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Physics 211 Lecture 16

Physics 211 Lecture 16. Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration. Your Thoughts.

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Physics 211 Lecture 16

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  1. Physics 211Lecture 16 Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration

  2. Your Thoughts translational motion + rotational motion = DANG, BRO.The race between the hoop, the big cylinder, and the little cylinder confuse me to no end!wow. That was a LOT of information... Are you guys trying to fry us before break or something?I honestly don't know how I feel about these concepts until I start doing the homework because that's when I see what stuck and made sense and what didn't.I would like to see some examples similar to the homework that is coming up .I remember doing this stuff back in high school, my teacher took a can of pop that had liquid in it while the other was frozen solid, both having hte same mass, and he rolled them down a ramp.can we work on a practice problem or two in class...I learn best if I can see something being worked through. I can understand concepts but still be completely lost when it comes to actually doing a problem.My thoughts on this lecture are best summed up in this quote from Ozzy: "SHAAAAAROOOON."I feel like I can visualize this stuff well just from living on this earth for my whole life... Every monday night from 7-9 there is a free improv show at the Illini Union Courtyard Cafe... if you yell out physics lab as a suggestion for the group that goes up at 8, you would get to see a pretty awesome improvised MUSICAL! about physics.... i dare you to go see it march 13th is my birthday. and there's no better way to celebrate than to go to physics lecture! yay!PLEASE play Dirty Deeds Done Dirt Cheap!Stelzer > Selen. True story.

  3. 2nd Exam is the week after break: Covers lectures 7-13 (not the stuff we are doing now) Sign up for the conflict if you need to before the end of this week. As before, email Dr. Jain if you have a double conflict (rdjain1@illinois.edu). Tuesdays class right after break will be a review(we will work out Fall 2010 exam) . Don’t forget that there are lots of extra office hours during exam week.

  4. If there is no slipping: Rolling v = R (prelecture slide 5)

  5. Two ways of looking at this Energy Conservation H Dynamics a F = ma f a Mg

  6. Objects of differentIrolling down an inclined plane: Energy Conservation K= 0 U=Mgh v=0 =0 K=0 R M U=0 h v=R

  7. Lets work out the general case… c c c c Hoop: c= 1 Disk: c= 1/2 Sphere: c= 2/5 etc... Use v = R and I = cMR2. v So: Doesn’t depend on M or R, just on c (the shape) Ramp demo

  8. Clicker Question A hula-hoop rolls along the floor without slipping. What is the ratio of its rotational kinetic energy to its translational kinetic energy? A) B) C) Recall that I = MR2for a hoop about an axis through its CM:

  9. CheckPoint A block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. Which one makes it furthest up the ramp? A) Block B) Ball C) Both reach the same height. v w v

  10. CheckPoint The block slides without friction and the ball rolls without slipping. Which one makes it furthest up the ramp? A) Block B) Ball C) Same • A) The block would have more distance as all the energy is put into moving it translationally, not rotationally. • B) Ball has more initial kinetic energy so goes up higher. • C) they both have the same total kinetic energy therefore they will have the same height. v w v

  11. CheckPoint A cylinder and a hoop have the same mass and radius. They are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first? A) Cylinder B) Hoop C) Both reach the bottom at the same time

  12. Which one reaches the bottom first? A) Cylinder B) Hoop C) Both reach the bottom at the same time A) They have the same potential energy but the hoop has a larger rotational inertia so more energy converted into rotational kinetic energy B) The moment of inertia is higher on the hoop and it will have more energy going down the ramp. C)Same mass and radius and both rotate, no difference.

  13. CheckPoint A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first? A) Small cylinder B) Large cylinder C) Both reach the bottom at the same time

  14. A) Small cylinder B) Large cylinder C) Both reach the bottom at the same time A) the large cylinder has a large moment of inertia. B)greater moment of inertia. C) Both cylinders will have the same equation for rotational inertia, so mass will end up being superfluous, so the fact that one is larger doesn't matter.

  15. Acceleration

  16. Acceleration

  17. Acceleration Acceleration depends only on the shape, not on mass or radius.

  18. Example Problem Suppose a cylinder (radius R, mass M) is used as a pulley. Two masses (m1>m2) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. What is the acceleration of the blocks? M  R T2 T1 a m2 m1 a

  19. Clicker Question Suppose a cylinder (radius R, mass M) is used as a pulley. Two masses (m1>m2) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. Compare the tension in the string on either side of the pulley: A) T1> T2 B) T1= T2 C) T1< T2 M  R T2 T1 a2 m2 m1 a1

  20. Atwood's Machine with Massive Pulley: A pair of masses are hung over a massive disk-shaped pulley as shown. Find the acceleration of the blocks. For the pulley use =I m2g (Since for a disk) m1g M For the hanging masses useF =ma  m1g - T1=m1a R • T2 - m2g=m2a T2 T1 a m2 m1 T1R - T2R a

  21. We have three equations and three unknowns (T1, T2, a). Solve for a. m1g - T1=m1a (1) T2- m2g = m2a (2) T1-T2(3) Atwood's Machine with Massive Pulley: m2g m1g M  R T2 T1 a m2 m1 a

  22. a v M R f

  23. a v M R f

  24. w v M a R v Oncev=wRit rollswithout slipping v = v0 - at v0 v0 t v = v0 - at wR= aRt w = at v = wR w t t

  25. w v M a R Plug in a and t found in parts 2) & 3)

  26. w v M a R Interesting aside: how v is related tov0 : Doesn’t depend on m We can try this…

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