So we ll start with Angular Measure. Consider a particle moving in a circular path. (p. 220, Figure 7.1)

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1 Lectue 17 Cicula Motion (Chapte 7) Angula Measue Angula Speed and Velocity Angula Acceleation We ve aleady dealt with cicula motion somewhat. Recall we leaned about centipetal acceleation: when you swing something aound in a cicle the net foce points towad the cente of the cicle. The centipetal acceleation can be calculated using the velocity of the object and the adius of the cicle: a c = v 2 We jumped into centipetal acceleation because we wee talking about how foces and acceleation wee elated. But now we e eady to fill out ou undestanding of cicula motion. So we ll stat with Angula Measue. Conside a paticle moving in a cicula path. (p. 220, Figue 7.1) (x,y) We have ou taditional Catesian (x-y) coodinate system. But this isn t optimal fo measuing something going in a cicle. (,θ) Thee is anothe coodinate system, pola coodinates, that is bette fo cicula motion. Fo something moving in a cicle, is constant. So while x and y constantly change fo a paticle moving in a cicle, only θ changes in pola coodinates. We can use and θ to find x and y: x = cos" y = sin"

2 Recall that we defined linea displacement Δx = x f x i. Fo cicula motion, we can define angula displacement Δθ = θ f θ i. Anothe vaiable impotant to cicula motion is ac length (s). Ac length is the distance between two points on a cicle. (Figue 7.2, p. 220) The fomula is: s = " The units of ac length ae metes, assuming that is measued in metes and θ is measued in adians. Thee ae 2π adians in 360. Thee ae π adians in 180. If we solve fo θ, we get: " = s which is the atio of two lengths. This makes a adian a pue numbe, dimensionless quantity. Let s ty an example: You measue the length of a distant ca to be subtended by an angula distance of 2.0. If the ca is actually 5.0m long, appoximately how fa away is the ca? s = " θ s = s " $ 2 o # ' & ) = ad % 180 o ( = 5.0m 0.035ad =143m Now that we undestand angula measue, we can move onto angula speed and velocity.

3 At this point, we ve gone ove speed and velocity a geat deal. We undestand that velocity is "x. Now fo cicula motion we have to define "t the angula analog: " = #$ # t The units ae typically adians/second, but evolutions/minute (pm) ae alos common units. The diection is defined based on whethe the speed is clockwise o counte-clockwise. The ule to follow is to use you ight hand and wap it in the diection of motion. If the angula speed is counteclockwise, you thumb points up (positive diection). This is the diection of the vecto associated with counte-clockwise motion. If the angula speed is clockwise, you thumb points down (negative diection). This is the diection of the vecto associated with clockwise motion. This is just a convention that has been defined abitaily because its easie to define vectos up and down than it is to define them clockwise and counte-clockwise. Let s ty an example: A gymnast on a high ba swings though two (clockwise) evolutions in a time of 1.90s. Find the angula velocity of the gymnast. The angula displacement of the gymnast is & 2% ad) "# = $2ev( + = $12.6 ad ' 1ev *, = "# "t = $12.6ad = $6.63 ad 1.90s s We can talk about the ate of angula change, but what if we want to know the linea velocity of the cicula motion? (Figue 7.6, pp.224) Thee is a vey easy elationship: v = " ω has to be in adians pe second. This means v will be in metes pe second. Let s ty an example: The tangential speed of a paticle on a otating wheel is 3.0m/s. If the paticle is 0.20 m fom the axis of otation, how long will it take fo the paticle to go though one evolution?

4 So what do we need to do? go fom tangential velocity to angula velocity, then use the fact that we know thee ae 2π adians in one evolution to get the amount of time it takes to complete one evolution. " = v = 3m s 0.2m =15 ad s 2#ad 15 ad s = 0.42s Of couse now that we ve coveed angula velocity the next logical thing to cove is angula acceleation. It is defined in tems of the change of angula velocity ove time: " = #$ #t the units of angula acceleation ae commonly adians/s 2. We can do a quick example to demonstate how to use angula acceleation: Duing an acceleation, the angula speed of an engine inceases fom 700 pm to 3000 pm in 3.0s. What is the aveage angula acceleation of the engine? " o = 700 ev $ 2# ad' $ min & ) 1min ' & ) = 73 % 1ev (% 60s ( ad s $ 2# ad' $ " = 3000ev min & ) 1min ' & ) = 314 ad % 1ev (% 60s ( s * = +" +t = 314 ad s, 73ad s 3s = 80 ad s 2 Each second the angula velocity of the engine is inceasing by 80 ad/s. Something to point out now is that just as we had a set of equations fo linea displacement, velocity, acceleation we have a set of kinematic equations fo angula displacement, velocity, and acceleation. (Table 7.2, p.236)

5 We ll come back duing the next lectue to wok a poblem utilizing these equations, but fo now we e going to segue into ideas of otational foce. Recall that we had Newton s second law which told us F net = ma. This implies that foce and acceleation ae connected. We ve just defined angula acceleation, theefoe thee must be some otational foce connected to it. In physics we do define a otational foce. We call it toque. The basis of toque is the idea that to poduce a change in otational motion, we need a otational foce. Let s think about what factos might be involved in a otational foce. If you think about tying to loosen a tight bolt, how do you maximize this effot? Fist, do you hold the wench close to the head o towad the end of the handle? Does it matte how had you push? These questions make sense when you conside the definition of toque: " = F sin# is called the leve am, F is the foce applied and θ is the angle between them. Obviously, the fist case is going to have the maximal effect. The leve am is longe and the angle between the foce and leve am is 90. The biggest that the sine function can be is 1. So by exeting the foce at 90, we e maximizing the toque. Also, by making the leve am () as long as possible, we e also maximizing the toque. Let s ty a poblem utilizing balanced toques: We have thee masses suspended fom a metestick. The question is how much mass must be suspended fom the ight side of the metestick to be in equilibium (no angula acceleation).

6 We have fou foces to deal with hee: Foce of gavity on the metestick, weight of m 1, weight of m 2, weight of m 3 (unknown). So what we can do is to daw a diagam showing all the potential toques: axis of otation

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