Physics H7A, Fall 2011 Homework 4 Solutions

Transcription

1 Physics H7A, Fall 20 Homework 4 Solutions. (K&K Problem 2.) A mass m is connected to a vertical revolving axle by two strings of length l, each making an angle of 45 with the axle, as shown. Both the axle and mass are revolving with angular velocity ω. Gravity is directed downward. (a) Draw a clear force diagram for m. (b) Find the tension in the upper string, T up, and the lower string, T low. Ans. clue. If lω 2 = 2g, T up = 2mg. (a) (Insert figure here.) (b) From the figure, we have Fx = T up / 2 T low / 2 = ma x = mω 2 l/ 2, Fy = T up / 2 T low / 2 mg = ma y = 0. Solving these equations gives T up = m(ω 2 l/2 + g/ 2), T low = m(ω 2 l/2 g/ 2). 2. (K&K Problem 2.3) Masses M and M 2 are connected to a system of strings and pulleys as shown. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of M. Ans. clue. If M = M 2, ẍ = g/5.

2 After drawing free body diagrams for M, M 2, and the lower pulley, we obtain the equations T M g = M a, T 2 M 2 g = M 2 a 2, T 2T 2 = 0. This provides three equations for four unknowns; the remaining equation comes from relating the accelerations of the two masses: a 2 = 2a pulley = 2a. Solving the four equations gives a = 2M 2 M g. M + 4M 2 3. *(K&K Problem 2.34) A mass m whirls around on a string which passes through a ring, as shown. Neglect gravity. Initially the mass is distance r 0 from the center and is revolving at angular velocity ω 0. The string is pulled with constant velocity V starting at t = 0 so that the radial distance to the mass decreases. Draw a force diagram and obtain a differential equation for ω. This equation is quite simple and can be solved either by inspection or by formal integration. Find (a) ω(t). Ans. clue. For V t = r 0 /2, ω = 4ω 0 (b) The force needed to pull the string. Hint: It might be convenient to use the expression for the acceleration vector a in polar coordinates, as on page 36 of K&K. In this context, ω(t) = θ(t). 2

3 The string is pulled downward with constant velocity V, so the radius of the mass must likewise decrease steadily at the rate ṙ = V. Since r(0) = r 0, we have r(t) = r 0 V t. Since the only force acting on the mass is from the tension in the string (remember we re neglecting gravity), we have F = Tˆr (the ˆr is our way of saying that the force acts radially inward). Recalling the expression for the acceleration vector in polar coordinates, a = ( r r θ 2 )ˆr + (r θ + 2ṙ θ)ˆθ, we see that F = ma implies the two equations T = m( r r θ 2 ), r θ + 2ṙ θ = 0. The first equations tells us that the applied force must be T = rω 2, while the second gives us a differential equation for ω: (r 0 V t) dω dt = 2V ω. This is a separable differential equation, which we can solve directly in just a few steps: ω(t) ω 0 ln ω(t) ω 0 dω ω = 2V dt r 0 V t dω ω = ω(t) = t 0 2V r 0 V t dt = 2 ln r 0 V t r 0 ω 0 r 2 0 (r 0 V t) 2 The force pulling on the string must therefore be F (t) = mω(t) 2 r(t) = mω2 0 r 0 (r 0 V t) 3. As an aside, note that the equation r ω + 2ṙω = 0 is equivalent to d dt (ωr2 ) = 0. 3

4 Thus, it is satisfied if and only if ωr 2 is constant, i.e. ω(t) r(t) 2. This is a special case of the law of conservation of angular momentum: in the absence of tangential forces acting on an point mass (torques), the angular momentum L = mωr 2 is conserved. 4. (K&K Problem 4.5) Mass m whirls on a frictionless table, held to circular motion by a string which passes through a hole in the table. The string is slowly pulled through the hole so that the radius of the circle changes from l to l 2. Show that the work done in pulling the string equals the increase in kinetic energy of the mass. Note: The previous problem should tell you that ω r 2. This also follows directly from angular momentum conservation, which we will discuss later in the course. Since ω r 2, we can say without loss of generality that ω = ω 0 r 2 0 r 2 for some values of ω 0 and r 0. Since the speed of a rotating object is v = ωr = ω 0 r 2 0 r, the change in kinetic energy of the mass is K 2 K = 2 m(v2 2 v 2 ) = 2 mω2 0r 4 0(l 2 2 l 2 ). From the previous problem, the force needed to pull the string at constant velocity is F = mω 2 r = mω 2 0 r4 0 r 3. The work done is therefore W 2 = 2 F dr = l l 2 mω 2 0r 4 0r 3 dr = 2 mω2 0r 4 0(l 2 2 l 2 ) = K 2 K. Notice that I swapped the integration limits in going from the second to the third expression above. Understanding the reason for this will prevent a lot of minus sign errors, and save you a lot of headaches. From a physicist s perspective, the line integral W = 2 F dr represents the following: we take our path from to 2, break it up into a connected series of path elements dr, compute the dot product F dr for each element, and sum the results. In this picture, it s clear that if F dr is positive for each path element, then W is obviously positive as well, since we just added up a bunch of positive numbers. However, for ordinary -D integrals, mathematicians like to define things in such a way that b a f(x)dx = a b f(x)dx, so that if a > b the integral of a positive function f(x) will come out negative. So to correct for this effect, when I express my line integral as an ordinary -D integral, I manually swap the integration limits so that my answer comes out positive. (An alternative and more robust approach is to use the formal procedure for evaluating line integrals, as explained in K&K pp This will always get you the right answer, but may take more lines of work.) 5. (K&K Problem 4.6) A small block slides from rest from the top of a frictionless sphere of radius R. How far below the top x does it lose contact with the sphere? The sphere does not move. Ans. R/3 4

5 Let θ represent the angular position of the block with respect to the vertical when it separates from the globe. From the figure, x = R( cos θ). Since the block starts from rest (or very close to it), the kinetic energy it picks up is equal to the change in potential energy: 2 mv2 = mgx = mgr( cos θ). With a free body diagram, we see that the forces acting on the block in the radial direction are the normal force N (outward), and the a component mg cos θ of the gravitational force (inward). So long as the block is sliding along the surface of the sphere, the radial component of the acceleration must be v 2 /R inward (it will obviously have tangential acceleration, since it s speeding up, but a r = v / R still holds). Thus N mg cos θ = mv 2 /R = 2mg( cos θ). At the point where the block separates from the sphere, the normal force vanishes (N = 0), which gives Hence x = R/3. mg cos θ = 2mg( cos θ) = cos θ =