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2 CHALLENGE PROBLEMS 6. Suppose the three coordinate planes are all mirrored and a light ra given b the vector a a, a, a 3 first strikes the z-plane, as shown in the figure. Use the fact that the angle of incidence equals the angle of reflection to show that the direction of the reflected ra is given b b a, a, a 3. Deduce that, after being reflected b all three mutuall perpendicular mirrors, the resulting ra is parallel to the initial ra. (American space scientists used this principle, together with laser beams and an arra of corner mirrors on the Moon, to calculate ver precisel the distance from the Earth to the Moon.) z a b v v t r FIGURE FOR PROBLEM 7 F 7. A particle P moves with constant angular speed around a circle whose center is at the origin and whose radius is R. The particle is said to be in uniform circular motion. Assume that the motion is counterclockwise and that the particle is at the point R, 0 when t 0. The position vector at time t 0 is rt R cos t i R sin t j. (a) Find the velocit vector v and show that v r 0. Conclude that v is tangent to the circle and points in the direction of the motion. (b) Show that the speed v of the particle is the constant R. The period T of the particle is the time required for one complete revolution. Conclude that T R v (c) Find the acceleration vector a. Show that it is proportional to r and that it points toward the origin. An acceleration with this propert is called a centripetal acceleration. Show that the magnitude of the acceleration vector is a R. (d) Suppose that the particle has mass m. Show that the magnitude of the force F that is required to produce this motion, called a centripetal force, is F m v R 8. A circular curve of radius R on a highwa is banked at an angle so that a car can safel traverse the curve without skidding when there is no friction between the road and the tires. The loss of friction could occur, for eample, if the road is covered with a film of water or ice. The rated speed v R of the curve is the maimum speed that a car can attain without skidding. Suppose a car of mass m is traversing the curve at the rated speed v R. Two forces are acting on the car: the vertical force, mt, due to the weight of the car, and a force F eerted b, and normal to, the road. (See the figure.) mg F cos mt The vertical component of F balances the weight of the car, so that. The horizontal component of F produces a centripetal force on the car so that, b Newton s Second Law and part (d) of Problem, FIGURE FOR PROBLEM 8 F sin mv R R (a) Show that vr Rt tan. (b) Find the rated speed of a circular curve with radius 400 ft that is banked at an angle of. (c) Suppose the design engineers want to keep the banking at, but wish to increase the rated speed b 50%. What should the radius of the curve be? 9. A projectile is fired from the origin with angle of elevation and initial speed v 0. Assuming that air resistance is negligible and that the onl force acting on the projectile is gravit, t, we showed in Eample 5 in Section 0.9 that the position vector of the projectile is

3 CHALLENGE PROBLEMS 3 rt v 0 cos t i [v 0 sin t tt ] j. We also showed that the maimum horizontal distance of the projectile is achieved when and in this case the range is R v 0t. (a) At what angle should the projectile be fired to achieve maimum height and what is the maimum height? (b) Fi the initial speed v and consider the parabola R R 0 0, whose graph is shown in the figure. Show that the projectile can hit an target inside or on the boundar 45 _R 0 R 0 D v a FIGURE FOR PROBLEM 0 of the region bounded b the parabola and the -ais, and that it can t hit an target outside this region. (c) Suppose that the gun is elevated to an angle of inclination in order to aim at a target that is suspended at a height h directl over a point D units downrange. The target is released at the instant the gun is fired. Show that the projectile alwas hits the target, regardless of the value v 0, provided the projectile does not hit the ground before D. 0. (a) A projectile is fired from the origin down an inclined plane that makes an angle with the horizontal. The angle of elevation of the gun and the initial speed of the projectile are and v 0, respectivel. Find the position vector of the projectile and the parametric equations of the path of the projectile as functions of the time t. (Ignore air resistance.) (b) Show that the angle of elevation that will maimize the downhill range is the angle halfwa between the plane and the vertical. (c) Suppose the projectile is fired up an inclined plane whose angle of inclination is. Show that, in order to maimize the (uphill) range, the projectile should be fired in the direction halfwa between the plane and the vertical. (d) In a paper presented in 686, Edmond Halle summarized the laws of gravit and projectile motion and applied them to gunner. One problem he posed involved firing a projectile to hit a target a distance R up an inclined plane. Show that the angle at which the projectile should be fired to hit the target but use the least amount of energ is the same as the angle in part (c). (Use the fact that the energ needed to fire the projectile is proportional to the square of the initial speed, so minimizing the energ is equivalent to minimizing the initial speed.). A projectile of mass m is fired from the origin at an angle of elevation. In addition to gravit, assume that air resistance provides a force that is proportional to the velocit and that opposes the motion. Then, b Newton s Second Law, the total force acting on the projectile satisfies the equation m d R dt mt j k dr dt where R is the position vector and k 0 is the constant of proportionalit. (a) Show that Equation can be integrated to obtain the equation dr dt k R v0 tt j m where v 0 v0 dr. dt 0 (b) Multipl both sides of the equation in part (a) b e kmt and show that the left-hand side of the resulting equation is the derivative of the product e kmt Rt. Then integrate to find an epression for the position vector Rt.. Find the curvature of the curve with parametric equations t sin( ) d 0 t cos( ) d 0

