lecturenotes-March15

lecturenotes-March15 - Question Answer Time for Chapters...

Info iconThis preview shows pages 1–12. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Question – Answer Time for Chapters 9.8 -11 Look at practice Exam #3
Background image of page 2
Quiz#4
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Quiz#5
Background image of page 4
Review: 1-D Collisions If system is closed and isolated, the total linear momentum P cannot change Inelastic collision : KE is not conserved (~ thermal energy ) P before = P after 1- D 2 particles m 1 v 1 i + m 2 v 2 i ( ) = m 1 v 1 f + m 2 v 2 f ( ) v 1 f = m 1 m 2 m 1 + m 2 v 1 i + 2 m 2 m 1 + m 2 v 2 i (Eqn. 9-75) v 2 f = 2 m 1 m 1 + m 2 v 1 i + m 2 m 1 m 1 + m 2 v 2 i (Eqn. 9-76) Elastic collision : TOTAL KE is conserved (~ Conservative forces ) P before = P after 1- D KE before = KE after 2 particles 1 2 m 1 v 1 i 2 + 1 2 m 2 v 2 i 2 ( ) = 1 2 m 1 v 1 f 2 + 1 2 m 2 v 2 f 2 ( )
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Relating Translational and Rotational Variables Rotational position and distance moved s = θ r (only radian units) Rotational and translational speed v = d r dt = ds dt = d dt r v = ω r Rotational and translational acceleration tangential acceleration a t = d dt r = α r a r = v 2 r = 2 r radial acceleration a tot 2 = a r 2 + a t 2 = 2 r 2 + r 2 v = × r a t = × r a r = × × r ( ) a tot = a t + a r
Background image of page 6
10.7--Some Rotational Inertias EACH OF THESE Rotational Inertias GO THROUGH THE Center of Mass ! rdv dr
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
10-7: Parallel-Axis Theorem Proof: I= r 2 dm = { x a ( ) 2 + y b ( ) 2 } dm I = x 2 + y 2 ( ) dm 2 a xdm 2 b ydm + a 2 + b 2 ( ) dm I = I COM + 0 + 0 + h 2 M I = I COM + Mh 2 If h is a perpendicular distance between a given axis and the axis through the center of mass (these two axes being parallel). Then the rotational inertia I about the given axis is
Background image of page 8
Rolling ( ch. 11 ) = Rotation ( ch. 10 ) + Translation ( ch. 1-9 ) 1 2 I com ω 2 + 1 2 Mv com 2 = KE tot Note: rotation about COM and translation of COM combine for total KE Wheel moving forward with constant speed v com s = θ R displacement: translation rotation v com = ds dt = d R ( ) dt = R a com = dv com dt = d R ( ) dt = α R Only if NO SLIDING [smooth rolling]
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Example: Rolling down a ramp with friction h θ L sin = h Object, with mass m and radius r , roles from top of incline plane to bottom. What is v , a , and Δ t at bottom KE rot + trans , COM ( ) final 0 ( ) init = 0 ( ) final mgh ( ) init [ ] 1 2 mv COM 2 + 1 2 I com ω 2 = mgL sin 1 2 mv COM 2 + 1 2 I COM v r 2 = mgL sin