Homework 7 # 4.22, 5.15, 5.21, 5.23

Homework 7: # 4.22, 5.15, 5.21, 5.23, Foucault pendulum 4.22 A projectile is fired horizontally along Earth’s surface. Show that to a first approximation the angular deviation from the direction of fire resulting from the Coriolis effect varies linearly with time at a rate ω cos θ where ω is the angular frequency of Earth’s rotation and θ is the co-latitude, the direction of deviation being to the right in the northern hemisphere. Answer: I’ll call the angular deviation ψ. We are to find ψ = ω cos θt We know ω is directed north along the axis of rotation, that is, sticking out of the north pole of the earth. We know θ is the co-latitude, that is, the angle from the poles to the point located on the surface of the Earth. The latitude, λ is the angle from the equator to the point located on the surface of the Earth. λ = π/2 − θ. Place ourselves in the coordinate system of whoever may be firing the projectile on the surface of the Earth. Call y ′ the horizontal direction pointing north (not toward the north pole or into the ground, but horizontally north), call x′ the horizontal direction pointed east, and call z ′ the vertical direction pointed toward the sky. With our coordinate system in hand, lets see where ω is. Parallel transport it to the surface and note that it is between y ′ and z ′ . If we are at the north pole, it is completely aligned with z ′ , if we are at the equator, ω is aligned with y ′ . Note that the angle between z ′ and ω is the co-latitude, θ.(θ is zero at the north pole, when ω and z ′ are aligned). If we look at the components of ω, we can take a hint from Goldstein’s Figure 4.13, that deflection of the horizontal trajectory in the northern hemisphere will depend on only the z ′ component of ω, labeled ωz′ . Only ωz is used for our approximation. It is clear that there is 1 no component of ω in the x′ direction. If we took into account the component in the y ′ direction we would have an effect causing the particle to move into the vertical direction, because the Coriolis effect is Fc = −2m(ω × v) and ωy ×v would add a contribution in the z direction because our projectile is fired only along x′ and y ′ , that is, horizontally. So following Goldstein’s figure, we shall only be concerned with ωz . Our acceleration due to the Coriolis force is ac = −2(ω × v) = 2(v × ω) The component of ω in the z ′ direction is ωz′ = ω cos θ. Thus the magnitude of the acceleration is ac = 2vωcos θ The distance affected by this acceleration can be found through the equation of motion, d= 1 2 ac t = vωcos θt2 2 And using a small angle of deviation, for ψ we can draw a triangle and note that the distance traveled by the projectile is just x = vt. xψ= d → ψ= d x 2 vωcos θt = ω cos θt vt Therefore the angular deviation varies linearly on time with a rate of ω cos θ. Note that there is no Coriolis effect at the equator when θ = π/2, therefore no angular deviation. 5.15 Find the principal moments of inertia about the center of mass of a flat rigid body in the shape of a 45o right triangle with uniform mass density. What are the principal axes? ψ= Answer: Using the moment of inertia formula for a lamina, which is a flat closed surface, (as explained on wolfram research) we can calculate the moment of inertia for the triangle, with it situated with long side on the x-axis, while the y-axis cuts through the middle. The off-diagonal elements of the inertia tensor vanish. 