4 4 CHALLENGE PROBLEMS 3. A disk of radius is rotating in the counterclockwise direction at a constant angular speed. A particle starts at the center of the disk and moves toward the edge along a fied radius so that its position at time t, t 0, is given b rt trt, where Rt cos t i sin t j (a) Show that the velocit v of the particle is v cos t i sin t j tv d where v d Rt is the velocit of a point on the edge of the disk. (b) Show that the acceleration a of the particle is a v d t a d where a d Rt is the acceleration of a point on the rim of the disk. The etra term v d is called the Coriolis acceleration; it is the result of the interaction of the rotation of the disk and the motion of the particle. One can obtain a phsical demonstration of this acceleration b walking toward the edge of a moving merr-go-round. (c) Determine the Coriolis acceleration of a particle that moves on a rotating disk according to the equation rt e t cos t i e t sin t j 4. In designing transfer curves to connect sections of straight railroad tracks, it s important to realize that the acceleration of the train should be continuous so that the reactive force eerted b the train on the track is also continuous. Because of the formulas for the components of acceleration in Section 0.9, this will be the case if the curvature varies continuousl. (a) A logical candidate for a transfer curve to join eisting tracks given b for 0 and s for s might be the function f s, 0 s, whose graph is the arc of the circle shown in the figure. It looks reasonable at first glance. Show that the function F s s if 0 if 0 s if s is continuous and has continuous slope, but does not have continuous curvature. Therefore, f is not an appropriate transfer curve. ; (b) Find a fifth-degree polnomial to serve as a transfer curve between the following straight line segments: 0 for 0 and for. Could this be done with a fourth-degree polnomial? Use a graphing calculator or computer to sketch the graph of the connected function and check to see that it looks like the one in the figure. =F() = 0 œ =0 0 transfer curve 3.5 ft FIGURE FOR PROBLEM 5 5. A ball rolls off a table with a speed of fts. The table is 3.5 ft high. (a) Determine the point at which the ball hits the floor and find its speed at the instant of impact. (b) Find the angle between the path of the ball and the vertical line drawn through the point of impact. (See the figure.) (c) Suppose the ball rebounds from the floor at the same angle with which it hits the floor, but loses 0% of its speed due to energ absorbed b the ball on impact. Where does the ball strike the floor on the second bounce? 6. A cable has radius r and length L and is wound around a spool with radius R without overlapping. What is the shortest length along the spool that is covered b the cable?

5 CHALLENGE PROBLEMS 5 ANSWERS S Solutions. (s3.5) m 3. (a) c cc z cc (b) t, z t (c) (a) v Rsin t i cos t j (c) a r 9. (a) 90, v 0 t. (b) Rt mk e ktm v 0 tmkmk e ktm tj 3. (c) e t v d e t R 5. (a) 0.94 ft to the right of the table s edge, 5 fts (b) 7.6 (c).3 ft to the right of the table s edge

