inkliing

Others have pointed out that A and B's rest frames are not inertial.

This isn't homework. I'm reviewing physics after many years of neglect. As with most of my posts, I made this problem up. Let object A have mass [math]m_A[/math] and object B have mass [math]m_B[/math]. One of A's surfaces is flat, as is one of B's. These flat surfaces are in contact and slide relative to each other in a straight line for a distance, x, experiencing kinetic friction. The kinetic friction, of magnitude f, is constant and is the only force acting on A or B in the direction of motion, and the only other forces acting on A and B are the normal forces pressing the surfaces together. The normal forces are assumed to be equal and opposite, so that the only acceleration of the objects is in the direction of friction. No net force acts on the center of mass. During a finite time interval, the speed of each surface relative to the other decreases from [math]v_i[/math] to [math]v_f[/math] due to friction. The nonconservative work done by each surface on the other is [math]W_{noncons} = -fx[/math] since each surface experiences the same displacement and the magnitude of the frictional force exerted by each surface on the other is the same since the two frictional forces comprise a force/reaction force pair. Therefore the change in kinetic energy of A, [math]\Delta K_A[/math] as seen from B's restframe is equal to the change in kinetic energy of B, [math]\Delta K_B[/math], as seen from A's restframe, since we are assuming that no other forces accelerate the objects. Therefore [math]\frac{1}{2}m_A v_{f}^{2} - \frac{1}{2}m_A v_{i}^{2} = \Delta K_A = \Delta K_B = \frac{1}{2}m_B v_{f}^{2} - \frac{1}{2}m_B v_{i}^{2} \Rightarrow m_A = m_B[/math], an absurd result. Something's wrong and it's driving me crazy! Please help me find the error in this reasoning, thanks.

This isn't homework...I'm reviewing physics after many years of neglect. Given 2 masses, [math]m_1, m_2[/math], connected by a rigid, massless rod, stationary with respect to a ramp which makes an angle of [math]\theta[/math] with the horizontal, with coefficients of static friction between the masses and the ramp = [math]\mu_{s1}, \mu_{s2}[/math] respectively, what is the magnitude of the tension or compression in the rod, and what are the magnitudes of the static friction, [math]f_{s1}, f_{s2}[/math], acting on each mass? This assumes [math]\theta[/math] is small enough that the masses do not lose traction, i.e., [math]\theta \leq \arctan \frac{\mu_{s1} m_1 + \mu_{s2} m_2}{m_1 + m_2}[/math] Note that if the masses are assumed to already be moving, then the problem is straightforward: [math]T = (\mu_{k1} - \mu_{k2})\frac{m_1 m_2}{m_1 + m_2}g\cos\theta[/math] [math]f_{k1} = \mu_{k1} m_1 g\cos\theta[/math] [math]f_{k2} = \mu_{k2} m_2 g\cos\theta[/math] Where T is the tension in the rod (T<0 implies compression). Such problems are common in basic physics, e.g., Halliday, Resnick, & Krane, 4th Ed., chap.6, problem 29. So I assumed it would be straightforward to do the same problem, but with the masses stationary. But I get 2 equations (sum of the forces for each mass) and 3 unknowns (T, [math]f_{s1}, f_{s2}[/math]). I find it surprising that such a simple situation is undetermined, and assume I've missed something simple. Note: I've looked through several physics texts and can find only problems in which the masses are already moving. Also, it's driving me crazy!

This isn't homework. I'm reviewing calculus and basic physics after many years of neglect. I want to show that a damped harmonic oscillator in one dimension is nonconservative. Given F = -kx - [math]\small\mu[/math]v, if F were conservative then there would exist P(x) such that [math]\small -\frac{dP}{dx} = F[/math]. I want to show that no such function, P(x), exists. The easy way would be to find a closed curve around which the integral of Fdx would be zero, but since Fdx is a 1-dimensional 1-form, this doesn't seem to be a meaningful way to do it. So I think brute force has to prevail. It should be true that: [math]\small W=\int_{x_1}^{x_2}Fdx = \int_{x_1}^{x_2}(-kx-\mu v)dx = \frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{x_1}^{x_2}\frac{dx}{dt}dx = \frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{t_1}^{t_2}\left(\frac{dx}{dt}\right)^2 dt[/math] So let [math]\small\omega_{\circ}=\sqrt{k/m}\mbox{ , }\zeta=\frac{\mu}{2\sqrt{mk}}\mbox{ , }\omega_1=\left\{\begin{matrix}\omega_{\circ}\sqrt{\zeta^2-1},&\zeta>1\\\omega_{\circ}\sqrt{1-\zeta^2},&\zeta<1\end{matrix}\right.[/math] For underdamped [math]\small\zeta<1\Rightarrow x=e^{-\zeta\omega_{\circ}t}(C_1 cos\omega_1 t + C_2 sin\omega_1 t)[/math] [math]\small\Rightarrow W=\frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{t_1}^{t_2}e^{-2\zeta\omega_{\circ}t}[(-\zeta\omega_{\circ}C_1+\omega_1 C_2) cos\omega_1 t + (-\omega_1 C_1-\zeta\omega_{\circ} C_2) sin\omega_1 t]^2 dt[/math] Therefore x(t) is not 1-1 [math]\small\Rightarrow \int_{x_1}^{x_2}vdx[/math] is multivalued implies W is not a function implies p(x) doesn't exist (since W=-[math]\small\Delta[/math]P) implies F is not conservative. Similarly for [math]\small\zeta=1[/math]. But in the overdamped case, [math]\small\zeta[/math]>1, x(t) is a non-oscillating decaying exponential which never crosses equilibrium, implying x(t) is 1-1, implying W is a function, implying F is conservative. But how can this be? How can a frictional damping force, which dissipates energy as heat, ever be conservative?

