Due: Friday, 3/14 Chapter 13: 5, 9, 41, 43, 59 Chapter 8: 3, 6, 7

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Name Period Date Homework 13A Due: Friday, 3/14 Chapter 13: 5, 9, 41, 43, 59 Chapter 8: 3, 6, 7 [1] A thin, uniform rod of mass M 1 and length, is initially at rest on a frictionless horizontal

1 Name Period Date Homework 13A Due: Friday, 3/14 Chapter 13: 5, 9, 41, 43, 59 Chapter 8: 3, 6, 7 [1] A thin, uniform rod of mass M 1 and length, is initially at rest on a frictionless horizontal surface. The moment of inertia of the rod about its center of mass is 1 12 M 2. As shown in Figure I, the rod is struck at point P by a mass m 2 whose initial velocity v is perpendicular to the rod. After the collision, mass m 2 has velocity -! v as shown in Figure 2. Answer the following in terms of the symbols given. a. Using the principle of conservation of linear momentum, determine the velocity v of the center of mass of this rod after the collision. b. Using the principle of conservation of angular momentum, determine the angular velocity ω of the rod about its center of mass after the collision. c. Determine the change in kinetic energy of the system resulting from the collision. Homework 13A 1 3/9/14

2 [2] A particle moves along a parabolic path with equation y = 1 2 x2 shown. a. Suppose the particle moves so that the x-component of its velocity has the constant value v x = C; that is, x = Ct i. On the graph, indicate the directions of the particle's velocity vector v and acceleration vector a at point R, and label each vector. ii. Determine the y-component of the particle's velocity as a function of x. y path followed by particle R 0 X Chain Rule: iii. Determine the y-component of the particle's acceleration. da dc = da db db dc b. Suppose, instead, that the particle moves along the same parabola with a velocity whose x-component is given by v x = C 1+x 2 i. Show that the particle's speed is constant in this case. y S 0 X ii. On the graph, indicate the directions of the particle's velocity vector v and acceleration vector a at point S, and label each vector. State the reasons for your choices. Homework 13A 2 3/9/14

3 Chapter We require the magnitude of force (given by Eq. 13-1) exerted by particle C on A be equal to that exerted by B on A. Thus, Gm A m C r 2 = Gm A m B d 2. We substitute in m B = 3m A and m B = 3m A, and (after canceling m A ) solve for r. We find r = 5d. Thus, particle C is placed on the x axis, to left of particle A (so it is at a negative value of x), at x = 5.00d. 9. The gravitational force from Earth on you (with mass m) is where F g GM m = = mg R E 2 E g = GM R = If r is the distance between you and a tiny black hole of mass 2 2 E / E 9.8 m/s. 11 b 1 10 kg same gravitational pull on you as the Earth, then F g GM m = = r b mg. 2 M = that has the Combining the two equations, we obtain mg = GM E m R E 2 = GM b m r 2 r = GM b g = ( m 3 /kg s 2 )( kg) 9.8 m/s m. 41. Let m = kg and d = m (the original edgelength, in terms of which the final edge-length is d/3). The total initial gravitational potential energy (using Eq and some elementary trigonometry) is U i = 4Gm2 d 2Gm2 2 d. Since U is inversely proportional to r then reducing the size by 1/3 means increasing the magnitude of the potential energy by a factor of 3, so 43. The period T and orbit radius r are related by the law of periods: T 2 = (4π 2 /GM)r 3, where M is the mass of Mars. The period is 7 h 39 min, which is s. We solve for M: M = 4π 2 r 3 GT 2 M = 4π 2 ( m) 3 ( m 3 / s 2 kg) s M = kg. ( ) 2 = U f = 3U i ΔU = 2U i = 2(4 + 2 ) Gm2 d J. = 4.82 Homework 13A 3 3/9/14