6 6 CHALLENGE PROBLEMS SOLUTIONS E Eercises. Since three-dimensional situations are often difficult to visualize and work with, let us first tr to find an analogous problem in two dimensions. The analogue of a cube is a square and the analogue of a sphere is a circle. Thus a similar problem in two dimensions is the following: if five circles with the same radius r are contained in a square of side m so that the circles touch each other and four of the circles touch two sides of the square, find r. The diagonal of the square is. The diagonal is also 4r +. But is the diagonal of a smaller square of side r. Therefore = r =4r + =4r + r = 4+ r r = 4+. Let us use these ideas to solve the original three-dimensional problem. The diagonal of the cube is + + = 3. The diagonal of the cube is also 4r + where is the diagonal of a smaller cube with edge r. Therefore = r + r + r = 3 r 3=4r + =4r + 3 r = 4+ 3 r.thus 3 r = 4+ 3 = 3 3. The radius of each ball is 3 3 m. 3. (a) We find the line of intersection L as in Eample 0.5.6(b). Observe that the point (,c,c) lies on both planes. Now since L lies in both planes, it is perpendicular to both of the normal vectors n and n, and thus parallel to their i j k cross product n n = c = c, c +, c. So smmetric equations of L can be written c c as + c = c c = z c, provided that c 6= 0, ±. c + If c =0, then the two planes are given b + z =0and =, so smmetric equations of L are =, = z. Ifc =, then the two planes are given b + + z = and + + z =, and the intersect in the line =0, = z. Ifc =,thenthetwoplanesaregivenb + + z =and + z =,and the intersect in the line =0, = z. (b) If we set z = t in the smmetric equations and solve for and separatel, we get += c = (t c)(c ) c + = ct +(c ) c + (t c)( c), c +, = (c )t +c. Eliminating c from these c + equations, we have + = t +. So the curve traced out b L in the plane z = t isacirclewithcenterat (0, 0,t) and radius t +. (c) The area of a horizontal cross-section of the solid is A(z) =π(z +),so V = 0 A(z)dz = π 3 z3 + z 0 = 4π 3.

7 CHALLENGE PROBLEMS 7 5. (a) When θ = θ s, the block is not moving, so the sum of the forces on the block must be 0,thusN + F + W = 0. This relationship is illustrated geometricall in the figure. Since the vectors form a right triangle, we have tan(θ s)= F N = μ sn n = μ s. (b) We place the block at the origin and sketch the force vectors acting on the block, including the additional horizontal force H, with initial points at the origin. We then rotate this sstem so that F lies along the positive -ais and the inclined plane is parallel to the -ais. F is maimal, so F = μ s n for θ>θ s. Then the vectors, in terms of components parallel and perpendicular to the inclined plane, are N = n j F =(μ s n) i W =( mg sin θ) i +( mg cos θ) j H =(h min cos θ) i +( h min sin θ) j Equating components, we have μ s n mg sin θ + h min cos θ =0 h min cos θ + μ s n = mg sin θ () n mg cos θ h min sin θ =0 h min sin θ + mg cos θ = n () (c) Since () is solved for n, we substitute into (): h min cos θ + μ s (h min sin θ + mg cos θ)=mg sin θ h min cos θ + h min μ s sin θ = mg sin θ mgμ s cos θ sin θ μs cos θ h min = mg cos θ + μ s sin θ tan θ μs = mg +μ s tan θ tan θ tan θs From part (a) we know μ s =tanθ s, so this becomes h min = mg and using a trigonometric +tanθ s tan θ identit, this is mg tan(θ θ s ) as desired. Note for θ = θ s, h min = mg tan 0 = 0, which makes sense since the block is at rest for θ s,thusno additional force H is necessar to prevent it from moving. As θ increases, the factor tan(θ θ s ), and hence the value of h min, increases slowl for small values of θ θ s but much more rapidl as θ θ s becomes significant. This seems reasonable, as the steeper the inclined plane, the less the horizontal components of the various forces