2 Z Z 2 Ix = σy dxdy= 2 Solving the algebra, Ix = 2M Z a 0 a−x 2M M 2 y dydx= A a 0 (a − x)3 dx 3 a2 0 3 2 2 3 Z a (−x + 3ax − 3a x + a )dx= M a2 8 1 6 [ − − ] = 4 6 4 4 2M a2 3 0 For Iy 3a Z 2 M 2 Z Z a Z a−y A x dxdy 0 Iy = σx2 dxdy= 2 This has the exact same form, so if 0 you’re clever, you won’t do the integral over again. Iy = M a2 6 For Iz Z 6 + M a2 1 )M a2 = 6 3 1 I = σ(x2 + y 2 )dxdy= Ix + Iy = ( We can zuse the parallel axis theorem to find the principal moments of inertia about the center of mass. The center of mass is ycm = 2 1 σ Z a Z 0 a−x ydxdy= 2 Z a 0 (a − x)2 dx 2 M 0 a2 x3 2 2 2 2 Z a (a − 2xa+ x )dx= a x − ax + a 1 a  a2 = 3 0  , 0). a a2 we can tell that the center of mass 3is (0, From symmetry Using the 3 parallel axis theorem, with r0 = a/3 ycm = 0 IX = Ix − M r 2 0 IY = Iy IZ = Iz − M r 2 0 These are IX = ( M a2 1 1 3 2 2 − )M a2 = ( − )M a = 18 6 9 18 18 IY = M a2 6 3 IZ = ( 1 1 2 2 − )M a2 = M a 9 9 3 5.21 A compound pendulum consists of a rigid body in the shape of a lamina suspended in the vertical plane at a point other than the center of gravity. Compute the period for small oscillations in terms of the radius of gyration about the center of gravity and the separation of the point of suspension from the center of gravity. Show that if the pendulum has the same period for two points of suspension at unequal distances from the center of gravity, then the sum of these distances is equal to the length of the equivalent simple pendulum. Answer: Looking for an equation of motion, we may equate the torque to the moment of inertia times the angular acceleration. lF= I ¨ θ The force is −M gsin θ, and the moment of inertia, using the parallel axis theorem is I = M r 2 g + M l2 where rg radius of gyration about the center of gravity, and l is the distance between the pivot point and center of gravity. The equation of motion becomes −lM gsin θ = (M r 2 g + M l2 ) ¨θ θ Using small oscillations, we can apply the small angle approximation sin θ ≈ −lgθ= (r2 g + l2 ) ¨θ lg θ+ ¨θ = 0 r2g + l2 This is with angular frequency and period + l2 2π s g → T = 2 + l2 rlg 2 lg g r ω= = 2π ω This is the same as the period for a physical pendulum s 2 g +l lg 2 r I T = 2π = 2π M gl If we have two points of suspension, l1 and l2 , each having the same period, T . Then we get s s 4 s 2π This is g + l21 r2 l1 g r2g + l21 l1 = 2π = + l22 l2 g g s r2 r2g + l22 l2 And in a more favorable form, add l1 to both sides, because we are looking for l1 + l2 to be equivalent to a simple pendulum length, r2g l1 + l1 + l1 = r2 g l2 + l2 + l1 r2g l1 l2 (l2 − l1 ) + 2l1 = l2 + l1 This is only true if r2g = l1 l2 Thus our period becomes g + l 21 1 s s l2 + l1 l1 l2 + l 2 Lg = 2π = 2π T = 2π = 2π l1 g of a simplel1 gpendulum equivalent. g where L is the length 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 speed. Obtain a formula for the time needed for the door to close if the acceleration 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 distance a from the hinges. Show that if f is 0.3m/s2 and the door is a uniform rectangle is 1.2m wide, the time will be approximately 3.04 s. s s r2 Answer: Begin by setting the torque equal to the product of the moment of inertia and angular acceleration. I ¨θ = aF The moment of inertia is I = mr 2 . The force is F = −mfsin θ. So we get 0 mr 20 θ¨ = −amfsin θ Our equation of motion is 5 θ̈ = − af sin θ r20 This is rough. In our case we can not use the small angle approximation. The door starts at 90o ! How do we go about solving this then? Lets try integrating it once and see how far we can get. Here is a handy trick, d dθ d ˙ θ d ˙ θ dθ d ˙ θ˙ θ = = = dt dt dθ dt dθ dt Plug this into our equation of motion θ̈ = af dθ̇˙ θ = − 2 sin θ r0 dθ This is separable, and may be integrated. θ˙2 af = 2 cos θ r0 2 ˙ s 2 cos θ 2afr 0 θ= The time may be found by integrating over the time of travel it takes for the door to shut. dθ s 0 √ 2 r2af cos θ dθ= = T = dθ Here is where the physics takes aθ backseat for a few, while the math runs 0 the show. If we throw in a − cos 90o we might notice that this integral is an elliptic integral of the first kind, denoted K . Z 2 π 0 dt Z 2 