Part 8: Rigid Body Dynamics

Document that contains homework problems. Comment out the solutions when printing off for students. Part 8: Rigid Body Dynamics Problem 1. Inertia review Find the moment of inertia for a thin uniform rod of mass M and length L: When it lies in the x y plane and is made to rotate about its midpoint (L/2) with ω pointing int the z-axis. When it is pivoted about one end with ω again pointing int the z-axis. For first part, using dm/dx = M/L: For the second part: I = I = r 2 dm = r 2 dm = L/2 L/2 L x 2 M L dx = ML2 /12 (1) x 2 M L dx = ML2 /3 (2) Note: you can define a radius of gyration, the radius at which the the rotation properties would be the same as a point mass of mass M, which for the second part is L/ 3. Problem 2. Rolling cube Consider a cube balancing perfectly on one edge, with sides of length a. At t = it is nudged and rolls over in the ˆx direction onto its side. Assuming initial speed is effectively, what is the angular speed when it hits the ground? You can assume that the inertia of the cube about its centre is Ma 2 /6 in this problem. As a challenging extension, what if the surface the cube rests on were frictionless? Conservation of energy is the easiest way to to this, with gravitational potential depending only on the height of the CoM: U g = mgh CM (3) where h CM (t = ) = a/ 2, and finishes at a/2. There is no kinetic energy beforehand, and after the motion we can use both the translation of the CoM and the rotational KE to give: ( ) a ( a mg = mg + 2 2) 1 2 mv2 CM + 1 2 Iω2 (4) 1

as the CoM is always a/ 2 (i.e. the axis of rotation doesn t change length) then the centre of mass velocity is v CM = ωa/ 2 (for rolling without slipping). Plugging everything in: mga( 2 1) = ma 2 ω 2 /2 + ma 2 ω 2 /6 = 2 3 ma2 ω 2 (5) so: 3g ω = 2a ( 2 1).8 g/a (6) If the surface is frictionless, then the cube will just drop straight down once the bottom gives way. As the cube falls, the CoM is at a height of: y = a 2 cos θ (7) where θ is the rotation angle of the cube (check: y(45 ) = a/2 as required). The vertical velocity is therefore: ẏ θ=π/4 = a θ sin θ θ=π/4 = 1 aω (8) 2 2 where ω = θ still represents the rotation frequency of the cube. Equating energies once again: ( ) a ( a mg = mg + 2 2) 2 mẏ2 (θ=45 )+ 1 2 Iω2 (9) subbing the same expression for inertia, and our new velocity term, this arranges to: 12g ω = 5a ( 2 1) g/a (1) which is faster than the non-slip version, as expected! Problem 3. Inertia of cone (needs I. tensor!) Find the inertia tensor for a spinning top that is effectively a uniform density cone (mass M, height h, radius of base R) spinning on its end. Use the axis of rotation as the ẑ direction. What is special about this inertia tensor? The volume of a cone is πr 2 h/3. Have to use integrals because of the solid nature of the body. going through the elements: I zz = σ(x 2 + y 2 ) dv (11) V where I will use σ as volume density. The easiest way to do this is via cylindrical polars: ρ, φ, z, as x 2 + y 2 = ρ 2, so: I zz = σ h dz 2π dφ 2 r=rz/h (ρ 2 )ρ dρ (12)

