PHY6426/Fall 07: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT #1 due by 9:35 a.m. Wed 09/05 Instructor: D. L. Maslov Rm.

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1 PHY646/Fall 07: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT # due by 9:35 a.m. Wed 09/05 Instructor: D. L. Maslov maslov@phys.ufl.edu Rm. 4 Please help your instructor by doing your work neatly.. Goldstein, Problem.4 Note that the rod can be at an arbitrary angle to the plane.) 0 points) Solution: or, in reverse, See Figure for notations. Introducing the c.o.m and relative coordinates we obtain for the kinetic energy R = r + r r = r r, r = R + r/ r = R r/, Now, T = m ṙ ) + ) m ṙ ) = m Ṙ + 4 m ṙ). ) Ṙ = a Φ) whereas r = sin θ cos φ, sin θ sin φ, cos θ) and Finally, ) ṙ = l θ cos θ cos φ sin θ φ sin φ, θ cos θ sin φ + sin θ φ cos φ, sin θ θ ) ) ṙ) = l θ + sin θ φ. ) T = ma ) ) Φ + 4 ml θ + sin θ φ. If the rode is constrained to move in the plane θ = π/, θ = 0), this expression simplifies to ) T = ma Φ ) + 4 ml φ.. A geodesic is a line that represents the shortest path between any two points when the path is restricted to a particular surface. Show that the geodesic on the surface of a right cylinder is a segment of a helix. Hint: The equation of a helix is z = Cφ, where C is a constant. 0 points) Solution: The element of distance along the surface is ds = dx + dy + dz. )

2 Φ θ φ FIG. : In cylindrical coordinates, x = ρ cos φ, y = ρ sin φ, z = z from which dx = ρ sin φdφ ) dy = ρ cos φdφ dz = dz. Substituting ) into ) and integrating along the entire path between points and, we find s = ρ dφ + dz = φ φ dφ ρ + ż, where ż = dz/dφ. If s is to be minimum, the function f = ρ + ż must satisfy the Euler equation because f z or = 0. Thus d d f dφ ż = f z = 0, f dφ ż = d ż dφ ρ + ż = 0 ż ρ + ż = C=constant Solving for ż, we find Since ρ is a constant, it means that ż = where A is a constant. This is the equation for a helix. C C ρ. z = Aφ,

3 3 3. Goldstein, Problem. 0 points) Solution: T = m x = l sin φ y = l cos φ x = l sin φ + l sin φ y = l cos φ + l cos φ U = m gl sin φ ẋ ) + ẏ ) ] = ) m l φ T = m U = m gl sin φ ẋ ) + ẏ ) ] = ) ) ] m l φ + l φ + l l φ φ cos φ φ ) Equations of motion L = T + T U U = m + m ) l m ) φ + l φ + m l l φ φ cos φ φ ) + m + m ) gl cos φ + m gl cos φ. m + m ) l φ + m l φ cos φ φ ) φ φ φ ) ] sin φ φ ) + m l φ φ sin φ φ ) + m + m ) g sin φ = 0 ) ] m + m ) l φ + m l φ cos φ φ ) + φ sin φ φ ) + m + m ) g sin φ = 0 and m l φ + m l l φ cos φ φ ) φ φ φ ) ] sin φ φ ) ml l φ φ sin φ φ ) + m gl sin φ = 0 l φ + l φ cos φ φ ) φ ) sinφ φ )] + g sin φ = 0. Linearizing the exact equations of motion for a double plane pendulum, obtained in the previous problem, find the characteristic frequencies of small oscillations. 0 points) olution: Approximating sin φ φ and cos φ in Eqs.??,??), we obtain m + m ) l φ + m l φ + m + m ) gφ = 0 l φ + l φ + gφ = 0 Substituting φ = A e iωt φ = A e iωt, we obtain a characteristic equation A m + m ) g ω ) l m l ω A = 0 A l ω + A m g l ω ) = 0

4 4 φ φ FIG. : The condition that the determinant is equal to zero yields { } g ω = m + m ) l + l ) ± m + m ) m + m ) l + l ) 4m l l ]. m l l. a) Goldstein, Problem.4. 5 points) Variational solution: Notations: distance to the center of mass of the hoop is ρ, the angular position of the c.o.m. os θ, and the angular position of a point on the rim is φ. Lagrangian L = m ρ) + ) mρ θ ) + mr φ mgρ cos θ Constraints: ρ = r + R 3) ρθ = rφ 4) The second constraint is the no-slipping condition ρ θ = r φ, integrated over time. Indeed, a general condition of a no-slipping contact of two bodies is that the velocities at the contact point are the same. When one of the bodies is stationary cylinder), this implies that the contact point is stationary as well. The velocity of the contact point is the sum of the velocity of the center of mass and the linear velocity of the rim. At the contact point, these two velocities are anti-parallel, hence On the other hand, Hence, and Eq.4). 0 = v c.m. r φ. v c.m. = ρ θ. ρ θ = r φ, Correspondingly, there are two constraint functions f f = ρ r R = ρθ rφ Since neither of the constraints functions depends on the velocities, only the spatial derivatives f α / q j appear. Lagrange s equations d L dt ρ L ρ = λ f ρ + λ f ρ = λ + λ θ

5 5 or d L L dt θ θ = λ f θ + λ d L L dt φ φ = λ f φ = λ r f θ = λ ρ Differentiating Eqs.3) and 4) over time, we find Eq.7) then gives m ρ mρ θ) + mg cos θ = λ + λ θ 5) d ) mρ θ mgρ sin θ = λ ρ dt 6) mr φ = rλ 7) ρ = ρ = 0 8) ρ θ = r φ. Substuting λ from the last equation into??), we find that Recalling that mρ θ) + mg cos θ = λ + λ θ 9) ρ θ g sin θ = λ /m 0) mρ θ = λ ) θ = g sin θ. ρ θ = d dθ θ and integrating the resulting equation with the initial conditions θ t = 0) = 0 and θ t = 0) = 0, we obtain Substuting this into Eq.9) θ = g cos θ). ρ mg cos θ mg = λ + λ θ The physical meaning of λ + λ θ is the normal force. λ + λ θ = 0 or when The hoop looses contact with the cylinder, when cos θ = θ = 60. b) Repeat Problem.4 for an object with inertia I rolling down the cylinder. In which limit does the answer reduce to the case of a point mass sliding down the cylinder? 5 points) SolutionThe variational solution can be carried out in the same way. Just to do things differently this time, I will solve this problem using energy conservation and nd Law. Once the constraint v c.m. = ωr, where ω is the angular velocity of rolling, is resolved, the kinetic energy can be written as for an object with inertia I T = mv c.m. + Iω = mv c.m. + I v c.m. r. The kinetic energy is equal to the change in the potential energy U = mgρ cos θ)

6 6 and Second law for the center of mass of the hoop says mv c.m. + I v c.m. r = mgρ cos θ). mv c.m. ρ = mg cos θ N, where N is the normal force. Condition N = 0, when the hoop loses the contact with the cylinder, correspond to or mv c.m. ρ = mg cos θ, mv c.m. = mgρ cos θ. Substituting the last relation into the energy conservation, we find For a hoop, I = mr and mgρ cos θ + I/mr ) = mgρ cos θ) cos θ = 3 + I/mr. cos θ = /. The limit I 0 correspond to a non-rolling object. In this case we recover cos θ = /3