General Physics I Honors
General Physics I Honors PHSX 213
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This 16 page Class Notes was uploaded by Dr. Baby Lebsack on Monday September 7, 2015. The Class Notes belongs to PHSX 213 at Kansas taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/182462/phsx-213-kansas in Physics 2 at Kansas.
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Date Created: 09/07/15
PHSX213 class Class Stuff Upcoming midterm is Wed Oct 17th on Ch 611 0 Same room as last time Malott 1001 8 PM 0 Practice test distributed last Wed Answer key online Rotation Mon Oct 15th 1 Exam Big Picture Energy Conservation Momentum Conservation Angular Momentum Conservation And understanding the necessary conditions for these to apply MonOc L 15th Torque DefinitMn fm a Pamigle y 1 V jpl IFQCQ 739 pmpcamd w laur it J x i h we p ifcmnmml L39r mp om wk Mm mm ml ly ma 2 TLovmu m Wm albamu mm Clearer Drawing with the Spool Example Spool may r011 translation of COM rotation about COM or rotation about the point of contact F I 139 r c is Ve into 1 is Ve out of T r x F 0 page page counter Mon Oct 15m clockw15e so clockWISe so rolls to right rolls to left Angular M0memtmm Definitim fm 3 Particle lrXp gw 23 lt gtAlt 7 xi U H 5 PQ f gt P m lglt m y mak m t fg k N Mm Om Imam mem f Anglar Mmemtmm Definitim Wm fm is JP Writ11 Angular Mamemum Cbnservati m Rotation axis Ir 0588 Mm HS I m Cluteh design 53m Angular Momentum Demo 75 76 Mon Oct lSth CheckPoint I Two uniform solid cylinders have different masses and different rotational inertias They simultaneously start from rest at the top of an inclined plane and roll without sliding down the plane The cylinder that gets to the bottom rst is A the one with the larger mass B the one with the smaller mass C the one with the larger rotational inertia D the one with the smaller rotational inertia E neither they arrive together MonOct 15th 10 CheckPoint I Two uniform cylinders have di erent masses and different rotational inertias They simultaneously start from rest at the top of an inclined plane and roll without sliding down the plane The cylinder that gets to the bottom rst is A the one with the larger mass B the one with the smaller mass C the one with the larger rotational inertia D the one with the smaller rotational inertia E neither they arrive together MonOctlSth Mon Oct lSth Looptheloop Forces in Rolling L C i CUE 1 x 3 Q Q 2 1 quotlt 1 09 BE if 9 k7 1 9 g E W ix 1 Q 1 x 3 35 E Co 69 i 69 La 9 Q E Q 7 2 g Aid A 19 no l l O 1 o 5 392 ex 9 O H W 6 E6 7 L J 9 gt 4 LOY O 7Q O 03 FF 93917 19 L0 Q3 Lg 5 1 3w EFL 5 ND EL Note I 11 de ne x down the incline Mm TL llSiiIm 113 Rolling 0 One can derive that fS 1 aR2 B13 mg sine 39 And that a g sine1 IMR2 g sine1H3 Where we have de ned 3 Via I B M R2 Bhoop 1 chlinder 03959 Bsolid sphere 0394 MonOcL 15th 14 Discuss rolling in terms of KE From energy conservation considerations Note the frictional force in this case doesn t oppose the angular motion and the work done by this nonconservative force is transformed into rotational kinetic energy W I T d9 Andwe ndthatK12mv2121032 12m13v2 mgh Since the contact point doesn t move there is 15 M l ih mm 5 no translational dzsplacement Another way to look at this MonOct lSth Stationary observer sees rotation about an axis at point P with co ZVCoMZR vCoMR Using the parallel axis theorem IP ICoM mR2 So K 1A II 02 l2IC0M 02 12mv2 just as before
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