Classical Mechanics - Homework Assignment 6 Alejandro G´ omez omez Espinosa
∗
November 7, 2012
Goldstein, Ch.5, 13 Two thin rods each of mass m and length l are connected to an ideal (no friction) hinge and a horizontal thread. The system rests on a smooth surface as shown in the figure. At time t = 0, thread thread is cut. Negle Neglecting cting the mass of the hinge and the threa thread, d, and consider considering ing only motion in the xy plane
Figure 1: Sketch of Problem 13. (a) Find the speed at which the hinge hits the floor. Let’s define the vector r as the position of the center of mass. Then, the kinetic energy for the system is the sum of the kinetic energy of the center of mass plus the rotation over this point: T = T cm cm + T rot rot . Therefore, the vector r and its kinetic energy is given by: l l r(x, y) = sin θx + cos θy 2 2
The kinetic energy due to the rotation is T rot rot = l/2
I =
l/2
−
T cm cm
⇒
I θ˙2 2
m = 2
l ˙ θ 2
2
ml2 θ˙2 = 8
(1)
, where the moment of inertia is:
mx2 m x3 dx = l l 3
l/2 l/2
−
ml2 = 12
(2)
Hence, using (1 (1) and (2 (2), the kinetic energy for one side of our system is: ml2 θ˙ 2 ml2 θ˙ 2 ml2θ˙2 T = + = 8 24 6 ∗
[email protected]
(3)
The potential energy of one side of the system is: V = mg 2l sin θ. For the final Lagrangi Lagrangian an of our system, we must multiply it by two to take into account the other side of the problem. Thus, ml2 θ˙2 L = T − V = (4) − mgl sin θ 3 Since the Lagrangian doesn’t depend explicitly on time, the total energy is conserved. At the beginning, we have only potential energy with θ = 30 and, at the end only kinetic energy: ml2 θ˙2 mgl sin( sin(30 30)) = 3 3g = θ˙2 2l 3g = θ˙ 2l
v = lθ˙ =
3gl 2
(b) Find the time it takes for the hinge to hit the floor. Using the results from (a): dθ dt
=
dt = t =
3g 2l 2l 3g
π 6
π/ 6
dθ
0
2l 3g
Goldstein, Ch.5, 23 An automobile is started from rest with one of its doors initially at right angles. If the hinges of the door are toward the front of the car, the door will slam shut as the automobile picks up spee speed. d. Obtain Obtain a formula formula for the time need needed ed for the door door to close if the acceler acceleration ation f f is constant, the radius of gyration of the door about the axis of rotation is r0 , and the center of mass is at a distan distancce a from the hinges. hinges. Show that that if f is 0.3 m/s2 and the door is a uniform rectangle 1.2 m wide, the time will be approximately 3.04 s. Let’s define θ as the angle between the car and the open door, then the torque is equals to τ =
dL d(I θ˙ ) = = I θ¨ = mr02 θ¨ dt dt
(5)
where here, the moment of inertia is given by I = mr02. But the torque torque is also define define as the cross cross product: τ = r × F = aF sin aF sin θ = a(−mf )sin mf )sin θ (6) where the force is negative due to is against the motion of the car. Replacing (5 (5) and (6 (6), we found the equation of motion of our system: mr02 θ¨ = −maf sin maf sin θ
⇒
af θ¨ = − 2 sin θ r0
(7)
dθ˙ dt
To solve this equation, we can use that θ¨ =
=
dθ˙ dθ dθ dt
=
dθ˙ ˙ dθ θ :
θ¨ =
−
dθ˙ ˙ θ = dθ
−
af sin θ r02
af sin θ r02 af θ˙ dθ˙ = − 2 sin θdθ r0 θ˙2 af = cos θ 2 r02 θ˙ =
π/ 2
0
r02 dθ = 2af cos af cos θ
2af cos θ r02 dt
t = 2.622
r02 2af
where the last integration integration was solve using Mathematica. Mathematica. Finally Finally, we need a relation relation betw b etween een the 2 radius of gyration r0 and the distance a to the center center of mass. Let’s Let’s use the mome moment nt of inertia. inertia. From tables, we can know that the moment of inertia of a uniform rectangle about the axis that bisects it, is equal to ma . Thus, 3 2
2 0
mr = I = I cm cm + I rot rot
ma2 4 = ma + = ma2 3 3 2
