Physics 1116 Fall 2011 Homework 1
Due: Wednesday, August 31st
Reading: Kleppner & Kolenkow Chapter 1.1-1.8 and Note 1.1 1. Vector Algebra. Compute:
ˆ and B = 4 ˆi − 6 ˆj + 3 k. ˆ Consider two vectors A = − ˆi + 2 ˆj + 3 k
a. A + B b. A − 2B c. A · B d. A×B e. B×A f. A · (A×B) and B · (A×B) Solution: ˆ a. 3 ˆi + −4 ˆj + 6 k ˆ b. −9 ˆi + 14 ˆj + −3 k c. -7 ˆ d. 24 ˆi + 15 ˆj + 8 k ˆ e. −24 ˆi − 15 ˆj − 8 k f. Both are zero. The vector A×B is perpendicular to both A and B. 2. Distance of closest approach. Consider an arbitrary point in space given by a vector ˆ . We can define a straight line r0 , and an arbitrary direction given by a unit vector n in space, passing through the given point in the given direction, by the locus of points ˆ for all values of a real number η. Use vector techniques to answer the r = r0 + η n following questions. a. What is the distance of closest approach of this line to the origin? b. If a new line is drawn from the origin to the point of closest approach, at what angle do these two lines intersect one another? c. List two possible conditions that, if either is true, the distance of closest approach will be zero. Comment. In experimental particle physics it is often important to know the distance of closest approach of a particle track to the point where the primary beams collided. In many cases this distance helps to measure some other particle’s lifetime.
Solution: ˆ. a. Evaluate the length, r, of r, and minimize with respect to η: r2 = r20 +η 2 +2ηr0 · n ˆ Minimum occurs for η = −r0 · n, so the point of closest approach is given by ˆ )ˆ rmin = p r0 − (r0 · n n, and the distance (to the origin) of closest approach is 2 ˆ )2 . rmin = r0 − (r0 · n ˆ = 0, so this implies that they intersect at right angles. b. You can show rmin · n ˆ = ˆr0 . c. rmin = 0 if either r0 = 0 or n 3. Kinematics in one dimension with contant acceleration. KK Problem 1.13 Solution: The elevator travels upward with (unknown) velocity v, starting from y = 0 at t = 0. So the elevator position is given by y(t) = vt. We are asked to find the height of the elevator at t = T1 , namely h ≡ y(T1 ) = vT1 . We will have to find v. Once dropped, the marble accelerates downward with acceleration g. Note that, being dropped from the moving elevator, its initial velocity is +v and its starting time is T1 ; thus vm (t) = v − g(t − T1 ). The position of the marble is given by Z ym (t) = h +
vm (t0 )dt0 .
The marble hits the ground when ym (t) = 0. Do the integral, solve for v: v=g
T2 2 . 2(T1 + T2 )
Plug back into h ≡ y(T1 ) = vT1 to find y1 = gT1 T2 2 /(2(T1 + T2 )). For T1 = T2 = 4s this yields h = 39.2m as promised in the hint. 4. Kinematics in two dimensions with contant acceleration. KK Problem 1.21 (You might want to work out the problem first for the traditional case of throwing the rock while standing on level ground.) Solution: The definition of “range” is left somewhat ambiguous. Do they mean the horizontal distance, or do they mean the distance along the sloping ground? Show that if either of those is maximized in your calculation, then the other one automatically is also. Also
note that they don’t specify the initial speed v of the rock. While the range will depend on that, the launch angle that gives you the maximum range does not. The parabolic trajectory of the rock is derived in KK, Example 1.10. In the context of this problem it may be written g x2 y(x) = x tan θ − 2v 2 cos2 θ Equate this y(x) to the line describing the sloping hill, y(x) = −x tan φ and solve for x: 2v 2 x= (tan θ + tan φ) cos2 θ. g Now you just have to find the value of θ that maximizes this expression. First simplify, using trig identities to convert to x=
v2 (sin 2θ + tan φ cos 2θ + tan φ) g
and then solve dx/dθ = 0 to get the result: the range is maximal when θ is chosen such that tan 2θ = 1/ tan φ = cot φ = tan(π/2 − φ). From this we find 2θ = π/2 − φ, or equivalently, θ + φ = 21 (φ + π/2). This says that the angle θ + φ, which is the angle of the launch as measured from the slope, bisects the angle between the slope and true vertical. Of course if φ = 0 we recover the standard case that Galileo found about 400 years ago, θ = 45◦ . 5. Kinematics of Uniform Circular Motion. The earth has a diameter of approximately d = 1.3×107 m; a day is about 86,400 seconds; and the latitude of Ithaca is about 42◦ N. a. What is the angular velocity ω of a tree (a) at the equator, (b) in Ithaca? b. What is the magnitude of the velocity (relative to a non-rotating earth) of a tree (a) at the equator, (b) in Ithaca? c. What is the direction of the velocity vector of a tree (a) at the equator, (b) in Ithaca? d. What is the magnitude of the acceleration (relative to a non-rotating earth) of a tree (a) at the equator, (b) in Ithaca? e. What is the direction of the acceleration vector of a tree (a) at the equator, (b) in Ithaca? Solution: a. ω = 2π/T and T = 24 hrs = 86, 400 sec, so ω = 7.3×10−5 radians/second.