outermeasure at http://www.sosmath.com/CBB/viewtopic.php?f=3&t=58677 explained it simply: Seems obvious now. Rotate to be able to integrate both of the iterated integrals with antiderivatives of elementary functions rather than power series. I should have seen that. :)

While reviewing calculus, I noticed that James Stewart's Calculus 3rd ed., chap 14, problems plus (at the end of the chapter), p.965, prob#2 asks the student to show that [math]\zeta(2)=\sum_{i=1}^\infty\frac 1{i^2}=\int_0^1 dx\int_0^1 dy\frac1{1-xy}=I_\circ[/math] by rotating the integration region by [math]-\frac {\pi}4[/math]. Mathworld also evaluates the same integral with the same rotation, http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html . Mathworld refers to Simmons (1992), which is Simmons, G. F. "Euler's Formula [math]\sum_{k=1}^\infty\frac 1{k^2}=\frac {\pi^2}6[/math] by Double Integration." Ch. B. 24 in Calculus Gems: Brief Lives and Memorable Mathematics. New York: McGraw-Hill, 1992, which is in Google Books, but as usual, the needed page is missing. 1. I assume both Stewart and mathworld rotate the integration region in an attempt to remove the pole at (x,y)=(1,1). Rotating by [math]\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix} \cos\frac {\pi}4&\sin\frac {\pi}4\\-\sin\frac {\pi}4&\cos\frac {\pi}4\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}[/math] gives [math]I_\circ=\int_0^{\frac 1{\sqrt2}} du\int_{-u}^u dv\frac 1{1-\frac 12(u^2-v^2)}+\int_{\frac 1{\sqrt2}}^{\sqrt2} du\int_{-\sqrt2+u}^{\sqrt2-u} dv\frac 1{1-\frac 12(u^2-v^2)}[/math] which still blows up at (u,v)=([math]\sqrt2[/math],0). 2. It seems to me that [math]I_\circ=\int_0^1 dx\int_0^1 dy\frac1{1-xy}[/math] [math]I_\circ=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\frac1{1-xy}[/math] [math]I_\circ=\lim_{b\to 1^{-}}\int_0^b dx\lim_{a\to 1^{-}}\int_0^a dy\sum_{i=0}^\infty x^iy^i=\zeta(2)[/math] So why rotate if the rotated integrand still blows up? Please be specific. Please stay as close as possible to the question. Thanks in advance.

