Stephen Martin PHYS 10 Homework #1 Question 1: A particle starts at rest and moves along a cycloid whose equation is [ ( ) a y x = ± a cos 1 + ] ay y a There is a gravitational field of strength g in the negative y direction. Obtain and solve the equations of motion. Show that no matter where on the cycloid the particle starts out at time t = 0, it will reach the bottom at the same time. Solution: We start by defining an angle θ: Defining x and y in terms of θ, ( ) a y θ = cos 1 a y =a(1 cos θ) x = ± a (θ + sin θ) The potential and kinetic energies are given by U =mgy =mga(1 cos θ) T = 1 m(ẋ + ẏ ) = 1 ] [a m θ (1 + cos θ) + a θ sin θ =ma θ (1 + cos θ) Continuing with the Lagrangian: L =T U =ma θ (1 + cos θ) mga(1 cos θ) L θ = ma θ sin θ mga sin θ d L dt θ =ma θ(1 + cos θ) ma θ sin θ 1
Thus, ma θ(1 + cos θ) ma θ sin θ + mga sin θ = 0 Noticing that θ = θ d θ dθ : ma d θ θ [ ] dθ (1 + cos θ) = ma θ mga sin θ θd θ θ g/a = sin θdθ 1 + cos θ ( ln θ a) g = ln (1 + cos θ) + C θ C = 1 + cos θ + g a Because the particle starts at rest, we can set the boundary conditions ẋ t=0 = 0 and ẏ = 0. Solving for C: C = g(1 + cos θ 0) a And our equation becomes θ = g ( 1 1 + cos θ ) 0 a 1 + cos θ We can now integrate to solve for θ: dθ = 1 1+cos θ 0 1+cos θ [ ] 1 + cos θ 1/ dθ = cos θ cos θ 0 g g [ sin ] 1/ θ dθ = (cos θ cos θ 0 )(1 cos θ) g sin θdθ (cos θ cos θ0 )(1 cos θ) = g
3 du g = u + (1 + cos θ 0 )u cos θ 0 [ ] sin 1 u + 1 + cos θ 0 = (1 + cos θ0 ) 4 cos θ 0 [ ] cos θ + 1 + cos sin 1 θ0 = 1 cos θ 0 Using our boundary conditions once more: 1 cos θ 0 = sin D 1 cos θ 0 g g Rearranging, we finally have D = π cos θ + 1 + cos θ 0 = (1 cos θ 0 ) sin cos θ = 1 cos θ 0 Changing back from θ to y ( g a t + π ) ( ) g cos a t + 1 + cos θ 0 a y = y ( ) 0 g cos a t + a y 0 y = y ( ) 0 g cos a t At the bottom of the arc, y = 0, so a t = π g which is independent of y 0. + y 0
4 Question : A point particle of mass m is constrained to move frictionlessly on the inside of a circular wire hoop of radius r, uniform density and mass M. The hoop is constrained to the xyplane, it can roll on a fixed line (the x-axis), but it does not slide, nor can it lose contact with the x-axis. The point particle is acted on by gravity exerting a force along the negative y-axis. At t = 0 suppose the hoop is at rest. At this time the particle is at the top of the hoop, and is a given velocity v 0 along the x-axis. What is the velocity v f, with respect to the fixed axis, when the particle comes to the bottom of the hoop? Simplify your answer in the limits m/m 0 and M/m 0. Solution: Using generalized coordinates x (center of hoop horizontal position) and θ (angle between particle and top of hoop), the coordinates of the particle are given by We then have x m =x + r sin θ y m =r(1 + cos θ) U =mgy m + Mgr =mgr(1 + cos θ) + Mgr The Lagrangian is T = 1 m(ẋ m + ẏm) + 1 M(ẋ M + ẏm) + 1 Iω M = 1 [(ẋ m + r cos θ θ) + (r sin θ θ) ] + 1 Mẋ + 1 (ẋ (Mr ) r = 1 [ m ẋ + ẋr cos θ θ ] + r θ + Mẋ L = 1 m [ẋ + ẋr cos θ θ + r θ ] + Mẋ mgr(1 + cos θ) Mgr The equation of motion for x is determined by But L does not depend on x, so L/ ẋ = C d L dt ẋ L x = 0 (m + M)ẋ + mr cos θ θ = C )
5 Solve for C using boundary conditions ẋ(0) = 0 and ẋ m (0) = r cos θ θ = v 0 : C = mv 0 (m + M)ẋ + mr cos θ θ mv 0 = 0 (1) We now take a look at the equation for θ: d dt L L θ θ =0 mẍr cos θ mẋr sin θ θ + mr θ = mẋr sin θ θ + mgr sin θ ẍ cos θ + r θ g sin θ =0 ẍ = g tan θ r sec θ θ () However, equation (1) gave us (m + M)ẋ mv 0 = mr cos θ θ Multiplying equations (1) and (): (m + M)ẋẍ mv 0 ẍ = mrg sin θ θ + mr θ θ Integrating with respect to t: 1 (m + M) d dt (ẋ ) mv 0 ẍ =mrg d dt (cos θ) + 1 mr d dt ( θ ) 1 (m + M)ẋ mv 0 ẋ = mrg cos θ + 1 mr θ + D We use our initial conditions to solve for D: Reducing our equation of motion to D = mrg mv 0 m + M m ẋ v 0 ẋ = rg cos θ + r θ rg v 0 (3)
6 Now, we simply look at equations (1) and (3) for the case of θ = π: m + M m ẋ v 0 ẋ + v0 =r θ 4rg m + M m ẋ v 0 =r θ This set of equations is solved by ẋ = m m + M [ v 0 + v 0 + rg ( m + M M ) ] θ = 1 r v 0 + rg ( m + M M (Note: there are solutions to this quadratic set of equations, but the solution was chosen so that θ > 0, which is the only physical solution that makes sense) From here, we can find ẋ m θ=π : ) ẋ m =ẋ r θ = mv 0 m + M M ( ) m + M v0 m + M + rg M In the limiting case m/m 0, ẋ m = v0 + 4rg which is what we would expect from a similar system with a fixed hoop. In the limiting case of M/m 0, ẋ m = v 0 which is simply a requirement of the conservation of linear momentum.
