Uniform Circular Motion

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Cordelia Gray
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1 Unifom Cicula Motion Intoduction Ealie we defined acceleation as being the change in velocity with time: a = v t Until now we have only talked about changes in the magnitude of the acceleation: the speeding up o slowing down of objects. Howeve, since velocity is a vecto, it has both a magnitude and a diection; so anothe way that velocity can change is by changing its diection...even while its speed emains constant. his is the type of acceleation that will be exploed in this chapte. A vey impotant case of changing diection while maintaining constant speed is called unifom cicula motion. In that case, the object s speed emains constant, hence unifom, while its diection is constantly changing, this is necessay to keep it moving in a cicle. If thee wee no acceleation it would tavel in a staight line. his type of motion occus in a numbe of instances: impotant examples being the motion of the planets aound the sun o the moon aound the eath. It was though an analysis of unifom cicula motion that Newton was able to develop this theoy of Univesal Gavitation; so it s natual that that will also be exploed in this chapte. Unifom Cicula Motion Fom Newton s fist law we know that if thee is no net foce acting on an object, it will tavel in a staight line at constant speed. Wheneve an object fails to tavel in this way it is, by definition acceleating. By Newton s second law we can also conclude that thee must be a net foce acting on it. Cicula motion is a special case of an object expeiencing a constant acceleation. In the case of cicula motion, an object is constantly changing diection. Rathe than taveling in a staight line, its path is always bent towads the cente of the cicle that defines its path. As shown below, its velocity is always tangent to the cicle that descibes its motion and its acceleation is always diected towads the cente of the cicle. By examining the velocity vecto of the object ove a small time diffeence, Δt, it can be seen that the change in velocity, the aow connecting the tips of the ealie velocity vecto to the tip of the late vecto is always diected towads the cente. his aow epesents the change in velocity ove the time, Δt: thus it epesents the diection of the acceleation. V V a he object s velocity is always tangential to its path about the cicle. a (Path of Motion) a he object s acceleation is always pointing towads the cente of the cicle, pependicula to the velocity vecto. V Unifom Cicula Motion 1 v by Goodman & Zavootniy

2 Also, it can be seen that the tiangle made up of v 1 and v 2 (shown below) is simila to the tiangle made up of the adii of the cicle and the distance taveled duing time Δt, vt (shown below). he magnitude of v 1 and v 2 ae the same, since the object s speed is constant, so we can set up the following popotion: his is a vey impotant elationship as it gives us the magnitude of the acceleation of any object taveling along a cicula path. he diection of that acceleation is towads the cente of the cicle. he acceleation of an object moving along a cicula path is given by: a = v2 towads the cente of the cicle his is a vey impotant esult as it can be added to a few othe special cases we have peviously exploed. Special Cases of Acceleation An object taveling in a staight line at constant speed: a = 0 An unsuppoted object nea the suface of the eath: a = g = 9.8 m/s 2 towads the cente of the eath An object moving along a cicula path: a = v2 towads the cente of the cicle Example 1 A 4.0 kg object is taveling in unifom cicula motion of adius 2.0m. he magnitude of its velocity, its speed, is 15 m/s. Detemine its acceleation. Detemine the net foce acting on it. Since the object is taveling in cicula motion the magnitude of its acceleation is give by a = v2 and the diection of its acceleation is towads the cente of the cicle. Unifom Cicula Motion 2 v by Goodman & Zavootniy

3 a = v2 a = (15m s )2 2m a = 225m2 s 2 2m a = m/s 2 towads the cente of the cicle he net foce acting on the object is esponsible fo its acceleation so: ΣF = (4kg)(112.5 m/s 2 ) F net = 450N Since the acceleation of an object is always diected in the same diection as the net foce, the net foce must also be diected towads the cente of the cicle. F net = 450N towads the cente of the cicle Example 2 How much net foce is equied to keep a 5.0 kg object taveling in a cicle of adius 6.0m with a speed of 12 m/s? Since the object is taveling in a cicle its acceleation must be v 2 /, so ΣF = m( v2 ) ΣF = (5kg) (12m s )2 6m ΣF = (5kg) 144m2 s 2 6m ΣF = 120 kg m/s 2 F net = 120 N towads the cente of the cicle Peiod and Fequency hee ae two closely elated tems that ae used in descibing cicula motion: peiod and fequency. he peiod of an object s motion is the time it takes it to go once aound a cicle. Peiod is a measue of time so the standad units fo peiod ae seconds and the symbol fo peiod is (easily confused with the symbol fo ension). If an object completes a cetain numbe of otations, n, in a given amount of time, t, then it follows that = t/n, since that is the time it must be taking to complete each otation. Fo example, if I go aound in a cicle ten times in five seconds, my peiod is the time it takes fo each tip aound the cicle and is given by: = t n Unifom Cicula Motion 3 v by Goodman & Zavootniy

