chp10 - Problem Section 10-1: Center of Mass ,,9. Find the...

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Unformatted text preview: Problem Section 10-1: Center of Mass ,,9. Find the center of mass of a pentagon of side a with one triangle missing, as shown in Fig. 10-24. Hint: See Example 10-3, and treat the pentagon as a group of triangles. _-Pro61~m1. A 28-kg child sits at one end of a 3.5-~~l~g seesaw. Where should her 65-kg father sit so the center of mass will be at the center of the seesaw? Solution I -- Solution Take the araxis along the seesaw in the direction of the father with origin at the center. The center of mass of the child and her father is at the origin, so Xcm = 0= mcxc + mlxf, where the masses are given, and Xc = -(3.5 m)/2 (half the length of the seesaw in the negative X direction). Thus, x I = -mcxc/mf = (28/65)(1. 75 m) = 75.4 em from the center. 1 Problem . - S:-"Threeequal masses lie at the corners of an equilateral triangle of side i. Where is the center of , mass? I j I Choose coordinates as shown. From symmetry,.xcm = O. If the fifth isosceles triangle (with the same assumed uniform density) were present, the center of mass of the whole pentagon would be at the origin, so 0= (mys + 4mYcm)/5m, where Ycm gives the position of the center of mass of the figure we want to find, and Ys is the position of the center of mass of the fifth \ triangle. Of course, the mass of the figure is four times the mass of the triangle. In Example 10-3, the center : of mass of an isoscelestriangle is calculated, so , Ys=' and from the geometry of a pentagon, tan36 =!a/l. Therefore, Ycm = = ~l = 112 a cot 36 = O.115a. i -ii, -hs Solution Take x-y coordinates with origin at the center of one side as shown. From the symmetry (for every mass at ' x, there is an equal mass at -x) Xcm = O. Since Y = 0 ; for the two masses on the araxis, and y = t sin 60 = tv'3/2 for the other mass, Equation 10-2 gives Ycm = , I m(iv'3/2)/3m = /2.../3 = 0.289. \ FIGURE 10-24 Problem 9 Solution. Problem--~~------18. Find the center of mass of the uniform solid cone. o~height h and constant density p sho~n in Fig. 10-28: Hint: Integrate over disk-shaped masS elements of thickness dy, as shown in the figure. m -~X Problem 5 Solution. x FIGURE 10-28 Problem 18 Solution. Solution Choose the y-axis along the axis of the cone and the origin at the center of the base. For mass el~ments take disks at height y, of radius r = R(l - y/h), , parallel to the base. Then dm = p1rr2 dy, where p is the density of the cone, and M = ~p1T R2 h is its mass. The eM is along the axis (from symmetry) at a height Yem 1 {h 3 {h , M Y dm = p1TR2h J Yp1TR2(1 - y/h)2 dy . 0 o y 2 =y--+dy h 0 h h2 2 2 2 = ~ (h _ 2h + h ) = !h h 2 3 4 4' Problem 27. During a heavy storm, rain falls at the rate of 2.0 em/hour; the speed of the individual ~aindrops is 25 m/s. (a) If the rain strikes a flat roof and then flows off the roof with negligible speed, what is the force exerted per square meter of roof area? (b) How much water would have to stand on the roof to exert the same force? The density of water is 1.0 g/cm3. = 31h ( 1 2 y3) Solution (a) Suppose the number of drops (each with mass 8m and speed v) falling per second is dN/dt. The momentum lost per second isdp/dt = (dN/dt)8m. v. This equals the magnitude of the force exerted on the , roof, so F/A = (dN/dt)fJm. viA, where A is the roof , area. The mass of water falling on the roof per unit time equals the volume rate of fall times the density, or (dN/dt)fJm = A(2 cm/h)(l g/cm3). Therefore, F/A = - (2 cm/3600 s)(l g/cm3)(25 m/s) = 0.139 N/m2. (b) Water standing on area A to a depth h has weight M 9 = Ahpg. Equating this to the force found in part (a), we find: Ahpg/A = 0.139 N/m3, or h = (0.139 N/m3)/(9.8 m/s2)(103 kg/m3) = 0.0142 rom. (Even during a heavy storm, the force of the raindrops is quite small.) Section 10-2: Momentum Problem 19. A popcorn kernel in a hot pan bursts into two pieces, with masses of 91 mg and 64 mg. The more massive piece moves horizontally at 47 cm/s. Describe the motion of the second piece. I i Solution Suppose the popcorn kernel was initially at rest, and not subject to a net external force while bursting apart. The momentum of the two pieces will still be zero immediately afterwards (conservation of momentum during the break-up). Then 0 = mivi + m2V2,where VI and V2are the final velocities of the two pieces, so V2= -(mIfm2)vI' If VI = (47 cm/s)i is the final velocity of the more massive piece, then V2= -(91 mg/64 mg)(47 cm/s)i = -(66.8 cm/s)ij i.e., the less