b. In uniform circular motion v = rω, but note that at the equator and in Ithaca the distance to the axis of rotation is different. (The distance to the center of the earth is approximately the same, but that is a different matter.) By elementary trigonometry (see figure below), the distance to the equator is r = (d/2) cos θ for a point at a latitude of θ (latitude is measured from the equator); at the equator cos θ = 1 while in Ithaca cos θ = 0.74. So the two velocities are 472 m/s and 351 m/s. c. The velocity vector points east in both cases. d. In uniform circular motion the magnitude of the acceleration is a = rω 2 . Therefore a = 0.034 m/s2 at the equator and a = 0.025 m/s2 in Ithaca. e. In uniform circular motion the acceleration vector always points inward toward the axis of rotation (ˆ a = −ˆr). At the equator, this means straight down. In Ithaca, at latitude 42◦ , the direction is a vector pointing to the axis of rotation, more or less downward, but tilted to the north by 42◦ . See figure.
r = (d/2) cosθ
ca ha t I
r = d/2
Figure 1: Geometry for rotating earth
6. More Kinematics of Uniform Circular Motion. A wheel of radius R has a dot painted on the rim at the position r = R, φ = 0. The wheel begins to roll to the left Page 4
(without slipping) at constant angular velocity ω, so the dot traces out a cycloid curve. We will choose ω > 0 so the wheel moves leftward as t increases. See Figure 2, below, for the geometry. a. Find r(t). Hint: decompose this into the sum of two vectors, one locating the center of the wheel with respect to the ground, and one locating the dot with respect to the center of the wheel. b. Find v(t). c. Find a(t). d. At the instant that the dot is at the top of the wheel, what is the radius of curvature of its path? (The radius of curvature is the radius of the circle that matches up locally with the path at a given point.) Hint: you know v(t) and a(t). y
Figure 2: Configuration for Problem 6. The wheel initially was in the position where the dot was at the φ = 0 axis. This snapshot shows the wheel shortly after rolling began, ie after φ has rotated about 40◦ . The small black arrow indicates the direction of rotation. The xy coordinate system is indicated by the red vectors.
Solution: a. Viewed from the xy coordinate frame, the center of the wheel has the coordinates rc = (rx , ry ) = (−Rφ, R). Notice that the height is constant (y = R) while the horizontal position is given by the amount of rolling that has occured, −Rφ. Viewed from the center of the wheel, the dot location is simply given by ud = (ux , uy ) = (R cos φ, R sin φ). So the position of the dot in the xy frame is just r = rc + ud = (−Rφ + cos φ, R + R sin φ). b. To get velocity, differentiate r with respect to time, using φ = ωt. This gives v = (−Rω − Rω sin ωt, Rω cos ωt).
c. To get acceleration, differentiate v with respect to time: A = (−Rω 2 cos ωt, − Rω 2 sin ωt). d. If we think of the dot as being momentarily in uniform circular motion when it passes through the top point, then the velocity and acceleration are related by the a = v 2 /ρ where ρ is the appropriate radius of curvature for the dot’s trajectory at that point. Turning this around, we can extract the radius of curvature from ρ = v 2 /a if we evaluate v and a at the point ωt = φ = π/2 (the top point). Using sin π/2 = 1 and cos π/2=0, we find v = (−2Rω, 0). A = (0, − Rω 2 ). Thus v = |v| = 2Rω and a = |A| = Rω 2 . Thus the radius of curvature of the overall path of the dot, evaluated at the topmost point, is ρ = v 2 /a = 4R. If you were expecting the result to be R, this may surprise you. Remember (a) the point of rotation is the bottom of the wheel, and (b) the wheel is moving horizontally. Both of these facts cause the actual radius of curvature of the path to be greater than you might otherwise expect.