⇒
4a2 r = 3 2 0
(8)
Plugging (12 (12)) in the previous equation of time:
t = 2.622 when f = 0. 0 .3 ms
2
−
4a2 = 2.622 6af
2a 3f
and a = 0.6 m :
t = 2.622
2(0. 2(0.6m) = 3.0276 0276ss 3(0. 3(0.3ms 2 ) −
Goldstein, Ch.5, 24 A wheel rolls down a flat inclined surface that makes an angle α angle α with the horizontal. The wheel is constrained so that its plane is always perpendicular to the inclined plane, but it may rotate about the axis normal to the surface. Obtain the solution for two-dimensional motion of the wheel, using Lagrange’s equations and the method of undetermined multipliers. Defining our system of coordinates as shown in Figure 2, the kinetic energy has two components in a two dimensional motion: mx˙ 2 mr2 θ˙ 2 T = T trans + (9) trans + T rot rot = 2 2 and the potential energy: V = mgy = mgx sin α (10)
Figure 2: Sketch of problem 24. A wheel rolls down a flat inclined surface. Therefore, the Lagrangian using (9 (9) and (10 (10)) is given by: mx˙ 2 mr2 θ˙2 − −mgx sin α + 2 2 The Lagrange’s equations and the method of undefined multipliers use the relation: L=
d dt
∂L ∂ ˙ ∂ q ˙
−
∂L ∂f = λq ∂q ∂q
(11)
(12)
Applying (12 (12)) in the Lagrangian (11 (11): ): ∂f (13) ∂x ∂f mr2 θ¨ = λθ (14) ∂θ where the constrain equation is rθ = x and therefore f = rθ − x = 0. Replaci Replacing ng this in (13 (13)) and (14 14), ), we found: mx ¨ − mg sin α = λx
mx ¨ − mg sin α = −λ mr2 θ¨ = rλ mr θ¨ = λ
(15) (16)
But, from the constrain we can found: rθ¨ = x ¨. Then, (16 (16)) is nothing but mx¨ = λ and the relations become: mg sin α λ = 2 mg sin α x¨ = (17) 2 mg sin α θ¨ = (18) 2r The general solution for the equation of motion (17 (17)) and (18 (18)) are: x(t) = θ(t) = where A,B,C,D are constants.
1 2 gt sin α + At + B 4 1 2 gt sin α + Ct + D 4r
Goldstein, Ch.5, 26 For the axially symmetric body precessing uniformly in the absence of torques, find analytical solutions for the Euler angles as a function of time. From Goldstein (5.47), we have the Euler’s equations for an axially symmetric body, with symmetry axis Lz : I 1ω˙ 1 = (I 1 − I 3 )ω3 ω2 I 2ω˙ 2 = (I 3 − I 1 )ω3 ω1 I 3ω˙ 3 = 0 where ω1 = A cosΩt cosΩt, ω2 = A sinΩt sinΩt and Ω = I I I ω3 . But, But, since since this angula angularr velocitie velocitiess can be written using Euler angles in the body fixed frame: 3− 1 1
ω1 = sin θ sin ψφ˙ + cos ψθ˙ ω2 = sin θ cos ψφ˙ − sin ψθ˙ ω3 = cos θφ˙ + ψ˙ we found the relations: sin θ sin ψφ˙ + cos ψθ˙ = A cos(Ωt cos(Ωt + α) sin θ cos ψ φ˙ − sin ψθ˙ = A sin(Ωt sin(Ωt + α) cos θφ˙ + ψ˙ = B
(19) (20) (21)
where A, B are constants. Multiplying (19 (19)) by cos ψ and (20 (20)) by sin ψ and subtracting subtracting them: sin θ sin ψ cos ψφ˙ + cos2 ψθ˙ − sin θ cos ψ sin ψφ˙ + sin2 ψθ˙ = A cos(Ωt cos(Ωt + α)cos ψ − A sin(Ωt sin(Ωt + α)sin ψ θ˙ = A cos(Ωt cos(Ωt + α + ψ ) The condition for uniform precession recalls θ˙ = 0, therefore therefore:: nπ ⇒ cos(Ωt cos(Ωt + α + ψ ) = 0 ⇒ Ωt + α + ψ = ψ = ψ0 − Ωt 2 where n = 0, 1, 2, .. and ψ0 = α + by cos ψ and add them:
nπ
⇒
ψ˙ = −Ω
(22)
is the initial angle of ψ. Next, multiply (19 (19)) by sin ψ and (20 (20))
2
sin θ sin2 ψφ˙ + cos ψ sin ψθ˙ + sin θ cos2 ψφ˙ − sin ψ cos ψθ˙ = A cos(Ωt cos(Ωt + α)sin ψ + A sin(Ωt sin(Ωt + α)cos ψ sin θφ˙ = A sin(Ωt sin(Ωt + α + ψ) A φ˙ = sin(Ωt sin(Ωt + α + ψ) sin θ Replacing this, and (22 (22), ), in (21 (21): ): cos θ sin(Ωt sin(Ωt + α + ψ) − Ω = A cot θ − Ω sin θ nπ A = (ω3 + Ω) Ω) tanθ tan θ ⇐⇒ Ωt + α + ψ = 2 Plugging the definition for Ω: ω3 = A
A=
I 3 − I 1 I 3 ω3 + ω3 tan θ = tan θ I 1 I 1
(23)
Finally, replacing (23 (23)) in the θ˙ : φ˙ =
I 3 I 3 tan θ sin(Ωt sin(Ωt + α + ψ) = I 1 sin θ I 1 cos θ
⇐⇒
Ωt + α + ψ =
nπ 2
(24) (25)
that has the solution: φ=
I 3 t + φ0 I 1 cos θ