π 0 dθ ˙ Z 2 π dθ dθ 0 0 0√ π s s Z π Z π 2K(sin 4 ) 0 0 2 r2 r22af r2 p √ T = = = 2af 2af cos θ − cosofπ elliptic integrals, at cosfrom θ mathworld’s This can be seen treatment http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html. Now we have s 2 2 0 s √ 2 ) 2 af = T √ K( 22 ) belongs to a group of functions called ‘elliptic integral singular values’, K(kr ) A treatment of them and a table of their values that correspond to gamma functions are given here: http://mathworld.wolfram.com/EllipticIntegralSingularValue.html. r2 K( 6 The ‘elliptic lambda function’ determines the value of kr . A table of lambda functions is here http://mathworld.wolfram.com/EllipticLambdaFunction.html. √ Our kr value of 22 corresponds to k1 . From the singular value table, K(k1 ) = Our time is now Γ2 ( 1 4 ) √ 4 π 2 1 0 Γ ( 4) √ r2af 4 π T = Fortunately, there are nice calculators that will compute gamma functions quickly. I used this one http://www.efunda.com/math/gamma/findgamma.cfm. I now have s 1 Γ( ) = 3.63 4 Back to the physics. The moment of inertia of a uniform rectangle about M 2 the axis that bisects it is 3 a . Move the axis to the edge of the rectangle using the parallel axis theorem. I = M r 2 0 = M a2 + M 3 a2 = 4 M a2 3 we now have 4 2 a 3 With a = .6m, that is, half of the length of the car door, assuming its mass is uniform. And with f = .3m/s2 we have r20 = r 1 4a3f s 1 √ (3.63)2 = 3.035 ≈ 3.04 s 3(.3) 4 3.14 4(.6) T = √ (3.63)2 = 4 π Foucault Pendulum Find the period of rotation as a function of latitude. Hint: neglect centrifugal force, neglect change in height, solve for ξ = x + iy Answer: The Foucault pendulum is a swinging weight that the wire’s upper support restrains the wire and the weight is set swinging with no lateral of the pendulum gradually rotates, demonstrating 7 supported by a long wire, so only in the vertical direction or circular motion. The plane the Earth’s rotation. Solve for the period of rotation of this plane. The equation of motion for acceleration takes into account the vertical acceleration due to gravity, the acceleration from the tension and the Coriolis acceleration. ar = g + T − 2ω × vr m In my system, I have x facing east, y facing north, and z facing to the sky. This yeilds ωx = 0 ωy = ω sin θ = ω cos λ ωz = ω cos θ = ω sin λ The only velocity contributions come from the x and y components, for we can ignore the change in height. The Coriolis acceleration is quickly derived ac = ẏωsin λˆ x − ẋωsin λˆ y + ẋωcos λˆ z Looking for the period of rotation, we are concerned only with the x and y accelerations. Our overall acceleration equations become ẍ = − g g l x + 2 ẏωsin λ y¨ = − y − 2 ẋωsin λ l The g/lterms were found using approximations for the tension components, that is, Tx = −T x l → Tx /ml= g/land the same for y. Introducing ξ = x + iyand adding the two equations after multiplying the second one by i ξ̈ + g l ξ = −2ω sin λ(−y˙ + i ẋ) ξ̈ + g ξ = −2iωsin λ ˙ξ l g ξ + 2iωsin λ ξ̇ = 0 l This is the damped oscillation expression. It’s solution is, using ω sin λ, the over damped case ξ̈ + √ √g t −i g l + Be ξ = e−iω sinλt (Aei The equation for oscillation of a pendulum is q¨ + g q= 0 l It has solution 8 l t ) g l >> √g √g l t q = Aei l t + Be−i We can simplify our expression then, using q ξ = qe−iω sinλt Where the angular frequency of the plane’s rotation is ω cos θ, or ω sin λ where λ is the latitude, and θ is the co-latitude. The period can be found using, ω = 2π/T. 2π Tearth 2π TEarth → TF oucault = cos θ TF oucault This can be checked because we know the pendulum rotates completely in 1 day at the North pole where θ = 0 and has no rotation at the equator where θ = 900 . Chapel Hill has a latitude of 36o , a Foucault pendulum takes cos θ = TF oucault = 24 hours sin 36o to make a full revolution. 9 ≈ 41 hours