8 8 CHALLENGE PROBLEMS affect the movement of the block, so we would need a much larger magnitude of horizontal force to keep the block motionless. If we allow θ 90, corresponding to the inclined plane being placed verticall, the value of h min is quite large; this is to be epected, as it takes a great amount of horizontal force to keep an object from moving verticall. In fact, without friction (so θ s =0), we would have θ 90 h min,andit would be impossible to keep the block from slipping. (d) Since h ma is the largest value of h that keeps the block from slipping, the force of friction is keeping the block from moving up the inclined plane; thus, F is directed down the plane. Our sstem of forces is similar to that in part (b), then, ecept that we have F = (μ s n) i. (Note that F is again maimal.) Following our procedure in parts (b) and (c), we equate components: μ s n mg sin θ + h ma cos θ =0 h ma cos θ μ s n = mg sin θ n mg cos θ h ma sin θ =0 h ma sin θ + mg cos θ = n Then substituting, h ma cos θ μ s (h ma sin θ + mg cos θ)=mg sin θ h ma cos θ h ma μ s sin θ = mg sin θ + mgμ s cos θ sin θ + μs cos θ h ma = mg cos θ μ s sin θ tan θ + μs = mg μ s tan θ tan θ +tanθs = mg = mg tan(θ + θ s) tan θ s tan θ We would epect h ma to increase as θ increases, with similar behavior as we established for h min, butwith h ma values alwas larger than h min. We can see that this is the case if we graph h ma as a function of θ,asthe curve is the graph of h min translated θ s to the left, so the equation does seem reasonable. Notice that the equation predicts h ma as θ (90 θ s). In fact, as h ma increases, the normal force increases as well. When (90 θ s) θ 90, the horizontal force is completel counteracted b the sum of the normal and frictional forces, so no part of the horizontal force contributes to moving the block up the plane no matter how large its magnitude. 7. (a) r(t) =R cos ωt i + R sin ωt j v = r 0 (t) = ωr sin ωt i + ωr cos ωt j,sor = R(cos ωt i +sinωt j) and v = ωr( sin ωt i +cosωt j). v r = ωr ( cos ωt sin ωt +sinωt cos ωt) =0,sov r. Sincer points along a radius of the circle, and v r, v is tangent to the circle. Because it is a velocit vector, v points in the direction of motion. (b) In (a), we wrote v in the form ωr u, whereu is the unit vector sin ωt i +cosωt j. Clearl v = ωr u = ωr. AtspeedωR, the particle completes one revolution, a distance πr, in time T = πr ωr = π ω. (c) a = dv dt = ω R cos ωt i ω R sin ωt j = ω R(cos ωt i +sinωt j),soa = ω r. This shows that a is proportional to r and points in the opposite direction (toward the origin). Also, a = ω r = ω R. (d) B Newton s Second Law (see Section 0.9), F = ma,so F = m a = mrω = m (ωr) R = m v R.

9 CHALLENGE PROBLEMS 9 9. (a) The projectile reaches maimum height when 0= d dt = d dt [(v0 sin α)t gt ]=v 0 sin α gt;thatis,when t = v 0 sin α v0 sin α and =(v 0 sin α) g g g v0 sin α = v 0 sin α. This is the maimum height g g attained when the projectile is firedwithanangleofelevationα. This maimum height is largest when α = π. In that case, sin α =and the maimum height is v 0 g. (b) Let R = v 0 g. We are asked to consider the parabola +R R =0which can be rewritten as = R + R. The points on or inside this parabola are those for which R R and 0 R + R. When the projectile is fired at angle of elevation α, the points (, ) along its path satisf the relations =(v 0 cos α) t and =(v 0 sin α)t gt,where0 t (v 0 sin α)/g (as in Eample 0.9.5). Thus v0 cos α v0 sin α = g v 0 sin α g v0 g = R. This shows that R R. For t in the specified range, we also have = t v 0 sin α gt = gt v0 sin α t 0 and g =(v 0 sin α) v 0 cos α g g = (tan α) v 0 cos α v0 cos α = R cos α + (tan α). Thus R + R = R cos α + R +(tanα) R = + (tan α) R R cos α = ( sec α)+r (tan α) R R = (tan α) +R (tan α) R R = [(tan α) R] R 0 We have shown that ever target that can be hit b the projectile lies on or inside the parabola = R + R.Nowlet(a, b) be an point on or inside the parabola = R + R.Then R a R and 0 b R a + R.Weseekanangleα such that (a, b) lies in the path of the projectile; that is, we wish to find an angle α such that b = R cos α a + (tan α) a or equivalentl b = R (tan α +)a + (tan α) a. Rearranging this equation we get a R tan α a tan α + a R + b =0or a (tan α) ar(tan α)+(a +br) =0 ( ). This quadratic equation for tan α has real solutions eactl when the discriminant is nonnegative. Now B 4AC 0 ( ar) 4a (a +br) 0 4a (R a br) 0 a br + R 0 b R (R a ) b R a + R. This condition is satisfied since (a, b) is on or inside the parabola = R + R. It follows that (a, b) lies in the path of the projectile when tan α satisfies ( ), that is, when tan α = ar ± 4a (R a br) a = R ± R br a. a