Some of the following is more or less Halliday & Resnick, 4th Ed., section 9.8 (not the whole section and I added a lot): Assume motion in a stright line. Therefore, for position, velocity, acceleratiom, force, momentum, etc., d/dt|vector|=|d/dt(vector)|. This is not true in general for motion along a curved path. Assume we have an idealized rocket of mass m and speed |v| at time t traveling in the forward direction relative to an inertial reference frame. The rocket then instantaneously begins to emit a constant stream of exhaust in the backwards direction. let: F_net_ext = net external force = net force on (rocket + exhaust) = sum of gravity, atmospheric drag, etc. F_net_r = net force on the rocket P = momentum of (rocket + exhaust) p = momentum of rocket u = velocity of exhaust (as it leaves the rocket) pointing backwards. v_rel = velocity of exhaust (as it leaves the rocket) relative to rocket = u - v = vector pointing back and is constant for an idealized rocket. Note F_net_ext, F_net_r, P, p, v, u, v_rel, dv, a, etc. are all vetors. Their lengths are |vector|. In time interval dt the rocket emits a mass -dm. Note that dm < 0 and |dm/dt| = constant for an idealized rocket = mass flow rate of exhaust backwards from the rocket. So at time t+dt we now have a bit of exhaust of mass -dm moving backwards with velocity u and the rocket, now of mass m+dm, moving forward at velocity v+dv. Note dv clearly points forward: the rocket's acceleration, a, points forward. (1) F_net__ext = dP/dt = [1/dt)][(m+dm)(v+dv)+(-dm)u - mv] = mdv/dt + vdm/dt - udm/dt (note (dv)dm/dt ->0 in the limit of the derivative) This equation is standard for rockets. There are generally 2 ways it is rewritten: (2) F_net_ext = mdv/dt + vdm/dt - udm/dt = dp/dt - udm/dt (3) F_net_ext = mdv/dt + vdm/dt - udm/dt = mdv/dt - (u-v)dm/dt = ma - (v_rel)dm/dt F_net_ext is not the force that propels the rocket. The force that propels the rocket is an internal force within the (rocket+exhaust) system. If F_net_ext = 0 then P remains constant but the rocket experiences a thrust which changes its momentum. The change in p in time dt is equal and opposite to the momentum, udm, carried away by the exhaust. for a rocket (u-v)dm/dt = thrust = (v_rel)dm/dt = forward pointing vector since u-v points back and dm/dt < 0. The thrust is the rate at which momentum enters the rocket. For an idealized rocket with constant v_rel and constant dm/dt, the thrust is a constant. This is pretty much the extent of Halliday & Resnick on this subject, but every time I read it I'm struck by the seeming inconsistency that thrust is clearly constant yet the rate of change of the rocket's momentum, dp/dt, is not. I hope you wont mind if I walk you thru my reasoning. By (3), F_net_ext=0 implies ma=thrust. thrust is constant and points forward implies ma is therefore a forward pointing constant vector. ma is constant and m is decreasing with time implies |a| is increasing with time. Therefore as long as thrust exits the rocket backwards, then the rocket will not only accelerate in the forward direction, but the rate at which it accelerates increases with time. dm/dt < 0 and v always points forward implies the term vdm/dt is a backwards pointing vector which is not constant but grows in length as v grows. At time t_0 when the exhaust initially begins to flow back from the rocket, u is a backwards pointing vector of length |v_rel| since v=0. u then shrinks in length as the rocket moves faster, but v_rel remains constant. At some point |v| = |v_rel| and u=0, and after this |v| > |v_rel| and u points forward, tho still shorter than v by |v_rel|. By (2), F_net_ext=0 implies dp/dt = udmdt. dm/dt<0 implies the term udm/dt points forward until |v|>|v_rel|, then it points back. Therefore dp/dt points forward and then, after |v|>|v_rel|, dp/dt points back. This is one of the major counterintuitive things about rockets that confuses me (and presumably most people, unless my calculations are incorrect). When |v|>|v_rel|, dp/dt of the rocket points back, meaning the net force on the rocket (not F_net_ext) points backwards, opposite the forward motion of the rocket. The momentum of the rocket always points forward in this example, but it's getting shorter at the rate of dp/dt, and tho the net force on the rocket points back after |v| exceeds |v_rel|, nevertheless the rocket not only continues to accelerate forward, but the rate of forward acceleration increases with time! Many books and websites claim that, in the absence of external forces such as gravity, friction, etc., the net force on a rocket is equal to the thrust. I've even seen this on some NASA websites. But the net force on a rocket does not seem to be equal to the thrust. Assuming F_net_ext = 0 then F_net_r = dp/dt = udm/dt by (2) = udm/dt + vdm/dt - vdm/dt = (u-v)dm/dt + vdm/dt = thrust + vdm/dt. So the net force on a rocket, absent external forces, does not seem to be equal to the thrust, but rather equal to the thrust + vdm/dt. Hopefully, at this point, you're saying to yourself "what's the problem?" If so, then there probably is no problem. But I'm bothered by the equations: (4) F_net_ext=0 implies F_net_r = thrust + vdm/dt = u dm/dt Both equations seem counterintuitive even tho I've reasoned thru the details. It bothers me that many books and websites set F_net_r = thrust + external forces and ignore the vdm/dt term. It makes me wonder whether eqn (4) is correct. I haven't found either of the equations (4) in Halliday & Resnick or Marion & Thornotn's Classical Dynamics or Goldstein's Classical Mechanics, but I haven't found anything in those texts that disputes eqns (4) either. I would like to know if eqns (4) are correct? Is there something simple here that I'm missing? Thanks in advance.

