Newtonian(2)

Newtonian Mechanics

Problem:

Three right, homogeneous cylinders have equal lengths, diameters, and weights. Two of these are placed side-by-side in contact with a horizontal surface, and the third is placed upon the other two with its axis parallel to the axes of the other two. The diagram shows the vertical plane containing one set of the three cylinder ends. The coefficient of friction between all contacting surfaces is 0.15.

(a) Is the stack stable?
(b) If the stack collapses, it does so by what mechanism?

Solution:

	Concepts, principles, relations that apply to the problem: Static friction: f_max = mN, N = normal force Static equilibrium: F = 0, t = 0, for a stable stack. Newton’s 2^nd law: F = dp/dt, t = dL/dt for an unstable stack.
	Why do they apply? The diagram shows the forces acting on a bottom cylinder. The frictional force acts on the rim, tangential to the rim, and therefore can result in a torque. In static equilibrium the total force and the total torque about the CM must be zero.
	How do they apply? F = F_xi + F_yj. F_x = F_scos(60^o) - f₂sin(60^o) - f₁. F_y = F_W - W - F_ssin(60^o) - f₂cos(60^o). In equilibrium F_W = (3/2)W. F_y = (1/2)W - F_ssin(60^o) - f₂cos(60^o) = 0. F_ssin(60^o) + f₂cos(60^o) = (1/2)W. t = k(f₂ - f₁)a, where a is the radius of the cylinder.
	Details of the calculation: (a) The stack is stable if F_x = 0 and t = 0. t = 0 --> f₁ = f₂ = f. F_x = 0 --> F_scos(60^o) - fsin(60^o) - f = 0, We need f = F_scos(60^o)/(1 + sin(60^o)) = 0.2679 F_s > mF_s. The stack therefore is not stable. (b) When f₂ takes on its maximum value, 0.15F_s, we have F_y = (1/2)W - F_ssin(60^o) - 0.15F_scos(60^o) = 0, F_s = 0.5313W = 0.3542F_W. F_x = F_scos(60^o) - 0.15F_ssin(60^o) - f₁ = 0.3542F_Wcos(60^o) - 0.15*0.3542F_Wsin(60^o) - f₁. F_x is zero when f₁ = 0.1311 F_W. This is much less that the maximum value for f₁, f_1max = mF_W = 0.15F_W. So when f₂ takes on its maximum value F_x is still 0. But t ¹ 0 since f₂ = 0.15Fs = 0.053F_W and f₁ = 0.1311F_W. Therefore the middle cylinder will drop and the other cylinders will roll away.

Problem:
Suppose a particle of unit mass (m = 1) to move in a double-well potential V(x) = (x²- x₀²)², as illustrated in the figure.

Consider the motion of this particle in imaginary time. (Make a change of variable t ® -it and consider t to be the new time variable.)

(a) Show that this transformation is equivalent to inversion of the potential for the classical trajectories. (Potential wells become potential barriers, and vice-versa).
(b) Show that x(t) = ±x₀tanh(Ö2x₀t) is a solution for classical trajectories of this system and interpret physically the corresponding motion.

Solution:

	Concepts, principles, relations that apply to the problem: F = dp/dt = mdv/dt = -ÑV. Here d²x/dt² = -dV(x)/dx. Change of variable.
	Why do they apply? We are asked to find the equation of motion for the system and then transform it by making a change of variable.
	How do they apply? (a) The equation of motion is d²x/dt² = -4x(x²- x₀²). Let t = it, then d²x/dt ² = 4x(x²- x₀²) = -d(-V)/dx = -dV'/dx. d²x/dt ² = 4x(x²- x₀²) is the equation of motion after a change of variable. We would have obtained an equation of the same form if we would have inverted the potential.
	Details of the calculation: (b) Let x(t) = ±x₀tanh(Ö2x₀t). Then dx/dt = ±x₀Ö2x₀sech²(Ö2x₀t). d²x/dt ² = ±Ö2x₀²2sech(Ö2x₀t)(-sech(Ö2x₀t)tanh(Ö2x₀t)Ö2x₀= ±4x₀³tanh(Ö2x₀t)(tanh²(Ö2x₀t) - 1) = 4(x³ - xx₀²) = 4x(x² - x₀²). x(t) = ±x₀tanh(Ö2x₀t) is a specific solution to the equation of motion of the system. Interpretation of this solution. As t = 0, x(t) = 0, dx/dt = ±Ö2x₀². The particle is moving through the stable equilibrium point of the inverted potential. For very large t tanh(Ö2x₀t) --> 1 and x --> ±x₀. So x cannot become greater than x₀. The particle starts at very large negative t on one of the hills of the inverted potential, moves through the equilibrium point at t = 0, and ends up on the other hill at very large positive t.

Problem:

Two identical springs with spring constant k = 1N/m and equilibrium length l = 0.25m are connected by a middle string of length L = (3/8)m and support a weight w = 0.5N. Two strings of length 1m are loosely connected as shown. Find the position of the weight below the support and show that it moves up when the middle string is cut.