10 0 CHALLENGE PROBLEMS (c) =(v 0 sin α)t gt = D tan α If the gun is pointed at a target with height h at a distance D downrange, then tan α = h/d. When the projectile reaches a distance D downrange (remember we are assuming that it doesn t hit the ground first), we have D = =(v 0 cos α)t,sot = fallen from height h to height h gt = D tan α. (a) m d R dt dr = mg j k dt d dt D v 0 cos α and gd. Meanwhile, the target, whose -coordinate is also D,has v0 cos α gd. Thus the projectile hits the target. v0 cos α m dr dt + k R + mgt j = 0 m dr dt + k R + mgt j = c (c is a constant vector in the -plane). At t =0,thissasthatm v(0) + k R(0) = c. Sincev(0) = v 0 and R(0) = 0,wehavec = mv 0. Therefore dr dt + k m R + gt j = v 0,or dr dt + k m R = v 0 gt j. (b) Multipling b e (k/m)t gives e (k/m)t dr dt + k m e(k/m)t R = e (k/m)t v 0 gte (k/m)t j or d dt (e(k/m)t R)=e (k/m)t v 0 gte (k/m)t j. Integrating gives e (k/m)t R = m mg k e(k/m)t v 0 k te(k/m)t m g k e(k/m)t j + b for some constant vector b. Setting t =0ields the relation R(0) = m k v 0 + m g j + b, sob = m k v 0 m g j. Thus e (k/m)t R = m k R(t) = m e kt/m k mg e (k/m)t v 0 v 0 + mg k k k te(k/m)t m g k m k ( e kt/m ) t j. 3. (a) Instead of proceeding directl, we use Formula 3 of Theorem 0.7.5: e (k/m)t j and r(t) =t R(t) v = r 0 (t) =R(t)+t R 0 (t) =cosωt i +sinωt j + t v d. (b) Using the same method as in part (a) and starting with v = R(t)+t R 0 (t), wehave a = v 0 = R 0 (t)+r 0 (t)+t R 00 (t) =R 0 (t)+t R 00 (t) =v d + t a d. (c)herewehaver(t) =e t cos ωt i + e t sin ωt j = e t R(t). So, as in parts (a) and (b), k v = r 0 (t) =e t R 0 (t) e t R(t) =e t [R 0 (t) R(t)] a = v 0 = e t [R 00 (t) R 0 (t)] e t [R 0 (t) R(t)] = e t [R 00 (t) R 0 (t)+r(t)] = e t a d e t v d + e t R Thus, the Coriolis acceleration (the sum of the etra terms not involving a d )is e t v d + e t R. 5. (a) a = g j v = v 0 gt j =i gt j s = s 0 +t i gt j =3.5 j +t i gt j s =t i gt j. Therefore =0when t = 7/g seconds. At that instant, the ball is 7/g 0.94 ft to the right of the table top. Its coordinates (relative to an origin on the floor directl under the table s edge) are (0.94, 0). At impact, the velocit is v =i 7g j,sothespeedis v = 4+7g 5 ft/s.

11 CHALLENGE PROBLEMS (b) The slope of the curve when t = and θ d is g d = d/dt d/dt = gt = g 7/g = 7g 7g. Thus cot θ = (c) From (a), v = 4+7g. So the ball rebounds with speed g.08 ft/s at angle of inclination 90 θ B Eample 0.9.5, the horizontal distance traveled between bounces is d = v 0 sin α, g where v 0.08 ft/sandα Therefore, d.97 ft. So the ball strikes the floor at about 7/g ft to the right of the table s edge.

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