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Physics 211 Lecture 15 Today’s Concepts: a) Parallel Axis Theorem b) Torque & Angular Acceleration Mechanics Lecture 15, Slide 1 Your Comments every thing was new to me, but not difficult to understand How many times can Superman run into a wall before he gets a headache? As long as I went through each step, I arrived at a sensible answer. This is all easy to do because you just have to walk through the steps. The one thing I am confused about is direction. So Torque will point along the axis in either direction. I am just confused on how I can see the apply in real life. Like does a force pointing along the axis of rotation do anything? I bet you're going to say this can all be figured out with F = ma huh? My wish for "To Make a Wish Foundation" was to get this put onto today's lecture... and it came true... I discovered that approaching the prelectures with the same zeal as the homework makes me understand things a lot better. Now, onto serious business. You don't have Maiden, which is outrageous, but what about Rush? There is no excuse for not having Rush. Actually, it is a sin. I am expecting Subdivisions on Thursday. Can i get a SHOUT OUT?! so i relaized something tonight...F=MA is that awkward friend that changes his personality depending on who he wants to impress. His been in 1/2mv^2 and now his in TORQUE?!!! liking this, seems cpmlicated but once you think about it its okay. Gotta love the prelectures and garrryyy! cheers, me Mechanics Lecture 15, Slide 2 Proven Fact: If you don’t prepare for lecture you don’t learn as much… You're not supposed to understand everything in the prelecture – that’s the whole point of having a lecture. Mechanics Lecture 15, Slide 3 Parallel Axis Theorem Smallest when D = 0 Mechanics Lecture 15, Slide 4 Clicker Question A solid ball of mass M and radius is connected to a thin rod of mass m and length L as shown. What is the moment of inertia of this system about an axis perpendicular to the other end of the rod? 2 1 2 2 2 A) I = MR ML mL 5 3 2 1 2 2 B) I = MR mL ML2 5 3 2 1 2 2 C) I = MR mL 5 3 1 D) I = ML2 mL2 3 R M m L axis Mechanics Lecture 15, Slide 5 CheckPoint A ball of mass 3M at x = 0 is connected to a ball of mass M at x = L by a massless rod. Consider the three rotation axes A, B, and C as shown, all parallel to the y axis. For which rotation axis is the moment of inertia of the object smallest? (It may help you to figure out where the center of mass of the object is.) A C B y x 3M M L Rotation is smallest around the objects center of mass, and the center of mass for this system is at L/4. 0 L/4 L/2 Mechanics Lecture 15, Slide 6 Right Hand Rule for finding Directions I learned that the right hand rule was for B fields about wires and B fields with E fields crossing them. Now you tell me there is another right hand rule. Seriously physics, be more creative. I think that with all the smart people who study you could at least give your rules more diverse names. Mechanics Lecture 15, Slide 7 Clicker Question A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular velocity vector point when the ball is rolling up the ramp? A) Into the page B) Out of the page C) Up D) Down Mechanics Lecture 15, Slide 8 Clicker Question A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is rolling up the ramp? A) Into the page B) Out of the page Mechanics Lecture 15, Slide 9 Clicker Question A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is rolling back down the ramp? A) into the page B) out of the page Mechanics Lecture 15, Slide 10 Torque I quite don't understand the concept of torque. So generally, if Torque is greater, what will happen other than the door will closing faster? t = rF sin(q ) Mechanics Lecture 15, Slide 11 Torque and F=ma Why in the new equation for torque analogous to Newtons's Second Law deals with angular acceleration only? Are these concepts separate? r r F = ma Fq = maq Fq = mr rFq r= mr r r r F = mr 2 r r t = I v F q r 2 Mechanics Lecture 15, Slide 12 Look at Prelecture slide 9 again It’s all there…. Mechanics Lecture 15, Slide 13 I know torque = rxF but what exactly is r? r F q r r r r r F = rF sinq q r F r F q q q r r r r =r(Fsinq) =F(rsinq) Mechanics Lecture 15, Slide 14 CheckPoint In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pushing at an angle of 30 degrees a distance L from the axis. In which case is the torque due to the force about the rotation axis biggest? A) Case 1 B) Case 2 C) Same axis axis L/2 90o Case 1 F L 30o F Case 2 Mechanics Lecture 15, Slide 15 CheckPoint In which case is the torque due to the force about the rotation axis biggest? axis A) Case 1 B) Case 2 C) Same 90o L/2 F A) Pushing perpendicular to the door is always best.. B) the length of the rotation arm is longer in case 2.. C) FL/2=FLsin30 Case 1 axis L 30o F Case 2 Mechanics Lecture 15, Slide 16 Similarity to 1D motion Mechanics Lecture 15, Slide 17 Mechanics Lecture 15, Slide 18 Clicker Question Strings are wrapped around the circumference of two solid disks and pulled with identical forces. Disk 1 has a bigger radius, but both have the same moment of inertia. Which disk has the biggest angular acceleration? A) Disk 1 w2 w1 B) Disk 2 C) same F F Mechanics Lecture 15, Slide 19 CheckPoint Two hoops can rotate freely about fixed axles through their centers. The hoops have the same mass, but one has twice the radius of the other. Forces F1 and F2 are applied as shown. How are the magnitudes of the two forces related if the angular acceleration of the two hoops is the same? A) F2 = F1 B) F2 = 2F1 F2 F1 C) F2 = 4F1 Case 1 Case 2 Mechanics Lecture 15, Slide 20 CheckPoint How are the magnitudes of the two forces related if the angular acceleration of the two hoops is the same? F2 A) F2 = F1 F1 B) F2 = 2F1 M, R C) F2 = 4F1 Case 1 M, 2R Case 2 t = I RF = MR F = MR 2 Mechanics Lecture 15, Slide 21 t = R F s in q q q = 90 0 q = 00 q = 90 36 = 54 Mechanics Lecture 15, Slide 22 t = R F s in q Direction is perpendicular to both R and F, given by the right hand rule tx = 0 ty = 0 t z = t F1 t F2 t F3 Mechanics Lecture 15, Slide 23 (i) 1 I DISK = MR2 2 (ii) t = I 1 2 (iii) K = Iw 2 Use (i) & (ii) Use (iii) Mechanics Lecture 15, Slide 24