PHYSICS 151 Notes for Online Lecture #20

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1 PHYSICS 151 Notes fo Online Lectue #20 Toque: The whole eason that we want to woy about centes of mass is that we ae limited to looking at point masses unless we know how to deal with otations. Let s evisit the metestick. Say I apply a foce of 10 N on one end and 10 N on the othe end as shown 10 N 10 N We can fist attack this poblem using what we leaned when we wee studying statics. Σ F = 0 Since thee ae only foces in the y diection, we can add them up diectly. Taking up as positive: -10 N + 10 N = 0 So accoding to ou old definition of equilibium, this system is in equilibium. Howeve, we know that the metestick is going to move, even though the net foce is zeo. This tells us that thee is something missing fom ou definition of equilibium. TRANSLATIONAL EQUILIBRIUM means that the vecto sum of the foces is zeo. Σ F = 0 Line of Action The eason ou definition has failed us is that we now ae allowing objects to have spatial extent and so they can otate. Thee ae two things that detemine the chaacteistics of the otation: the magnitude of the foce and how fa fom the pivot the foce acts. Axis of otation Applied Foce Lectue 20 Page 1

2 Impotant Toque Definitions AXIS OF ROTATION: The axis about which the object moves (o would move if the object is in equilibium.) LINE OF FORCE o LINE OF ACTION: A staight line unning diectly though the applied foce. LEVER ARM: The leve am is the pependicula distance fom the axis of otation to the line of action. Finding the leve am is usually one of the hadest pats of toque poblems. Stat by finding, which is the distance fom the axis of otation to the point whee the foce is applied. Note that is always going to lie along the object that the foce is acting upon. We can then daw the leve am by dawing a line that is pependicula to the line of foce and passes though the axis of otation. We can call the leve am. The angle between and F is called θ. Note, then that = b g sin θ I said that the details of the otation wee detemined by the magnitude of the foce and the distance fom the CM that the foce acts. The poduct of these two quantities is called the toque. Toque descibes otation. τ = F τ = Fsinbθg Note: The units of toque ae N-m Line of even though we called that a Joule when we Action dealt with enegy and wok, we don t call it a Joule when we e using it to descibe toque. Axis of otation θ θ F Lectue 20 Page 2

3 Execise: Identify the foce, show the line of foce and the leve am fo each of the following situations in the box next to the dawing. The axis of otation is denoted by a dot. Lectue 20 Page 3

4 Diection: We use the ight hand ule to detemine the Counteclockwise = positive diection of the toque. If you cul you finges fom to F, Clockwise =negative you thumb will point in the diection of the toque. F F In the left pictue, the toque will point into the page. In the ight-hand pictue, the toque will point out of the page. Ex 20-1: A civil enginee needs to hang a taffic light ove an intesection fom a pole, as shown. The light weighs 70 N. 5 m 60 a) Daw the leve am. b) What is the magnitude of the leve am assuming the base of the suppot as the axis of otation? (sin 60 = 3/ 2, cos 60 = 1/2) c) What is the diection of the toque? d) What is the magnitude of the toque? axis of e) Descibe diffeent ways that the enginee could decease the amount of toque on the suppot. Lectue 20 Page 4

5 Ex. 20-2: An atist is designing a mobile, as shown. Whee must the mobile be hung to keep the od hoizontal and what is the tension in the hanging sting? Assume that the od is massless. An atist is designing a mobile as shown. Find whee along the od the mobile must be hung to keep the od level. a) Assume the od is massless and b) Assume the od has a mass of 1.0 kg m A = 0.30 kg 10.0m 8.0 m m B = 0.50 kg a) Assuming fist that the od has no mass. Let s assume that the point at which the mobile is to be hung is a distance x fom the ight end of the od. Fist, daw a fee-body diagam to show all of the foces acting on the body. We can then stat by applying the condition fo tanslational equilibium - the math fo tanslational equilibium is often easie, so that's the fist one we'll ty applying. ΣFy = 0 T mag mbg = 0 T = mag+ mbg T = m + m g b A Bg b ge j T = 03. kg+ 05. kg 98. T = 784. N = 79. N m 2 s This tells us the tension, but not whee the suppot must be placed. Fo that, we have to apply the toque equation. We can take toques about any point. Let s take them about the sting fom which the mobile balances. This has the advantage that, if we couldn t figue out the tension in the sting, it won t ente the poblem. Lectue 20 Page 5

6 8-x T x 8.0 m m A g m B g 10.0m Στ = 0 mbgx+ magb8 xg= 0 mbx = mab8 xg mbx = 8mA xma bma + mbgx = 8mA 8mA x = bma + mbg 803 (. kg) x = 03. kg kg b g = 3m Note that we could also take the toques about a diffeent point. Let s calculate them about the ight end of the od. The sum of the foces emains the same. Lectue 20 Page 6

7 Ex. 20-3: How does the distance change if the od has a mass of 1.0 kg? We now have to include the weight of the od. The weight of the od acts at the CM. Since the od is unifom, this will be the geometic cente of the od. 8-x T x 5m m A g m g 8.0 m 10.0m m B g ΣF T m g m g m g = 0 A B T = m + m + m g T = 03. kg+ 05. kg+ 10. kg 98. T = 17. 6N = 18N So the tension inceases, as we would expect. I m going to take the axis of otation to be at the ight end again. y Σ τ = 0 ightend mg(8) + mg(5) Tx= 0 A mg(8) + mg(5) = Tx A = 0 b A B g b ge j ( ) ma m g x T m [ 8(0.3 kg) + 5(1.0 kg) ]( ) m 2 s s x = = 4.0m 18 N Lectue 20 Page 7

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