HW 7 Help. 60 s t. (4.0 rev/s)(1 min) 240 rev 1 min Solving for the distance traveled, we ll need to convert to radians:

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1 HW 7 Help 30. ORGANIZE AND PLAN We ae given the angula velocity and the time, and we ae asked to ind the distance that is coveed. We can ist solve o the angula displacement using Equation 8.3: t. The distance taveled is just the ac length om Equation 8.: s. Known: 4.0 ev/s, t min, and 0.79 m. SOLVE Fist, ind the angula displacement: 60 s t (4.0 ev/s)( min) 40 ev min Solving o the distance taveled, we ll need to convet to adians: s (0.79 m)(40 ev) 00 m ev REFLECT This implies the wheel is spinning ast enough to go ove a kilomete in one minute, o moe pecisely 7 kilometes pe hou (43 mph). That would be awully ast o a bicycle, but since the wheel is simply spinning o o the gound, it seems easonable that it could tun that ast.

2 38. ORGANIZE AND PLAN We e given the initial angula velocity and the inal angula velocity (zeo in this case, since she stops spinning), and how long it takes. Theeoe, Equation 8.8 can be used to solve o the angula acceleation. Fo the second pat, we want to ind, 0 so eithe Equation 8.9 o 8.0 should wok. Known: 0.50 ev/s, 0 ev/s, t.75 s. SOLVE (a) Fist, conveting the initial angula velocity to ad/s: Plugging this into Equation 8.8: π ad 0.50 ev/s 5.7 ad/s ev 5.7 ad/s 8.98 ad/s t.75 s 0 (b) We will solve o 0 ist with Equation 8.9: t t (5.7 ad/s)(.75 s) ( 8.98 ad/s )(.75 s) 3.7 ad 0 As an execise, let s use Equation 8.0: 0 (5.7 ad/s) = 3.7 ad ( 8.98 ad/s ) So they agee as they should, but we want the answe in evolutions: ev 3.7 ad.8 ev π ad REFLECT The acceleation is negative because the skate slows down. I the skate had kept spinning at the same ate o the allotted time, then she would have completed (.5 ev/s)(.75 s) = 4.37 ev. Ou answe should be less than that (which it is), since the skate slows to a stop. In act, she completes hal as many tuns in the stopping case as in the constant spinning case.

3 40. ORGANIZE AND PLAN We have the initial and inal angula velocity and the time, so Equation 8.8 will be needed to ind the angula acceleation. Fo the second pat, we can use eithe Equation 8.9 o 8.0 to ind. 0 Known: ev/s, 5.50 ev/s, t.30 s. SOLVE (a) Let s ist convet the angula velocities into ad/s: Plugging these values into Equation 8.8: π ad ev/s.4 ad/s ev π ad 5.50 ev/s 34.6 ad/s ev 34.6 ad/s.4 ad/s 0. ad/s t.30 s 0 (b) We will use Equation 8.0 (but 8.9 could be used as well): 0 (34.6 ad/s) (.4 ad/s) = 36. ad (0. ad/s ) REFLECT The acceleation is positive, since the pulley inceases its speed when the gas pedal is pushed. 44. ORGANIZE AND PLAN We ae given the peiod (the time it takes to make a evolution), so we need to use Equation 8.5 to get the angula velocity. Once we have that, we can ind the tangential speed using Equation 8.: vt. Known: T 3.0 s, d.75 m. SOLVE The angula velocity o the wagon wheel is: π π.96 ad/s T 3.0 s Plugging this into Equation 8. gives: vt (.75 m)(.96 ad/s).7 m/s REFLECT The speed at the im o the wheel is also the speed that the wagon will move at (assuming the wheel is not slipping). Conveting the answe to a moe amilia quantity: v 3.43 m/s 3.87 mph. That cetainly seems a easonable speed o a wagon. t

4 47. ORGANIZE AND PLAN As we see in the igue below, the weight s acceleation induces a tangential acceleation on the pulley though the tension in the sting (i.e., ay at ). We can convet this into an angula acceleation with Equation 8.: at /. Fo pat (b), the alling weight achieves a velocity, v y, by the time it eaches the gound. This linea velocity is equal to the tangential velocity o the pulley: vy vt. Thee ae dieent ways to go about this, but we will solve ist o the weight s velocity using Equation.0: and then use this to ind the angula velocity om Equation 8.: /. v v 0 a y, y y y v t a t = 3.40 m/s a y = 3.40 m/s v t = v y v w = t Dy Known: positive y-diection cm, a 3.40 m/s, v0 0, y.30 m, y y v y whee we have deined down as the SOLVE (a) The angula acceleation comes diectly om the weight's acceleation: a y 3.40 m/s 00 cm 97. ad/s 3.50 cm m (b) The ist thing is to solve o the inal velocity o the weight: v v 0 a y (3.40 m/s )(.30 m).97 m/s y y y The pulley tuns at the same speed, so the angula velocity is: v y.97 m/s 84.9 ad/s m REFLECT You may wonde why the weight doesn t all with the gavitational acceleation o 9.8 m/s. The sting povides a tension that slows the weight's all. Notice, too, that we chose the down diection to be positive, but we can choose it to be negative, in which case all the answes would have a negative sign.