While reviewing basic calculus, I noticed that the curve (1+t^2,t^2,t^3), which clearly has a cusp at (1,0,0), has a derivative curve (2t,2t,3t^2) which is clearly smooth. This struck me as odd since differentiation usually seems to turn cusps into discontinuities, whereas integration smoothes out a curve, especially a curve described by polynomials. In fact, in general I have always taken a curve to be smooth iff it has a continuous derivative, which this curve has, and yet a cusp cannot be smooth in any sensible sense. I suspect the explanation is relatively simple - just something I'm missing. Thx in advance.

All of your examples are very clear. I understand now why both statements are necessary in the test. Thank you.

There's something that's confusing me about what appears to be the standard form of stating the alternating series test in basic calculus. The four sources I looked up were James Stewart's CALCULUS, Howard Anton's CALCULUS, wolfram alpha's mathworld, and wikipedia. All four had essentially the same statement for the alternating series test: If the sum from i=0 to infinity of [(-1)^n][b_n], with b_n>0 for all n, satisfies a) b_(n+1)<=b_n for all n ({b_n} is a decreasing sequence), and B) the limit as n goes to infinity of b_n = 0, then the series is convergent. What I'm confused about is this: since all four sources made it clear that all of the b_n were strictly greater than zero and that b_n->0 as n goes to infinity, what is the point of also adding part a)? Why add that {b_n} must be monotonically decreasing (in the less-than-or-equal-to sense)? It seems to me that b_n>0 for all n and b_n->0 as n goes to infinity implies that {b_n} must be monotonically decreasing (in the less-than-or-equal-to sense). So it seems to me that all four sources should have left out the part about b_(n+1)<=b_n for all n. Am I missing something simple here? It seems to me that the needed counterexample is that of a sequence of real numbers, all greater than zero, which go to zero, but which are not eventually decreasing, which I'm pretty sure is impossible. I would appreciate it if some1 could clearly show that B), coupled with b_n>0 for all n, doesn't imply a) so I can be confident that it is necessary to state a) in the statement of the test. Thanks in advance.

the equation is: r = (g/(2(omega)^2))(sinh((omega)t)-sin((omega)t)) and shouldn't the acceleration in the r direction be: d^2r/dt^2=gsin((omega)t)-r(omega)^2 ? but you're right: the equation for r from the text satisfies the equation you give for the acceleration in the radial direction. This seems odd to me, since it seems that the centripetal acceleration should point radially inward. clearly, I have forgotton much concerning the signs of radial accelerations.

This is neither homework nor a question from an exam. I am an aging engineer and I am reviewing all of the calculus that I have forgotton. On p.410 of 'Calculus with analytic geometry,' 5th ed. by Howard Anton, Chap. 7 technology exercises, problem #15, paraphrased slightly to be (hopefully) a little more clear: Suppose that an initially horizontal hollow tube rotates in a vertical plane with a constant angular velocity of omega radians/sec about a horizontal axis, perpendicular to the length of the tube, through one end of the tube. The direction of omega is such that the tube initially rotates down from its horizontal starting point. Assume that an object within the tube, initially at rest at the pivot-point-end of the tube, is free to slide without friction through the tube while the tube rotates, and that the object's diameter is equal to the inside diameter of the tube so that the object can't rattle around at all as it slides down the tube. Let r be the distance of the object from the pivot point for all time t >= 0 as the object slides through the length of the tube (r=0 when t=0). It can be shown that r = g/(2(omega)^2)(sinh((omega)t)-sin((omega)t)) during the period that the object is in the tube. But it can't be shown by me! I tried. Forget the actual problem - it's easy. I just wanna be able to derive the formula, or I won't be able to sleep. Let the origin be at the pivot point, the positive y direction up, the positive x direction toward the other end of the tube. Let theta be the angle between the tube and the horizontal and assume that the sliding object has mass m. A free body diagram of the object in the tube involves 2 forces: gravity, mg, pointing straight down, and what I'm calling a normal force, N, acting perpendicular to either the upper or the lower inside surface of the tube, depending on t and possibly omega. N can also be labeled as a force in the theta direction, F_theta, since it clearly has no radial component. When t is small but greater than zero, then the object has moved from its initial position but is not falling very fast, and so the upper inside surface of the tube will be moving downward at a faster speed than the falling object and will then push the object with a normal force perpendicular to that surface, forcing the object to slide parallel to the surface and down the tube. When t exeeds some value, then the object will be falling faster than the speed at which the tube is moving downward. Now the falling object is held up by the lower inside surface of the tube. This lower inside surface exerts a force on the object in a direction perpendicular to the surface, and the object is forced to slide without friction parallel to that surface and down the tube. It may also be true, though it isn't initially clear, that at low omega the upper inside surface never exerts a force on the object, and at high omega the lower surface never exerts a force. Let us consider the sum of the forces and the sum of the torques on the sliding object at a time after which the normal force has switched from the upper to the lower inside surface of the tube, since I suspect that this will lead to the formula in the text: sum of the forces in the y direction = Ncos(theta)-mg = ma_y = mdv_y/dt sum of the forces in the x direction = Nsin(theta) = ma_x = mdv_x/dt sum of the torques = mgrcos(theta)-Nr = I_object(alpha_object) = 0 since omega_tube = constant implies angular acceleration_tube =0 implies angular acceleration_object = 0 since the object is constrained to remain in the tube. So we have: dv_y/dt = g((cos((omega)t))^2-1) dv_x/dt = gcos((omega)t)sin((omega)t) Integrating twice with v_y(0) = 0, v_x(0)=0, x(0)=0, y(0)=0 gives: y= (g/(4(omega)^2))(sin((omega)t))^2-(g/4)t^2 x= (g/(4(omega)))t-(g/(8(omega)^2))cos((omega)t)sin((omega)t) Now, r=(x^2+y^2)^1/2 should give the desired equation. Good luck with that. If there are simplifying assumptions that will produce the desired equation, they weren't mentioned in the problem statement. Help me Obi-wan Kanobi! You're my only hope.