4 59. Each star is attracted toward each of the other two by a force of magnitude GM 2 /L 2, along the line that joins the stars. The net force on each star has magnitude 2(GM 2 /L 2 ) cos 30 and is directed toward the center of the triangle. This is a centripetal force and keeps the stars on the same circular orbit if their speeds are appropriate. If R is the radius of the orbit, Newton s second law yields (GM 2 /L 2 ) cos 30 = Mv 2 /R. Chapter 8 3. (a) The force of gravity is constant, so the work it does is given by, where is the force and is the displacement. The force is vertically downward and has magnitude mg, where m is the mass of the flake, so this reduces to W = mgh, where h is the height from which the flake falls. This is equal to the radius r of the bowl. Thus W=mgr =2x10-3 kg 9.8 m/s 2 22x10-2 m = 4.31x10-3 J (b) The force of gravity is conservative, so the change in gravitational potential energy of the flake-earth system is the negative of the work done: ΔU = W = J. (c) The potential energy when the flake is at the top is greater than when it is at the bottom by ΔU. If U = 0 at the bottom, then U = J at the top. (d) If U = 0 at the top, then U = J at the bottom. The stars rotate about their center of mass (marked by a circled dot on the diagram above) at the intersection of the perpendicular bisectors of the triangle sides, and the radius of the orbit is the distance from a star to the center of mass of the three-star system. We take the coordinate system to be as shown in the diagram, with its origin at the left-most star. The altitude of an equilateral triangle is, so the stars are located (e) All the answers are proportional to the mass of the flake. If the mass is doubled, all answers are doubled. at x = 0, y = 0; x = L, y = 0; and x = L/2,. The x coordinate of the center of mass is x c = (L + L/2)/3 = L/2 and the y coordinate is. The distance from a star to the center of mass is. Once the substitution for R is made Newton s second law becomes. This can be simplified somewhat by recognizing that, and we divide the equation by M. Then, GM/L 2 = v 2 /L and. Homework 13A 4 3/9/14

5 7. Height of the ball (relative to the low point of the swing) h = L L cos θ (for angle θ measured from vertical as shown in Fig. 8-34). Once this relation is established, then the principal results of this problem follow from Eq (for W g ) and Eq. 8-9 (for U ). (a) The vertical component of the displacement vector is downward with magnitude h, so we obtain W g = F g d = mgh W g = mgl(1 cosθ) = (5.00 kg)(9.80 m/s 2 )(2.00 m)(1 cos30 ) = 13.1 J W g = (5.00 kg)(9.80 m/s 2 )(2.00 m)(1 cos30 ) = 13.1 J (b) From Eq. 8-1, we have ΔU = W g = mgl(1 cos θ ) = 13.1 J. (c) With y = h, Eq. 8-9 yields U = mgl(1 cos θ ) = 13.1 J. (d) As the angle increases, we intuitively see that the height h increases (and, less obviously, from the mathematics, we see that cos θ decreases so that 1 cos θ increases), so the answers to parts (a) and (c) increase, and the absolute value of the answer to part (b) also increases. 6. Let be the normal force of the ice on him and m is his mass. The net inward force is mg cos θ F N and, according to Newton's second law, this must be equal to mv 2 /R, where v is the speed of the boy. At the point where the boy leaves the ice F N = 0, so g cos θ = v 2 /R. We wish to find his speed. If the gravitational potential energy is taken to be zero when he is at the top of the ice mound, then his potential energy at the time shown is U = mgr(1 cos θ ). He starts from rest and his kinetic energy at the time shown is. Thus conservation of energy gives or v 2 = 2gR(1 cos θ ). We substitute this expression into the equation developed from the second law to obtain g cos θ = 2g(1 cos θ ). This gives cos θ = 2/3. The height of the boy above the bottom of the mound is h = R cos θ = 2R/3 = 2(13.8 m)/3 = 9.20 m., Homework 13A 5 3/9/14

6 First Worksheet Question Homework 13A 6 3/9/14

7 Homework 13A 7 3/9/14

8 Homework 13A 8 3/9/14

PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?

PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true? 1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always