Physics 1114: Unit 5 Hand-out Homework (Answers)

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1 Physics 1114: Unit 5 Hand-out Homewok (Answes) Poblem set 1 1. The flywheel on an expeimental bus is otating at 420 RPM (evolutions pe minute). To find (a) the angula velocity in ad/s (adians/second), we need to convet evolutions to adians (2π adians pe evolution) and minutes to seconds: 420 ev min ( ) 2πad ev ( min 60s ) = 44 ad/s (b) The linea (i.e., tangential) speed of a point 16 cm fom the cente of the flywheel is v t = ω = 7.04 m/s. 2. A helicopte has blades that ae 10 ft long extending fom the axle. The tip of the blade should not exceed 1100 ft/s (the speed of sound). (a) The maximum angula velocity (ω) is ω = v t = 1100 ft/s 10 ft = 110 ad/s (Note that ac length pe sec divided by adius (length) gives adians pe second, no need to convet to metes. The convesion won t hut, but it cancels out because you would have the same facto in the numeato and denominato.) (b) 110 ad/s = 1050 RPM (invet the convesion fom poblem 1). 3. (a) A ball on the end of a sting is whiled in a hoizontal cicle with 0.5 m adius at a ate of one evolution evey two seconds (0.5 ev/s). The ball s centipetal acceleation (a c ) is a c = v2, so we find v fom v t = ω, but ω has to be ad/s, not ev/s: ω = 0.5 ev/s (2πad/ev) = 3.14ad/s (o π ad/s). Then v t = ω = (0.5 m)(3.14 ad/s) = 1.57 m/s. So that a c = v2 Altenatively, using v t = ω, and plug in fo and ω. = (1.57 m/s)2 0.5 m a c = v2 = (v tω) 2 = 4.93 m/s 2 = ω 2 (b) If the ball has a mass of 0.65 kg, The hoizontal component of the tension in the sting comes fom Newton s 2nd Law in the centipetal (adial) diection: F c = ma c = mv 2 /. The only foce acting in the centipetal diection is the hoizontal component of the tension because the only othe foce acting is the weight (gavity, F g, staight down): F T F T,y F T,hoiz. F g

2 Physics 1114: Unit V Poblem Answes 2 So, F T,hoiz = ma c = (0.65 kg)(4.93 m/s 2 ) = 3.2 N. The vetical component of the tension has to balance the weight (a y = 0, so F T,y F g = 0). Then F T,y = mg = (0.65 kg)(9.8 m/s 2 ). Then combine the components of tension by the Pythagoean theoem and get F T = 7.13 N. (c) Now the 0.65 kg ball is whiled in a vetical cicle of 0.5 m adius. The maximum tension the sting can withstand is 40 N. The maximum allowed speed of the ball is found fom the maximum a c (= v 2 /) that the ball can get fom the foces. The only foces acting on the ball ae the tension and the weight. i) At the top, tension and weight both act downwad (the centipetal diection), so F T +F g = mv 2 max / (v max = 6.0 m/s). ii) At the side (i.e., when the sting is hoizontal), only the tension acts in the centipetal diection, so F T = mv 2 max/ (v max = 5.5 m/s). and iii) at the bottom of the cicle, F T again acts in the centipetal diection, but F g acts in the anti-centipetal diection, so F T F g = mv 2 max / (v max = 5.1 m/s). 4. A 1550 kg ca is taveling at 12 m/s on a level oad whee the coefficient of static fiction between the ties and the oad is (a) Fo a flat suface, it is the fictional foce between the ties and the oad that povides the net foce to cause the change in diection (i.e., the centipetal component of the net foce). Fc = ma c F f = mv 2 / µf N = mv 2 / µmg = mv 2 / Note that the mass divides out. Then solving fo : = v2 µg = 18.4 m (b) When it ains, the coefficient of fiction on the oad dops to The maximum speed with which the ca can safely negotiate a tun with the same adius (18.4 m) is found fom the fomula above: v 2 = µg, o v = µg = 4.25 m/s 5. A wheel stats fom est (ω i = 0 ad/s) and otates with constant angula acceleation (α). Afte t = 6.0 s have elapsed, it has otated though θ = 25 ad. (a) The angula acceleation comes fom θ = ω i ( t) α( t)2 So that α = 1.39 ad/s ad = α(6.0 s)2 2 (b) The angula velocity at t = 6.0 s can be found a couple ways. Since we found α, we could use ω f = ω i + α( t) = 8.33 ad/s. O, we could use θ = 1 2 (ω f + ω i ) t, solving fo ω f.