4 = 10s 5 = 2s Sometimes it s helpful to think about the numbe of times I go aound pe second athe than the numbe of seconds it takes me to go aound once. his is called the fequency of otation since it descibes how fequently I complete a cycle. he fequency of an object s motion is the numbe of times that it goes aound a cicle in a given unit of time. he symbol fo fequency is f (easily confused with fiction). If an object completes a cetain numbe of otations, n, in a given amount of time, t, then it follows that f = n t, since that is the numbe of times I go aound in a given time. he units fo fequency must eflect the numbe of epetitions pe unit time. If time is measued in seconds, as will typically be done in this book, the unit of fequency is 1/s o s -1. his unit, s -1, has also been named the Hetz (Hz). his tem is commonly used in descibing adio stations: when you tune you adio to 104.3MHz you ae tuning it to a adio signal that epeats its cycle million times pe second. Similaly, the station at 880 khz has a signal that epeats 880 thousand times pe second. Since = t n and f = n t it follows that: = 1 f and f = 1 Fo example, if the peiod,, of an object s motion is 0.2 seconds then its fequency, f, is given by: f = 1 1 f =.2s f = 5 s -1 o 5 Hz Similaly if an object s fequency is 20 Hz then its peiod is given by = 1 f 1 = 20Hz = 0.05s hee is also a diect connection between peiod, fequency and speed. Since the distance aound a cicle is given by its cicumfeence: an object must tavel a distance of 2π in ode to complete one cicle. hat yields impotant elationships between speed, peiod and fequency. s = d t s = 2π s = 2π v = 2π in ode to tavel aound one cicle I must move a distance of 2π by the definition of peiod, the time it takes to do that is since the speed is the magnitude of the instantaneous velocity of an object, this means that Unifom Cicula Motion 4 v by Goodman & Zavootniy

5 v = 2πf Also, since f = 1, this can also be witten as Relationships between peiod, fequency and velocity = t 1 and = f = n 1 and f= n f v = 2π v = 2πf he units of peiod,, ae seconds. he units of fequency, f, ae s -1, o Hz. Example 3 An object is taveling in a cicle of adius 4m and completes five cycles in 2s. What ae its peiod, fequency and velocity? = t n = 2s 5 = 0.4s f = 1 f = 1.4s f = 2.5 Hz o f = n t f = 5 2s f = 2.5 Hz v = 2π o v = 2πf v = 2(3.14)(4m) v = 2(3.14)(4m)(2.5Hz).4s v = 62.8 m/s v = 62.8 m/s Example 4 A foce of 250 N is equied to keep an 8.0 kg object moving in a cicle whose adius is 15m. What ae the speed, peiod and fequency of the object? F net = m( v2 ) F net m = v2 v = F net m v = (250N)(15m) 8kg Unifom Cicula Motion 5 v by Goodman & Zavootniy

6 v = 3750kgm2 s 2 8kg v = 470 m2 s 2 v = m/s Since the object is moving in a cicle, its velocity must be tangent to the cicle v = m/s tangent to the cicle Next, we find the peiod of the object s motion. v = 2π = 2π V = 2(3.14)(15m) m s = 4.35s And finally, we find its fequency. f = 1 1 f = 4.35s f = 0.23Hz One impotant aspect of the equations fo unifom motion is that they apply even fo the bief peiods of time that an object s motion can be consideed cicula. Fo instance, as you dive aound a cuve in a oad, that cuve can be appoximated as a pat of a cicle. You don t have to tavel aound a complete cicle fo these equations to be useful: just like you didn t have to dive 60 miles in one hou fo you speed to be 60 miles pe hou. hese equations can be applied anytime the motion of an object is even biefly cicula in natue. Example 4 An object is attached to a sting which is supplying a ension that keeps it moving in a cicle of adius 0.50m as it slides along a fictionless table. he object has a mass of 2.0 kg and its motion has a peiod of 0.63s. What is the tension in the sting? F = m( v2 ) F = m( F = 4mπ2 2 4π R ) but v = 2π so v = 4π substituting Unifom Cicula Motion 6 v by Goodman & Zavootniy