Question 3: A double plane pendulum consists of a simple pendulum (mass m 1, length l 1 ) with another simple pendulum (mass m, length l ) suspended from m 1, both constrained to move in the same vertical plane. (a) Describe the configuration manifold Q of this dynamical system. Say what you can about TQ. 7 (b) (c) Write down the Lagrangian of the system in suitable coordinates. Derive Lagrange s equations. Solution: (a) The configuration manifold Q lies on the surface of a torus defined by x =(l 1 + l cos θ ) cos θ 1 y =(l 1 + l cos θ ) sin θ 1 z =l sin θ That is, the distance from the center of the ring to the center of the pipe is l 1 and the radius of the pipe is l. The tangent bundle TQ is 4-dimensional, so it is difficult to visualize, but it simply maps the tangent plane at each point on torus Q. (b) The natural choice for the generalized coordinates for this system are θ 1 and θ, the angles made by each pendulum with the y-axis. x 1 =l 1 sin θ 1 y 1 = l 1 cos θ 1 x =l 1 sin θ 1 + l sin θ y = l 1 cos θ 1 l cos θ ẋ 1 =l 1 cos θ 1 θ1 ẏ 1 =l 1 sin θ 1 θ1 ẋ =l 1 cos θ 1 θ1 + l cos θ θ ẏ =l 1 sin θ 1 θ1 + l sin θ θ
8 The Lagrangian is then: L =T U = 1 m 1l 1 θ 1 + 1 m = 1 m 1l 1 θ 1 + 1 m [ ( l 1 cos θ 1 θ1 + l cos θ θ ) + (l 1 sin θ 1 θ1 + l sin θ θ ) ] [ l1 θ 1 + l θ ] + l 1 l θ1 θ cos(θ 1 θ ) + m 1 gl 1 cos θ 1 + m g (l 1 cos θ 1 + l cos θ ) + (m 1 + m )gl 1 cos θ 1 + m gl cos θ (c) We start with the equation for θ 1 : d L dt θ L =0 1 θ 1 (m 1 + m )l 1 θ 1 + l 1 l θ cos(θ 1 θ ) l 1 l θ ( θ 1 θ ) sin(θ 1 θ ) + (m 1 + m )gl 1 sin θ 1 = 0 The equation for θ is given by d L dt θ L =0 θ m l θ + l 1 l θ1 cos(θ 1 θ ) l 1 l θ1 ( θ 1 θ ) sin(θ 1 θ ) + m gl sin θ = 0
Question 4: A cartesian coordinate system with axes x,y,z is rotating relative to an inertial frame with constant angular velocity ω about the z-axis. A particle of mass m moves under a force whose potential is V (x, y, z). Set up the Lagrange equations of motion in the coordinate system x, y, z. Show that these equations are the same as those for a particle in a fixed coordinate system acted on by the force V and a force derivable from a velocity-dependent potential U, and find U. Solution: In the coordinates x, y, z, the Lagrangian is given by So the equations of motion are The coordinate transformations are: L = 1 m[ẋ + ẏ + ż ] V (x, y, z) mẍ + V x =0 mÿ + V y =0 m z + V z =0 x =x cos ωt y sin ωt y =x sin ωt + y cos ωt z =z 9 ẋ =(ẋ ωy) cos ωt (ẏ + ωx) sin ωt ẏ =(ẋ ωy) sin ωt + (ẏ + ωx) cos ωt The Lagrangian for these coordinates is L = 1 m[(ẋ ωy) + (ẏ + ωx) + ż ] V (x, y, z) = 1 m[ẋ + ẏ + ż ] + 1 m[ω x + ω y ωyẋ + ωxẏ] V (x, y, z)
10 The equations of motion are This shows us that mẍ mωẏ mω x mωẏ + V x =0 mÿ + mωẋ mω y + mωẋ + V y =0 m z + V z =0 mẍ mω x mωẏ + V x =0 mÿ mω y + mωẋ + V y =0 m z + V z =0 F =(mω x + mωẏ V x )ˆx + (mω y mωẋ V V )ŷ y z ẑ =F V We can now consider a potential energy U where F = U U = (mω x + mωẏ)ˆx (mω y mωẋ)ŷ U = 1 mω x 1 mω y mωxẏ + mωyẋ U = 1 mω (x + y ) + mω(yẋ xẏ)