Physics 201 Homewok 4 Jan 30, 2013 1. Thee is a cleve kitchen gadget fo dying lettuce leaves afte you wash them. 19 m/s 2 It consists of a cylindical containe mounted so that it can be otated about its axis by tuning a hand cank. The oute wall of the cylinde is pefoated with small holes. You put the wet leaves in the containe and tun the cank to spin off the wate. The adius of the containe is 0.12 metes. When the cylinde is otating at 2.0 evolutions pe second, what is the magnitude of the centipetal acceleation at the oute wall? When we ae told the cylinde is otating at 2.0 evolutions pe second, we ae being told the peiod. This is because the peiod is the time it takes to make one evolution. It is the ecipocal of the numbe we ae given. The numbe of cycles pe second is the fequency of the motion. So, in symbols we have: T = 1/f = 1/(2.0) = 0.50 Now that we have the peiod we can use the altenate fomula fo centipetal acceleation, a = 4π 2 /T 2 : a = 4π2 (0.12) (0.5) 2 = 18.950 In fact, we can combine ou fomulas in this poblem and get a second altenative fomula fo centipetal acceleation: a = 4π 2 f 2 But it is so simple to get the peiod diectly, that this fomula is not quite as useful. 2. A jet flying at 123 m/s banks to make a hoizontal cicula tun. The adius of 2.11 10 6 metes the tun is 3810 metes, and the mass of the jet is 2.00 10 5 kilogams. Calculate the magnitude of the necessay lifting foce. Notice that we ae not told the angle involved. So, let s tead a bit caefully hee. The foces involved ae weight and lift. The motion is hoizontal, so we should align ou coodinate system as such. The net foce in the hoizontal is nothing but the hoizontal component of the lift: and the net foce in the vetical is F net,x = L x F net,y = L y W Since the motion is in the hoizontal, thee is no net foce in the vetical. Thus, So, L y mg = 0 L y = (2.00 10 5 )(9.8) = 1.9600 10 6 But we don t know the angle involved. Well, what about the hoizontal motion? We do know the hoizontal net foce is esponsible fo the centipetal acceleation, F = mv 2 /. Thus, L x = mv 2 / = (2.00 10 5 )(123) 2 /(3810) = 7.9417 10 5 1
Now we know the two components of the lifting foce: L x = 7.9417 10 5 L y = 1.9600 10 6 The magnitude of this vecto is given by mag = L 2 x + L 2 y, so: L = (7.9417 10 5 ) 2 + (1.9600 10 6 ) 2 = 2.1148 10 6 3. Mas has a mass of 6.46 10 23 kilogams and a adius of 3.39 10 6 metes. (a) 3.75 m/s 2 ; (b) 244 newtons (a) What is the acceleation due to gavity on Mas? (b) How much would a 65-kilogam peson weigh on this planet? (a) This is an application of the Newton s law of gavity, F gav = GMm/ 2, and Newton s second law of motion, F net = ma. When calculating the weight, we have F net = F gav which allows us to combine these two. ma = GMm/ 2 Cancelling m gives us the geneal fomula fo the acceleation due to gavity. a = GM/ 2 Notice that when we plug in the mass and the adius of the eath, we get the numbe we expect: Fo Mas, we get: O, about 38% of that on eath. a = (6.672 10 11 5.974 1024 ) (6.378 10 6 ) 2 = 9.8016 a = (6.672 10 11 6.46 1023 ) (3.39 10 6 ) 2 = 3.7517 (b) The weight on Mas is still the same fomula, but diffeent numbes: W = mg = (65)(3.7515) = 243.86 4. A ca tavels at a constant speed aound a cicula tack whose adius is 0.79 m/s 2 2600 metes. The ca goes once aound the tack in 360 seconds. What is the magnitude of the centipetal acceleation of the ca? The centipetal acceleation equied to maintain unifom cicula motion is given by a = v 2 /. The v in this equation is the speed aound the cicle. We ae told the peiod is 360 seconds, but not the speed. The missing piece is the distance. But the distance is simply the cicumfeence of the cicle which is given by C = 2π: C = 2π(2600) = 16336 This is the distance coveed in one cycle. Divide by the time to get the aveage speed: v = d t = (16336) = 45.379 (360) Finally, we can plug this into the fomula fo centipetal acceleation: a = v2 = (45.379)2 = 0.79200 (2600) 2
A diffeent (but simila) way to do this is to use an altenate fomula fo centipetal acceleation: a = 4π 2 /T 2 This fomula is the combination of v = 2π/T and a = v 2 /, so using it is essentially the same as the pevious solution: a = (4π2 )(2600) (360) 2 = 0.79200 5. In a skating stunt known as cack-the-whip, a numbe of skates hold hands 606 newtons and fom a staight line. They ty to skate so that the line otates about the skate at one end, who acts as the pivot. The skate fathest out has a mass of 80.0 kilogams and is 6.10 metes fom the pivot. He is skating at a speed of 6.80 m/s. Detemine the magnitude of the centipetal foce that acts on him. The centipetal foce equied to maintain the motion is given by Newton s second law: F net = ma, but in this case the a is given by a = v 2 /. So, F = (80.0)(6.80)2 (6.10) = 606.43 In this case the centipetal foce is povided by the tension between the skates hands. 