where we have to ensure we pass on a variable from the ρ integral to that in z as they are dependent, and the cone has a radius r = Rz/h at a height z. This gives: Inspection should indicate that I xx = I yy, and: I xx = σ (y 2 + z 2 ) = σ V I zz = 3 1 MR2 (13) V y 2 dv + σ z 2 dv (14) V You could try and work this out directly, but a simple check is that we already worked out the y 2 part in I zz, and you should be able to see that the x 2 and y 2 parts in I zz should be the same due to symmetry. As such, the first part is simply I zz /2 = 3MR 2 /2. The second part can be evaluated via polars: so: σ z 2 dv = σ V h z 2 dz 2π dφ r=rz/h ρ dρ = 3 5 Mh2 (15) I xx = 3 2 MR2 + 3 5 Mh2 = 3 2 M(R2 + 4h 2 ) (16) As the cone is symmetric about the x = and y =, making all the products of inertia. This gives our inertia tensor as: I = 3 R 2 + 4h 2 2 M R 2 + 4h 2 (17) 2R 2 This gives a diagonal inertia tensor. The importance is seen if we try and find the angular momentum about some arbitrary axis, in which case we obtain: L = I ω = (λ 1 ω x, λ 2 ω y, λ 3 ω z ) (18) where λ are the diagonal elements of our inertia tensor. So, if ω points along any of our coordinate axes then so will the angular momentum vector! This is an important property of diagonal inertia tensors. Problem 4. Rolling cube: better! Perform the same calculation as before, but now do NOT assume the inertia for the cube is simply Ma 2 /6, i.e. explicitly define a rotation about the edge of a cube, and show that you obtain the same result. You do not need to derive the inertia of a cone from scratch (use your notes!). 3

Our approach from before was OK, but we cheated a little bit. First, kinetic energy should really be calculated from the inertia tensor as: T = 1 2 ωt I ω (19) where ω = (,, ω). Now, to use this expression for the full kinetic energy, we need to re-compute the inertia tensor about an edge. To do so, we will shift our parallel axis by a/2 in x and y, but not in z. We can then re-construct our new tensor via: I ij = I ij + M(d 2 δ ij d i d j ) (2) where we have let d be our translation vector: d = (a/2, a/2, ). We will use our rotation tensor for CoM rotations: I = Ma 2 1/6. So, for the moments (d 2 = (a/2) 2 + (a/2) 2 = a 2 /2): I xx = I xx + M(d 2 δ xx d x d x ) = Ma 2 /6 + M(a 2 /2 a 2 /4) = 5 12 Ma2 I yy (21) I zz = I zz + M(d 2 δ zz d z d z ) = Ma 2 /6 + M(a 2 /2 ) = 2 3 Ma2 (22) I xy = I xy + M(d 2 δ xy d x d y ) = + M( a 2 /4) = 1 4 Ma2 I yx (23) I xz = I xz + M(d 2 δ xz d x d z ) = + M( ) = I zx I yz I zy (24) so finally our inertia tensor for rotating about the z-edge: 5/12 1/4 I = Ma 2 1/4 5/12 (25) 2/3 Evaluating our kinetic energy term we now have: T = 2/3Ma 2 ω 2 (26) 4

Something important here is that we can separate kinetic energy into rotation about the CoM and motion of the CoM: T = T CM + T rot (27) where rotation here is defined wrt the CoM, where we effectively used this in the previous question. Now though, our new inertia tensor includes the motion of the CoM wrt the rotational axis, so that this is the only KE term we need: ( ) a ( a mg = mg + (2/3Ma 2 2) 2 ω 2 ) (28) such that: as before! Problem 5. ω = Loaded dice 3g 2a ( 2 1).8 g/a (29) A loaded die is one that looks like a uniform cube, but in fact has a density that is weighted in a certain direction, such that: ( ρ(z) = ρ 1 + z ) (3) where ẑ is perpendicular to the weighted face and the cube centre is at z =. The centre of mass is now not at the geometric centre of the dice (hence why it is now not fair). The cube has sides of length and a mass of M. (a) Find the new centre of mass of the dice with respect to the cub centre. (b) Find the inertia tensor of the loaded dice with respect to the centre of mass. (c) Can you think of any ways to test for a loaded die in practice? HINT: For b) there are two approaches; either calculate I from the centre of mass with the correct integration limits, or from the cube s centre and then use the parallel axis theorem. The second approach is easier I think! a) The equation for the CoM is: R = 1 M r dm = 1 M (xî + yĵ + zˆk)ρ dv (31) as new CoM will be symmetric still in x, y so we can ignore those (an odd integral over a symmetric range, [, +b] = ). As for ˆk: R z = 1 b b b ( dx dy ρ 1 + z ) zdz (32) M b 5