3 Physics 1114: Unit V Poblem Answes 3 Poblem set 2 1. A wheel on a moving ca slows unifomly (α is constant and is negative) fom 72 ad/s (ω i ) to 44 ad/s (ω f ) in 7.0 seconds ( t). (a) The angula acceleation is α = ω f ω i t = 4 ad/s 2 (b) The angle though which the wheel tuns in the 7.0 s inteval is θ = 1 2 (ω f + ω i ) t = 406 ad. O, using ou value fo α, we could use θ = ω i ( t) α( t)2. 2. A motocycle wheel tuning at 0.25 ad/s (ω i ) is bought to est (ω f = 0) by the bakes in exactly two evolutions ( θ = 2 ev = 4π ad). The angula acceleation of the wheel comes fom ωf 2 = ω2 i + 2α θ. Solve fo α and plug in the numbes to get α = ad/s The sun has a mass of kg, while the mass of Mas is only kg. The adius of Mas s obit is km ( m). The foce of attaction between these two bodies is F G = Gm 1m 2 2 1,2 F G = ( Nm 2 kg 2 ) ( kg)( kg) ( m) 2 F G = N 4. The toque poduced by a 70 lb foce (311.5 N) applied to the end of a 9 inch (0.229 m) wench and the angle between the foce and wench is 80 : τ = F = F sin θ F, = 70.2 N m 5. To find the tension in the cable, analyze the foces and toques. The foces include the tension (F T ), the weight of the load (F g,l ) and the weight of the beam (F g,b ). Thee ae also foces at the end of the beam whee it meets the wall, but we will put ou toque efeence point thee so that thei toques will be zeo (moment am of zeo length). Now we use τ = I α and the equilibium condition of α = 0. The two weights make toques (τ b and τ L ) acting in a F T 30 o 4m 8m F g,b F g,l clockwise diection, so we ll make them positive, wheeas the toque (τ T ) due to the tension acts in the counte-clockwise diection and will be negative: τ = I α = 0

4 Physics 1114: Unit V Poblem Answes 4 τ b + τ L τ T = 0 Now use τ = F sin θ F, fo each: τ T = τ b + τ L F T (8 m) sin 30 = (20 kg)(9.8 m/s 2 )(4 m) sin 90 + (40 kg)(9.8 m/s 2 )(8 m) sin 90 and solve fo F T = 980 N. Poblem set 3 1. As in poblem 2-5, fo the biceps muscle foce poblem we use τ = I α and the equilibium condition of α = 0: τ = I α = 0 So that F = 221 N. τ am + τ L τ F = 0 τ F = τ am + τ L F(0.05 m) sin72 = (9 N)(0.14 m) sin90 + (28 N)(0.33 m) sin90 2. The moment of inetia is I = 0.21 ML 2, whee L is the am length. To find the angula acceleation of an 8.0 kg am, 0.66 m long, when subject to a (net) toque of τ = 27 N m, we stat with the elationship between net toque, moment of inetia, and angula acceleation (Newton s 2nd Law in toque fom): τ net = Iα Then α = 37 ad/s 2 α = τ net I 27 N m α = 0.21(8.0 kg)(0.66 m) 2 3. A solid steel disk (I = 1 2 MR2 ) has a adius of 0.52 m and a mass of 290 kg. To get an acceleation of 1.20 ad/s 2, we find the net toque (τ net ) as above: τ net = τ net = Iα [ 1 2 (290 kg)(0.52 m)2 ] (1.20 ad/s 2 ) = 47 N m

5 Physics 1114: Unit V Poblem Answes 5 25 o 1.2 m h = (1.2m) sin 25 o 4. A ball of mass 0.78 kg and adius of 0.06 m (I = 2 5 MR2 ) stats fom est and olls without slipping fo a distance of 1.2 m down a 25 incline. We can detemine the speed at that point by using the wok-enegy theoem: W net = K tot,f K tot,i, whee K tot is the sum of the linea and otational kinetic enegies. The ball stats fom est, so K tot,i = 0: W net = K tot,f 0 The only wok is done by gavity (W g = mgh). The nomal foce is always pependicula to the motion so it does zeo wok. mgh = K ot,f + K lin,f mgh = 1 2 Iω2 f mv2 f mgh = 1 ( ) m2 ωf mv2 f Since the ball is olling without slipping, ω = v/, so the adius will cancel out afte squaing. We can also divide out the mass: gh = 1 5 v2 f v2 f = 7 10 v2 f Using g = 9.8 m s 2 and h = (1.2 m) sin 25 = m: 10gh 10(9.8 m/s2 )(0.507 m) v f = = = 2.66 m/s 7 7 Note that the final linea speed is less than fo an object sliding without fiction because pat of the enegy goes into otation. If we wite the moment of inetia as I = xmr 2, whee x changes fo diffeent solid objects (sphees, disks, hoops), we see that v f deceases as x inceases. So the sphee s speed will be faste than a disk s speed fo the same situation (I disk = 1 2 MR2 ) because 1/2 > 2/5. And a hoop will be even slowe (I hoop = MR 2 ). 5. This is a consevation of (angula) momentum poblem. The exta tension to pull the ball inwad does not exet a extenal toque, so τ net = 0 and L f = L i. Using L = Iω, I f ω f = I i ω i Whee I i = m 2 i, I f = m 2 f, and ω i = v i / i = (2.2 m/s)/(1.2 m) = 1.83 ad/s. Also, i = 1.2 m and f = 0.7 m. m 2 fω f = m 2 i (1.83 ad/s) The mass cancels out, and we solve fo ω f = 5.38ad/s. So that v f = ω f f = 3.76 m/s. (The exta tension does do wok, howeve, so K is not conseved!)

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet Linea and angula analogs Linea Rotation x position x displacement v velocity a T tangential acceleation Vectos in otational motion Use the ight hand ule to detemine diection of the vecto! Don t foget centipetal