6. The moon obits the eath at a distance of 3.85 10 8 metes. Assume that this 27.5 days distance is between the centes of the eath and the moon and that the mass of the eath is 5.98 10 24 kilogams. Find the peiod fo the moon s motion aound the eath. Expess the answe in days and compae it to the length of a month. The centipetal foce is povided by gavity. The magnitude of the foce is given by F gav = GMm/ 2. Since we ae inteested in the peiod, we can combine this with the second fomula fo centipetal acceleation, F cent = 4π 2 m/t 2. Thus, Which simplifies to GMm 2 = 4π2 m T 2 GMT 2 = 4π 2 3 This is sometimes called the 1-2-3 law and is also known as Keple s thid law of planetay motion. Histoically, this fomula was one of the main clues Newton had in uncoveing his law of gavity. We can use it to solve ou poblem: (6.674 10 11 )(5.98 10 24 )(T ) 2 = 4π 2 (3.85 10 8 ) 3 = T = 2.3759 10 6 This is in seconds. Thee ae 86,400 seconds in one day, so the peiod in days is T = 2.3759 10 6 s Which is just shot of one full month. 1 dy 86400 s = 27.498 7. The National Aeonautics and Space Administation (NASA) studies the 33 m/s physiological effects of lage acceleations on astonauts. Some of these studies use a machine known as a centifuge. This machine consists of a long am, to one end of which is attached a chambe in which the astonaut sits. The othe end of the am is connected to an axis about which the am and chambe can be otated. The astonaut moves on a cicula path, much like a model aiplane flying in a cicle on a guideline. The chambe is located 15 metes fom the cente of the 3
cicle. At what speed must the chambe move so that an astonaut is subjected to 7.5 times the acceleation due to gavity? This will involve the fomula a = v 2 /. We know is equal to 15 metes and the acceleation is 7.5 times g. Thus, (7.5)(9.8) = (v)2 15 = v = 33.204 8. The hamme thow is a tack-and-field event in which a 7.3-kilogam ball (the 3500 newtons hamme ) is whiled aound in a cicle seveal times and eleased. It then moves upwad on the familia cuving path of pojectile motion and eventually etuns to eath some distance away. The wold ecod fo this distance is 86.75 metes, achieved in 1986 by Yuiy Sedykh. Ignoe ai esistance and the fact that the ball is eleased above the gound athe than at gound level. Futhemoe, assume that the ball is whiled on a cicle that has a adius of 1.8 metes and that its velocity at the instant of elease is diected 41 above the hoizontal. Find the magnitude of the centipetal foce acting on the ball just pio to the moment of elease. Okay. Let s stat with the end: the centipetal foce. Its fomula is F = mv 2 /. Now we know m and. So we need to find v. What we have emaining is the ange of the hamme. We can extact this fom the ange fomula, R = (v 2 /g) sin 2θ. Notice that we know R and θ, so we can solve fo the initial velocity. Thus, (86.75) = (v)2 (9.80) sin 2(41 ) = v = 29.300 We can take this velocity and calculate the centipetal foce: F = (7.3)(29.300)2 1.8 = 3481.7 9. A ca moving at 5.0 m/s ties to ound a cone in a cicula ac of 8.0 metes 0.32 adius. The oadway is flat. How lage must the coefficient of ficton be between the wheels and oadway if the ca is not to skid? Thee ae thee foces in this poblem: weight, suppot (fom the oadway) and fiction. The fee-body diagam is in Figue 1. Side View Top View N F = µn F cent W = mg Figue 1: Fee body diagam Since thee is no motion up and down, the suppot foce is equal to the weight, mg. The fomula fo the fiction involves this suppot foce, so we have F = µn = µmg 4
On the othe hand, this fiction povides the centipetal foce F cent. The fomula fo the centipetal foce is F cent = mv 2 / So, We have µmg = mv 2 / = µ = v 2 /g µ = (5.0)2 (8.0)(9.8) = 0.31888 10. A satellite obits the Eath at a height of 200 kilometes. Find (a) the speed (a) 7.8 km/s; (b) 88 minutes of the satellite and (b) the time taken (in minutes) to complete one evolution. (a) The adius of the satellite obit is the 200 kilometes plus the adius of the eath 6371 kilometes. Afte conveting to metes, we have: = 6.571 10 6 Since the foce of gavity F = GMm/ 2 povides the centipetal foce F cent = mv 2 /, we can cancel the mass m to yield: GMm 2 = mv2 Fo ou poblem we have = v = GM v = (6.673 10 11 )(5.974 10 24 ) 6.571 10 6 = 7789 (b) Now that we have the velocity we can use the fomula v = 2π/T 7798 = (2π)(6.571 106 ) T = T = 5301 But this answe is in seconds. Divide by 60 to convet to minutes: T = 88.34 min We could also use Keple s 3d law GMT 2 = 4π 3 to answe pat (b) without pat (a). 5