= ρ [ ] z 2 b M ()() 2 + z3 = 4 ρb 4 6b 3 M = b 6 such that our new CoM is shifted by R = (, b/6) wrt the geometric centre. b) Two things to worry about here. One is the non-uniform density, and the other the shifted CoM. Let s first do the inertia tensor about the CoM and then move it to R = (, b/6). So firstly the cube is still completely symmetric about xy, only varying density in z. As such, integrations over symmetric domains over odd functions will vanish. i.e., all off-diagonals (which contain either an odd function of x or y) will vanish: I xy = dv ρ(z)( xy) = (34) (33) I xz = I yz = dv ρ(z)( xz) = (35) dv ρ(z)( yz) = (36) Then we are left with a diagonal matrix. Evaluating the entries: I xx = b dv ρ(z)(y 2 + z 2 ) = b = ρ () dydz b = ρ () dz [ = ρ () 3 z3 + 3 3 z ( dxdydz ρ 1 + z ) (y 2 + z 2 ) (37) ( y 2 + z 2 + y2 z ) + z3 ) (z 3 + z 2 + b2 3 z + 3 3 ] b (38) (39) = 16 3 ρ b 5 = 2 3 Mb2 (4) where I simply discarded odd integral parts in the last step for speed! implies this is exactly the same for I yy = 2 3 Mb2. The last term is zz: I zz = b dv ρ(z)(y 2 + x 2 ) = b b = ()ρ dxdy (x 2 + y 2 ) = ()ρ dx Symmetry in xy ( dxdydz ρ 1 + z ) (x 2 + y 2 ) (41) ] b b [x 2 y + y3 = ()ρ dx 3 ) (2x 2 b + 3 3 (42) [ ] b = ()ρ 3 x3 + 3 3 x = 16 3 ρ b 5 = 2 3 Mb2 (43) So, about the geometric centre, there is no change compared to the uniform cube from lectures! This is due to our linear change in density, which is not changing symmetry in xy, so extra contributions to inertia tensor components vanish about centre. HOWEVER, the 6

loaded dice is still cheating, as it will move about the centre of mass when in free fall, not the geometric centre. We can now simply shift our inertia tensor to our new CoM, where our displacement vector is d = (,, b/6), and: I ij = I CoM,ij + M(d 2 δ ij d i d j ) (44) as d x = d y =, the we will drop the diagonal elements: I CoM,xy = I CoM,xz = I CoM,yz =. So the diagonals are thus: I CoM,zz = I zz M(b 2 /36 b 2 /36) = I zz = 2 3 Mb2 (45) which should have been expected, as the ẑ rotation axis is the same after our shift in coordinates. Finally: I CoM,xx = I xx M(b 2 /36 ) = 2 3 Mb2 Mb2 36 = 23 36 Mb2 (46) and again the same is true for I CoM,yy. So, our inertia tensor for our loaded die about the CoM is: 23/36.639 I CoM = Mb 2 23/36 Mb 2.639 (47) 2/3.667 NOTE: If you decided to do this the other way around, i.e. shift first and the find inertia tensor, then you can do as above and change the limits to 7b/6 and 5b/6, BUT you then need to take care with the density function, which will need to be shifted to our new coordinate system, z = z b/6: ( ) ρ CoM (z ) = 1 + z + b/6 (48) c) There are a few ways to do this. One is simply using buoyancy, as the die should be orientated with loaded side up if placed in water. Unfortunately, most dice will sink rather than float in water, but in principle it would work! You could also try balancing on a table at the midpoint, changing orientation until you see one way round causes it to fall because the CoM is over the edge. It would be pretty hard to exactly tell when the shift is only b/6 though. (e.g. a standard 12mm dice, b=6mm, so you are looking for a 1mm shift!) What I am looking for is that you should see changes in the rotational motion. A fair die should rotate as a spherical top, with equal inertia elements about the CoM, whereas our loaded die is a symmetric top, which will have ω precess/wobble if not perfectly rotating about the principal axes (when torque free). So if you simply toss the die in the air a few times, you should see the motion is a bit weird compared to fair dice. 7