7 F = 4(2kg)(3.14)2 (.5m) (.63s) 2 F = kg m/s 2 F = N towads the cente of the cicle Multiple foces and cicula motion Often moe than one foce is acting on an object...including an object taveling in unifom cicula motion. In that case, you teat this case the same way you did in any dynamics poblem, the sum of the foces mattes...not any one foce. So fo instance, if we changed the pio example by having the object moving in a vetical cicle, athe than a hoizontal one, we have two foces acting on the object to keep its motion cicula, the weight of the object will always be down, but the tension foce would always point towads the cente of the cicle. Since the acceleation of the object is always given by v 2 /, then the sum of the foces will always equal ma o mv 2 /. hus as long as the velocity is constant the net foce must be as well. Howeve, the tension foce will sometimes be opposed by the weight of the object, at the bottom of the cicle, and will sometimes be in the same diection, at the top of the cicle. hat means that the tension foce will have to vay since the sum of the foces is constant and the weight can t vay. Example 5 An object is attached to a sting which is supplying a ension that helps keeps it moving in a vetical cicle of adius 0.50m. he object has a mass of 2.0 kg and is taveling at a constant speed of 5.0 m/s (impactical to do, but let s use that as an assumption). What is the tension in the sting in the following situations? a. When the object is at the top of the cicle. b. When the object is at the bottom of the cicle. c. When the sting is hoizontal. a. At the top of the cicle, both the weight and the tension point downwads. So does the acceleation of the object, since the acceleation of an object in unifom cicula motion is always diected at the cente of the cicle. If we define down as negative: - - W = -ma but since all the signs ae negative, we can multiply by negative one on both sides and make them all positive. + W = ma = ma - W = m v2 - mg = m( v2 g) = (2.0 kg) ( (5m s )2 ) 9.8.5m m/s2 ) = (2.0 kg)( ( 25m2 s 2.5m )) 9.8 m/s2 ) = (2.0 kg)( (50 m 2 /s m/s 2 ) = 80.4 kg m/s 2 = 80.4 N downwads b. At the bottom of the cicle, the weight points downwad and the tension points upwads, towads the cente of the cicle. he acceleation of the object also point upwads towads the cente of the cicle, since Unifom Cicula Motion 7 v by Goodman & Zavootniy

8 the acceleation of an object in unifom cicula motion is always diected at the cente of the cicle. If we define down as negative: - W = ma = ma + W = m v2 + mg = m( v2 + g) = (2.0 kg) ( (5m s )2 ).5m m/s2 ) = (2.0 kg)( 25m2 s 2.5m m/s2 ) = (2.0 kg)( (50 m 2 /s m/s 2 ) = kg m/s 2 = N upwads Note that in this case and ma ae positive and only W is negative. c. When the sting is hoizontal, the weight points downwad and the tension points sideways, towads the cente of the cicle. he acceleation of the object, due to its cicula motion, also point sideways towads the cente of the cicle, since the acceleation of an object in unifom cicula motion is always diected at the cente of the cicle. In this case, the weight does not contibute to keeping the object moving in a cicle...it will seve to speed the object up. But since we ae only consideing the foces causing cicula motion, the weight will not affect the ension in the sting. his then becomes the same as Example 4, shown above...w will not play a ole in maintaining the cicula motion of the object. = m( v2 ) = (2.0 kg)( (5m s )2.5m ) = (2.0 kg)( 25m2 s 2.5m ) = ( 50kgm2 s 2 ).5m = 100 kg m/s 2 = 100 N towads the cente of the cicle In example 5a, it can be seen that thee will be a minimum speed that the object can tavel in a vetical cicle. At that minimum speed, the tension in the sting becomes zeo at the top of the cicle and only the weight of the object supplies the foce necessay to maintain cicula motion. If an object moves any slowe than that, it will depat fom cicula motion as it falls towads the cente of the cicle. In Example 5, the elationship that gives us this speed is seen in the equation: = m( v2 g). Fom this it can be seen that when v2 g = 0, the tension in the sting will be zeo. Since the tension in a sting cannot be negative, this is the lowest possible velocity fo cicula motion; and this velocity is independent of the mass of the object. Solving fo v gives us: v 2 g = 0 Unifom Cicula Motion 8 v by Goodman & Zavootniy

