| Chapter 8 : Do listed problems. | ||||||||||
| Simulations | ||||||||||
| Quiz 8 | ||||||||||
| Problem hints with asterisks * were covered in class or recently added to this page. | ||||||||||
| 4 | 5 | 8 | 9 | *10 | 11 | 13 | 18 | 35 | 38 | 39 |
| 41 | 46 | *47 | *48 | |||||||
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| TIP FOR THIS CHAPTER: We define vector-FNET as the centripetal force. As fig 8.17 suggests, vector-FNET points toward the center. In other words, at any position on a circle, we define the positive direction to be toward the center. We use the same method as in Chapters 5, 6 and 7. Mar = pos - neg, where ar = V2/r is the magnitude of the centripetal acceleration pointing to the center; "pos" and "neg" denote the magnitude of the positively and negatively directed force, respectively. V = magnitude of ( tangential ) velocity. Remember the positive direction is toward the center of the circle ! Mar = pos - neg, where ar = V2/r . |
| 4. Centripetal force always points toward the center of the circle of motion. For the cause in this case, see class notes and also problem 5 below. |
| 5. MV2/r = | friction force |. Solve for the magnitude fs of the friction force. Note that fs is not necessarily at maximum value. However, when it reaches that value , the car will "spin out" . Drive safely. |
| 8. See example 8.5 |
| 9. MV2/r = | gravitational
force |.
M and r are listed in table at back of book. You need V = 6.28r/T', where T' = period = 27.3 days. Convert to seconds by remembering there are 24 hours/day, and 3600 seconds/hour. The magnitude of gravitational force = "tension" force in the imaginary ( and long !) string. |
| 10. The force on the object rounding the circle
is centripetal; that force always points to the circle center. The
vertically upward normal force balances the downward weight mg, so the net
vertical force is zero. But the net horizontal normal force pointing to
the center supplies the centripetal force necessary to move in the
circle. MV2/r = pos - neg = N - 0 = N MV2/r = N, where N is the magnitude of the horizontal normal force. Get V from 60 rpm. The angular velocity (in rad /s) = w = (60 rev/min)*(6.28 rad/rev)*(1/60) (min/s) = 6.28 rad/s. Now, V = w*r. Compute MV2/r. Remember to convert to kg and to meters (m). |
| 11. V2/r =| gravitational acceleration | . Now V = 6.28r/T', where T' = period = 110 min. Convert to seconds by remembering there are 60 seconds/min. For r, see table at back of book |
| 13. See "Stop To Think 8.4", page
226. Also see fig. 8.17 which is for a slightly different problem.
But both problems reduce to this: The centripetal force points to the
center of the circle. We define vector-FNET as the centripetal force. As fig 8.17 suggests, vector-FNET points toward the center. In other words, at any position on circle, we define the positive direction to be toward the center. Thus, in the case of "Stop To Think 8.4", the positive direction is down toward the center of the circle defined by the hump in the road. We use the same method as in Chapters 5, 6 and 7. Mar = pos - neg, where ar = V2/r is the magnitude of the centripetal acceleration pointing to the center; "pos" and "neg" denote the magnitude of the positively and negatively directed force, respectively. Remember the positive direction is toward the center of the circle ! Thus for this problem: Mar = pos - neg MV2/r = mg - N. Solve for V when N = 0. Refer to "Stop To Think 8.4," p 226. |
| 18. Use equations 8.25 , page 226. The initial tangential velocity is zero. Find the change in time under the following condition: Vft2/r = at = 1.5 m/s2. In other words, Vft2 = (100 m)·(1.5 m/s2) . Solve for Vft and use one of equations 8.25 to find the change in time. Note you could also use equations 8.26 and 8.27. |
| 35. This classic, but not like coca-cola.
It's better than that: First of all, the horizontal circle's
radius is r = 0.20 m. Mar = pos - neg MV2/r = TsinØ - 0, where T = tension magnitude and TsinØ is the horizontal component pointing toward the center. Thus we are using the angle Ø = sin-1 (0.20/1). MV2/r = TsinØ. We can write: (A) MV2/r = T·(0.20) We want to eliminate T by finding another equation. We can do this be summing the forces in the vertical i.e. z direction: FNETz = pos - neg 0 = TcosØ - mg, where cos2Ø = 1 - sin2Ø = 1 - 0.04 = 0.96. Thus: (B) mg = TcosØ Solve equations (A) and (B) for T and V. From V you can find the angular speed w = V/r; Convert to rpm's by dividing by 6.28 and converting seconds to minutes. |
| 38. mat = pos - neg mat = 0 - fk mat = -µkN. You have to find the normal force magnitude in each case as discussed with problems on "humps" and "dips" in the road. For the dip N > mg; for the hump N < mg as long as the car is in contact with the ground. For the horizontally flat section, N = mg. Find at in each case. |
| 39. (a) Compute mg in Newtons. (b) and (c) Mar = pos - neg At top, pos is down: MVtop2/r = mg + Ttop - 0 MVtop2/r = mg + Ttop At bottom, pos is up: MVbot2/r = Tbot - mg Find the tensions at the top and bottom. |
| 41. See figures 8.17, 8.18 and 8.19. These should
explain it well. First of all get the speed V = 6.28r/T', where T' = 4.5
seconds. (a) Find the normal force at the top and bottom: Mar = pos - neg At top, pos is down: MV2/r = mg + Ntop - 0 MV2/r = mg + Ntop At bottom, pos is up: MV2/r = Nbot - mg Find the N at the top and bottom. (b) At top, pos is down: MV2/r = mg + Ntop - 0 MV2/r = mg + Ntop Set the normal force at top to zero. The solve for the speed. Then find the new period T'. |
| 46. The vertical Newton's equation for the hanging mass gives: m2g = T The horizontal Newton's equation for the orbiting mass on table gives: Mar = pos - neg, where ar = V2/r . Thus, m1V2/r = T Solve for V in terms of symbols. |
| More hints later !!! |