5 57. ORGANIZE AND PLAN We e only given the saw s initial otational kinetic enegy (Equation 8.5): K0 I 0. The otation ate dops in hal ( 0), while the otational inetia, I, does not change. Known: K0 44 J, 0. SOLVE The otational kinetic enegy ate the otation ate dop: K I I( ) ( I ) K (44 J) J This says that the otational kinetic enegy dops to one-outh its initial value. Relect Because the otation kinetic enegy is popotional to the angula velocity squaed, any incease o decease to the angula velocity will be squaed in the otational kinetic enegy. 68. ORGANIZE AND PLAN The tanslational kinetic and otational enegy ae the same as in the peceding poblem. The only thing that changes is the otational inetia: is: I MR (om Table 8.4). 5 SOLVE As beoe, let s calculate the otational kinetic enegy ist: K ( MR )( v / R) Mv ot 5 cm 5 cm The actions o kinetic enegy in tanslational and otational motion ae: K K K tans tans Kot 5 Kot 5 tans Kot 5 Whee the tem Mv cm was actoed out o the equations. REFLECT As was aleady discussed in Example 8., the solid sphee has slightly less otational inetia than the solid cylinde, which implies that a smalle action o its enegy goes into otation. This leaves moe o tanslation, so a sphee will beat a cylinde in olling down an incline

6 74. ORGANIZE AND PLAN The toque is given by Equation 8.6: F sin. We e told the adius, the oce applied and the angle. In the ist case, the angle is 90º (tangential means in the same diection as the motion), wheeas in the second case it s 45º. Known:.05 m, F 3.0 N, 90, 45. SOLVE Plugging the values in o the two cases: (a) (b) (.05 m)(3.0 N)sin90 4. Nm (.05 m)(3.0 N)sin N m REFLECT It s clea that pushing in the tangential diection is the most eective way to geneate toque. Any oce exeted in the diection pependicula to the motion will be wasted. 8. ORGANIZE AND PLAN We need to edo what we did in the peceding poblem, but this time include the mete stick s weight, which is a downwad oce at the midpoint (50-cm mak). Known: x 35 cm, m 0.0 kg, x 75 cm, m 0.40 kg, xs 50 cm, Ms 0.5 kg. SOLVE Deining the positive diections as beoe, the net oce and net toque ae: Fnet n mgmgmsg 0 mgmgn 0 net s s We have chosen the axis to be the 35-cm mak, so the toque o the ist mass is zeo, and the adii ae: 40 cm, s 5 cm, and x 35 cm. Solving o n in the ist equation and substituting it into the second equation gives: mgmg (40 cm)(0.40 kg) (5 cm)(0.5 kg) s s mg mg Mg s (0.0 kg) (0.40 kg) (0.5 kg) This means the ulcum needs to be put at: x 35 cm 59 cm. 4 cm REFLECT By including the mass o the mete stick in this poblem, we incease the amount o mass to the let o the ulcum (as it was placed in the pevious poblem). It makes sense that the ulcum will have to be moved to the let to compensate (i.e., om the 6-cm mak to the 59-cm mak).

7 94. ORGANIZE AND PLAN Imagine that the electon is cicling its poton in a counteclockwise diection as seen om above. The velocity we e given is the tangential velocity, so we ll need to use Equation 8.: vt / to ind the angula velocity. The electon can be teated as a point paticle moving in a cicle, so the otational inetia is: I m. The mass o the electon can be ound in the Appendix. 6 3 Known: vt.8 0 m/s, m, m 9. 0 kg. SOLVE The magnitude o the angula momentum is: L I ( m )( vt / ) mvt (9.0 kg)(.80 m/s)(5.90 m).050 J s The diection will be up i the electon is going in a counte-clockwise diection. REFLECT The angula momentum o the electon is an impotant quantity in quantum mechanics. It comes as speciic multiples o the so-called educed Planck constant: J s. The electon in this example has the lowest angula momentum: L.

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

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),