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Solution:

	Concepts, principles, relations that apply to the problem: Hooke's law: F = -kx, k = spring constant
	Why do they apply? Consider two isolated springs constants k. Connect them in series. The effective spring constant is k' = k/2. (The same force produces twice the displacement.) Connect them in parallel The effective spring constant is k'' = 2k. (The same force produces half the displacement.)
	How do they apply? Let the length of the unstretched springs be l(m). The distance from the support to the weight w in case (a) is d(m) = 2l + 3/8m + w/k' = 2l + 3/8m + 1m = (1/2 +3/8 +1)m = (15/8)m The distance from the support to the weight w in case (b) is d'(m) = l + 1m + w/k'' = (1/4 + 1 + 1/4)m = (12/8)m d - d' = (3/8)m, d > d'.
	Details of the calculation: None

Problem:

A uniform ladder of length 2a and mass M is made of aluminum, so that its electrical resistance is R between its ends. The ladder is leaning against a wall, with an initial angle of 60^o relative to the horizontal. The wall and floor are frictionless and have negligible electrical resistance. A constant magnetic field B is directed into the plane of the paper.

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(a) Find the differential equation of motion of the ladder.

(b) Show explicitly that the action of the magnetic field on the ladder is always such as to slow down the fall of the ladder. Consider separately the two cases in which the ladder makes an angle of more than 45^o with the horizontal and less than 45^o with the horizontal respectively.

Solution:

	Concepts, principles, relations that apply to the problem: Newton’s laws, the Lorentz force, the flux rule
	Why do they apply? As long as the ladder remains in contact with the wall, the wall and the ladder form a circuit. If the magnetic flux through this circuit changes, an emf will be induced and a current will flow. The magnetic field then exerts a force on the current carrying ladder.
	How do they apply? (a) As long as the ladder remains in contact with the wall, we have the constraints x = acosa and y = asina for the coordinates of the CM. If no magnetic field is present, then the total force on the ladder is F_x= F₁= Md²x/dt² and F_y= F₂- Mg = Md²y/dt². We have F₁= Md²x/dt² (i) and F₂= Md²y/dt²+ Mg. (ii) F₁ and F₂ are forces of constraint which exert a torque t = -tk about the CM of the ladder. t = aF₁sina - aF₂cosa = Id²a/dt². (iii) (Note the definitions are made such that if t is positive d²a/dt² is positive.) Here I = Ma²/3 is the moment of inertia of the ladder about an axis through its CM parallel to the z-axis. . Using the equations of constraints we find dx/dt and dy/dt in terms of da/dt. dx/dt = -asina(da/dt), d²x/dt²= -acosa(da/dt)² - asina(d²a/dt²). dy/dt = acosa(da/dt), d²y/dt²= -asina(da/dt)²+ acosa(d²a/dt²). Inserting equations (i) and (ii) into (iii) we now can obtain the equation of motion for a. (Ma²/3)d²a/dt²= aM(-acosa(da/dt)²- asina(d²a/dt²))sina -a(g-asina(da/dt)²+ acosa(d²a/dt²))cosa. (1/3)d²a/dt²= -d²a/dt²-( g/a)cosa. d²a/dt²= -(3g/4a)cosa. This is the equation of motion as long as the board remains in contact with the wall. (b) If a magnetic field is present, and a current is flowing through the circuit, there will be an additional force acting on the ladder. Assume that B = -Bk and a current is flowing clockwise through the circuit. F_mag = Il ´ B. F_mag= I2aB(sinai + cosaj). The magnetic force on the ladder produces no torque. When the ladder is moving, we can find the induced emf and therefore the induced current I using the flux rule. I = e/R, e = -dF/dt. F = BA, A = 2a²cosasina = a²sin2a is the area of the current loop. The direction of the current is given by Lenz’s rule. I = 2Ba²cos2a(da/dt)/R. For decreasing a I flows counterclockwise if a > 45^o and clockwise if a < 45^o. We now have F_x= F₁+ I2aBsina = Md²x/dt², F₁= Md²x/dt²- I2aBsina (i) F_y= F₂- Mg + I2aBcosa = Md²y/dt², F₂= Md²y/dt²+ Mg - I2aBcosa . (ii) I = 2Ba²cos2a(da/dt)/R. Inserting the new equations (i) and (ii) into equation (iii) yields the equations of motion.
	Details of the calculation: The new terms in (i) and (ii) change the torque by Dt = -aI2aBsinasina + aI2aBcosacosa = -2IBa²(sin²a - cos²a) = 2IBa²cos2a. Dt = 4B²a⁴[(da/dt)/R]cos²(2a) < 0. The equation of motion for a becomes d²a/dt²= -(3g/4a)cosa. - (12/M)B²a²[(da/dt)/R]cos²(2a). The magnitude of d²a/dt² is smaller when a magnetic field is present, the field slows the fall of the ladder.