9 v 2 = g v 2 = g v = g So in example 5, the lowest possible velocity fo the object to maintain cicula motion would be: v min = g v min = 9.8 m s2 (.5m) v min = 4.9 m2 s 2 v min = 2.2 m/s Example 6 A bucket of wate is spun in a vetical cicle such that the bucket is upside down with the wate in it at the top of the cicle. he peson wate in the bucket is spinning in a cicle of adius 0.80 m. a. What is the minimum velocity that the wate must maintain to stay in the bucket and not dench the peson below? b. If the wate has a mass of 0.25kg and the velocity of the bucket is the same at the bottom as it is at the top of the cicle, what nomal foce must the bucket povide at the bottom of the cicle? he wate expeiences two foces as it tavels though along its cicula path: its weight down and the nomal foce of the bottom of the bucket diected towads the cente of the cicle. he sum of these two foces must always equal ma, o m v2. a. At the top of the cicle, the weight, nomal foce and acceleation all point down. So, -F N - W = -ma but since all the signs ae negative, we can multiply by negative one on both sides and make them all positive. F N + W = ma Unifom Cicula Motion 9 v by Goodman & Zavootniy

10 W = ma mg = ma a = g v 2 = g v 2 = g v = g he minimum speed will be when F N = 0. At that point, the wate will appea to be weightless...in fee fall. v min = 9.8 m s2 (.8m) v min = 7.84 m2 s 2 v min = 2.8 m/s Note that the mass of the wate didn t matte, only the adius of the cicle. b. At the bottom of the cicle, the bucket and acceleation ae pointed upwads, while the weight of the wate is pointed down. So, F N - W = ma F N = W + ma F N = mg + ma F N = m(g + a) F N = m(g + v2 ) but since all the signs ae negative, we can multiply by negative one on both sides and make them all positive. We then use the velocity we calculated fo the top of the cicle F N = (0.25 kg)(9.8 m/s 2 + (2.8m s )2.8m ) F N = (0.25 kg)(9.8 m/s m/s 2 ) F N = (0.25 kg)(19.6 m/s 2 ) F N = 4.9 N Note that this nomal foce is double what it would be if that wee holding the wate is stationay. he appaent weight of the wate is double at the bottom and is zeo at the top of the cicle. Sometimes the foce keeping an object in cicula motion is due to fiction. In that case, thee ae eally thee dimensions involved in solving the poblem: the two dimensions in which the cicula motion is defined plus the nomal foce, which is pependicula to the plane of the cicula motion. It will be seen that in these cases the mass of the object does not affect the outcome. Example 7 A ca is ounding a cuve with a speed of 20 m/s. At that location, the cuve can be appoximated by a cicle of adius 150m. What is the minimum coefficient of static fiction that will allow the ca to make the cuve without sliding off the oad? Fom the top, a sketch of this poblem would show the ca taveling in a cicle. Howeve, fom that pespective it s not possible to daw a fee body diagam showing all the foces necessay to solve this Unifom Cicula Motion 10 v by Goodman & Zavootniy

11 poblem. So it s also impotant to make a sketch fom the pespective of someone standing on the oad with the ca diving away. op Down Fee Body Diagam Side view Fee Body Diagam adial-diection vetical-diection F sf = ma F N mg = 0 μ s F N = m( v2 F N = mg μ s mg = m v2 μ s = v2 g (20 m s )2 μ s = 9.8 m s 2 (150m) μ s = 0.27 Note that the mass of the ca doesn t matte. So when a oad is being designed, it s not impotant to know the mass of the cas that will be using it when detemining the maximum cuvatue of the oad. Unifom Cicula Motion 11 v by Goodman & Zavootniy

Describing Circular motion

Describing Circular motion Unifom Cicula Motion Descibing Cicula motion In ode to undestand cicula motion, we fist need to discuss how to subtact vectos. The easiest way to explain subtacting vectos is to descibe it as adding a