Problem 6. The wobbly space station A space station is freely floating in space. It is axially symmetric such that e 3 is a principal axis, and λ 1 = λ 2. Rockets mounted on the space station are fired to make it spin, and exert a constant torque Γ on the station about the axis of symmetry ( e 3 ). Solve Euler s equations to find how ω (angular frequency relative to the body axis) evolves due to the applied torque. Take the initial conditions to be ω(t = ) = (ω 1,, ω 3 ), where ω 1 and ω 3 are some constants. Describe the motion of the station. HINT: Solve for ω 3 first, and then use that to solve the other two, and recall some tricks we have used for coupled equations of motion! Euler s equations with non-zero torque are: with our system, these simplify to: λ 1 ω 1 (λ 2 λ 3 )ω 2 ω 3 = Γ 1 (49) λ 2 ω 2 (λ 3 λ 1 )ω 3 ω 1 = Γ 2 (5) λ 3 ω 3 (λ 1 λ 2 )ω 1 ω 2 = Γ 3 (51) λ 1 ω 1 + (λ 3 λ 1 )ω 2 ω 3 = (52) λ 1 ω 2 (λ 3 λ 1 )ω 3 ω 1 = (53) λ 3 ω 3 = Γ (54) we will define some inertia ratio to make things cleaner: β = (λ 3 λ 1 )/λ 1 so that: ω 1 + βω 2 ω 3 = (55) ω 2 βω 3 ω 1 = (56) λ 3 ω 3 = Γ (57) the easiest to solve is the 3 component, such that: ω 3 = Γ λ 3 dt = Γ λ 3 t + ω 3 (58) putting this into the other Euler equations: ( ) Γ ω 1 + β t + ω 3 ω 2 = (59) λ 3 ω 2 β ( ) Γ t + ω 3 ω 1 = (6) λ 3 8

We can define a new constant here again: α = βγ/λ 3, and an additional time constant as t = ω 3 λ 3 /Γ such that: ω 1 + α(t + t )ω 2 = (61) ω 2 α(t + t )ω 1 = (62) The easiest way to solve this thing is to find a solution of the form (we ve use this several times already!): η = ω 1 + iω 2 (63) This means we can re-arrange everything to: η = iα(t + t )η (64) which is easy to solve! so our solution is: ( ) t 2 ln η = iα 2 + t t + C (65) η(t) = ω 1 + iω 2 = exp(iα(t 2 /2 + t t + C)) = A exp(iα(t 2 /2 + t t) (66) boundary conditions state that at t =, η = ω 1 +, therefore A = ω 1, so: ω 1 + iω 2 = ω 1 exp(iα(t 2 /2 + t t)) = ω 1 cos(α(t 2 /2 + t t)) + i sin(α(t 2 /2 + t t)) (67) so reading off Re and Im parts: ω 1 (t) = ω 1 cos(α(t 2 /2 + t t)) (68) ω 2 (t) = ω 1 sin(α(t 2 /2 + t t)) (69) ω 3 (t) = Γ λ 3 (t + t ) (7) Checking BCs: ω 1 () = ω 1, ω 2 () = and ω 3 () = ω 3 as required [can check also satisfies original equations]. The angular motion precesses in the 1,2 directions (in the body frame, these components of ω will precess about e 3 ). The frequency about e 3 increases with time (i.e. period decreases) which is expected as the torque is never deactivated and there are no dissipative forces in free space! Recall in class we looked at torque-free motion with an initial kick in e 1, the same as here, which resulted in precession but a constant frequency in e 3. Note all this precession is just because ω 1 was not zero d properly when the system was set in motion! The figure below shows this motion for ω(t = ) = (.3,, 1), λ = (2., 2., 1.) and Γ = 1. 9

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