This is not homework. I am reviewing all of the calculus that I have forgotten. Howard Anton's Calculus w/Analytical Geometry, 5th ed., section 4.1, prob #32: A police helicopter is flying due north at 100km/hr and at a constant altitude of 1/2mi. Below, a car is traveling west on a highway at 75km/hr. At the moment that the helicopter passes over the highway the car is 2mi east of the helicopter. a)How fast is the distance between the car and the helicopter changing at the moment the helicopter crosses the highway? The Complete Solutions Manual to Anton's 5th ed. Calculus, Prepared by Albert Herr, provides the following solution for prob 32, sec4.1. I quote it verbatim, except I describe the diagram: Let x, y, and z be the distances shown in the figure. [the diagram labels the point on the highway where the helicopter crosses as O, the point 1/2mi above O as P, the position of the helicopter as H, the position of the car as C, x=CO, y=HP, z=HC]. Find dz/dt evaluated at x=2, y=0 given that dx/dt=-75 and dy/dt=-100. Because triangle OPC is a right triangle, it follows that PC has length (x^2+(1/2)^2)^1/2; but triangle HPC is also a right triangle so z^2=((x^2+(1/2)^2)^1/2)^2+y^2=x^2+y^2+1/4 and 2zdz/dt=2xdx/dt+2ydy/dt+0, dz/dt=(1/z)(xdx/dt+ydy/dt). Now, when x=2 and y=0, z^2=2^2+0^2+1/4=17/4, z=(17^1/2)/2 so dz/dt [at x=2, y=0] = (2/(17^1/2))(2(-75)+0(-100))=-300/(17^1/2). This published solution doesn't seem right to me. The helicopter's 100km/hr northerly rate should contribute to the rate of change of the distance between the helicopter and the car: Let the point O in the above solution be the origin and let the positive x-axis extend from O to C and beyond. Let the z-axis be vertical. Therefore, C=(2,0,0) and the car is moving in the negative x direction at 75km/hr and H=(0,0,1/2) and the helicopter is moving in the positive y direction at 100km/hr. The distance between the two points H and C seems to me to be irrelevant. The rate at which the distance between H and C is changing with time, d/dt(C-H)=dC/dt-dH/dt. let C=(x_c,y_c,z_c) and H=(x_h,y_h,z_h). Then dC/dt=(dx_c/dt,dy_c/dt,dz_c/dt)=(-75,0,0) and dH/dt=(dx_h/dt,dy_h/dt,dz_h/dt)=(0,100,0). dC/dt-dH/dt=(-75,-100,0)= the rate of change of the position of the car relative to the helicopter, or in the frame of reference in which the helicopter is stationary. If we had started with H-C rather than C-H then we would have (75,100,0) as the rate of change of the position of the helicopter with respect to the car. Either way, this velocity, (-75,-100,0) in the helicopter's frame, or (75,100,0) the the car's reference frame, =the rate of change with time of the distance between the two objects, and has a magnitude=(75^2+100^2)^1/2=125km/hr and is decreasing in both reference frames. Yet, since published solutions manuals are usually checked by several people, I'm hesitant as to whether or not my solution is correct. Please let me know whether you think my solution, the published solution, or some other solution is correct. No opinions, please. Supply precise reasoning. Thanks in advance to all responses.