of inertia about the CM, with respect to the z-axis, perpendicular to the plane of the figure. To find I we will calculate it with respect to the center of disk (since it is easier!) and use Steiner s theorem to obtain it with respect to the CM. I = ρ R drr π dθ(x + )+ρ R drr = ρ R4 4 π +ρr4 4 π = ρr4 4 π = MR CM I = I MȳCM = MR M 6 8 = [ MR ]. 8π R π π π dθ(x + ) x

13 Then we can calculate the kinetic energ T = T trans +T rot = M(ẋ CM +ẏ CM )+ I θ CM = [ ( MR θ 4 ) ( ) ] 4 9π cosθ + 9π sinθ + 4 MR θ = [ MR θ + 6 8π 8 9π cosθ + 6 ] 8π = [ MR θ 8 ] 9π cosθ. The potential energ must also be defined with respect to the center of mass, [ U = Mg CM = MgR 4 ] 9π cosθ, and the Lagrangian is the sum of these two terms, L = T V = MR θ [ ] 8π [ 8 ] [ 9π cosθ mgr 4 ] 9π cosθ. ρ ρ x (x CM, CM ) x θ ȳ CM x Figure :