massive piece moves with speed 66.8 cm/s. in the opposite direction to the more massive piece. ~ ----- L . . : Problem 31. A 1600-kg automobile is resting at one end of a 4500-kg railroad flatcar that is also at rest. The automobile then drives along the flatcar at 15 km/h relative to the flatcar. Unfortunately, the flatcar brakes are not set. How fast does it move? Problem 23. A 680-g wood block is at rest on a frictionless \table, when a 27-g bullet is fired into it. If the block with the embedded bullet moves off at 19 mis, what was the original speed of the bullet? Solution Suppose that the track is horizontal and that the horizontal component of the net external force on the flatcar/automobile system is negligible (e.g. rolling friction and air resistance). Then the total momentum along the track, initially zero, is conserved, or . Ptot = O.Let Vc be the velocity of the flatcar relatIve to the track, and VA = 15 km/h + Vc be the velocity (positive in the direction of the auto:s motion) of the auto relative to the track (see EquatIOn 3-10). Then \ 0 = mcvc + mAVA = mcvc + mA(15 km/h) + mAvc, or Vc = -mA(15 km/h)/(mA + mc) =~-(16/61)X / (15km/h) = -3.93 kmLh. . . - -- - ~ f Solution Since the table's surface is frictionless, there are no horizontal external forces on the block and bullet, and the component of their combined momentum, in the direction of the bullet's initial horizontal velocity vo, is conserved. As in Example 10-6, mvo = (m + M) V, so Vo = (680 + 27)(19 m/s)/27 = 498 mfs. ----------- _ -_ .. .. Problem - 41. Determine the center of mass and internal kinetic energies before and after decay of the lithium nucleus of Example 10-7. Treat the individual nuclei as point particles. Solution Before the decay, the system consists of one particle (the 5Li-nucleus), so Kint,i = 0, and Kern = Ki = 4mLivli = !<5x1.67xlO-27 kg)(1.6x106 m/s)2 = 1.07xlO-14 J = 66.8 keY. Afterwards, Kem is the -same (since momentum is conserved) while K- t f = I I 2' K f - K em = '2mHvH2 + '2MHeVHe - Kern = In, ~(1.67xlO-27 kg)[(4.5x106 m/s)2 + 4(1.4x106 m/s)2J- 66.8 keY = 79.8 keY. 44 A sealed can of air contains 10.21air molecules . moving randomly, on average, at 340.mfs. If_the can is thrown at 10.mfs, compare the . center-of-mass and internal energies of the au. Problem Problem 62. A drunk driver in a 16o.O-kg plows into a car 13o.O-kg parked car with its brake set. Police measurements show that the two cars skid together a distance of 25 m before stopping. If the effective coefficient of friction is 0..77,how fast was the drunk going just before the collision? Solution The random velocities of the molecules averag~ to K - !(~ mo)(1o. mfs)2, where rna IS the zero, so em - 2 L..J 11 ixture of mass of an air "molecule." (Air is rea y. am gases so mo is the average molecular weIght.) The average speed relative to the center ~f mass (o~, . mterna 1 sp eed) l'S' the same for each molecule, so fK _ Kint = !(Emo)(34o. mfs)2. Evidently, Kint em (34)2 = 11~6:----~ ~ ~ ~ ---~- ~ .... Solution If the wreckage skidded on a horizontal road, the work-energy theorem requires that the work done by friCtion be equal to the negative of the kinetic energy of both cars just after the collision, Wnc = -J1.k(ml +m:i)gx = t1K = o.-! x (ml +m2)y2. Thus, the speed of the cars just after the collision is Y = ..j2J1.kgX. During the instant of time in which the , collision occurred, momentum was conserved (see , Section 11-2), so if VI, was the speed of the drunk driver's car just before the collision (and V2 = o..forthe; parked car), mlVI = (mi + m2)V, or VI = / [(160.0.+ 13o.o.)fI6o.o.] V2(.77)(9.8 m/s2)(25 m)= -- Problem 57. Figure 10-32 shows a paraboloidal solid of constant density p. It extends a height h along the z-axis, and is described by z = ar2, where the units of a are m'-I, and.J' is the radius in a plane perpendicular to the z-axis. Find expressions for (a) the mass of the solid and (b) the z coordinate of its center of mass. ,'----~--~------ 35.2 mfs. (This is ~79 mi/h, so the,driver was speeding as well as intoxicated.) y FIGURE 10-32 Problem 57. Solution For mass elements, take disks parallel to the :vy plane, of radius r = and thickness dz, as shown in Fig. 10-32 and also in Fig. 10-28 for Problem 18. Then: dm = p1rr2 dz = p1r(zfa) dz. (a) The total mass is M = I h Io~dm = Io (p1r/a)z dz = p1rh2f2a. (b) Zcm = h Io zdmfM == (2afp7rh2) Ioh(p1r/a)z2dz = (2/h2)X (h3 f3) = 2h/3. (Since the paraboloid is symmetrical . about the z-axis, Xcm = Ycm = 0..) / ..;zra -'" ...
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