Your Thoughts Suppose a ball rolls down a ramp with a coefficient of friction just big enough to keep the ball from slipping. An identical ball rolls down an identical ramp with a coefficient of friction of 1. Do both balls reach the bottom at the same time? (y/n) Does the moment of inertia play a part in determining the amount of time it takes an object to get to the bottom of an incline. For the first checkpoint, I know that when a ball is slipping down, the KE is bigger than the block, since there is some rotational energy but if they are moving up the ramp, should the rotational energy be negative or positive? Mechanics Lecture 16, Slide 1 2nd Exam Monday July 15th Covers Lectures 7-13 only 90 Minutes. 1:30-3:00pm In Here: 141 Loomis Mechanics Lecture 9, Slide 2 Physics 211 Lecture 16 Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration Physics 211 Lecture 16, Slide 3 Rolling What is meant by without slipping? v = R 2R2 = v2 If there is no slipping: Mechanics Lecture 16, Slide 4 Two ways of looking at rolling down ramp PE = MgH Energy Conservation KE = 0 PE = 0 H KE = Ktrans Krot Dynamics a f a F = ma τ = Iα Mg Mechanics Lecture 16, Slide 5 Energy Conservation Objects of different I rolling down an inclined plane: for the first checkpoint, I know that when a K=0 U = Mgh R M v=0 =0 K=0 h ball is slipping down, the KE is bigger than the block, since there is some rotational energy but if they are moving up the ramp, should the rotational energy be negative or positive? 1 1 I 2 Mv 2 2 2 U=0 K= v = R K U = 0 1 2 1 2 MgH = I Mv 2 2 Mechanics Lecture 16, Slide 6 Lets work out the general case… Does the moment of inertia play a part in determining the amount of time it takes an object to get to the bottom of an incline? K= 1 2 1 2 I Mv 2 2 K= v Use v = R and I = cMR2 . 1 1 2 1 2 2 MR Mv = c 1Mv 2 c 2 2 2 Hoop: c = 1 Disk: c = 1/2 Sphere: c = 2/5 etc... So: 1 2 c 1Mv 2 = Mgh v= 2 gh 1 c 1 Doesn’t depend on M or R, just on c (the shape) Ramp demo Mechanics Lecture 16, Slide 7 Clicker Question A solid cylinder of mass M rolls along the floor without slipping with speed V. What is its total kinetic energy? 1 2 A) K = MV 2 Recall that I cm 3 B) K = MV 2 C) K = MV 2 4 1 = MR 2 for a solid cylinder: 2 Mechanics Lecture 16, Slide 8 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 v Mechanics Lecture 16, Slide 9 CheckPoint The block slides without friction and the ball rolls without slipping. Which one makes it furthest up the ramp? v A) Block B) Ball C) Same v A) The block does not have kinetic energy wasted by rolling. [It actually has less energy!] B) The ball has both translational KE (equal to that of the block) and rotational KE. Because it has more KE to start with, it must have more PE to finish with, and therefore it makes it further up the ramp. [Yes] C) since they have same mass and same initial velocity, so they should reach the same height of the ramp. [Energy is not just kinetic- mass & velocity] Mechanics Lecture 16, Slide 10 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 Mechanics Lecture 16, Slide 11 Which one reaches the bottom first? A) Cylinder B) Hoop C) Both reach the bottom at the same time A) At the bottom both objects have a combination of kinetic translational and rotational motion, which are equal because they started with the same potential energy. Since the hollow cylinder's moment of inertia is greater, its rotational kinetic energy is greater, and therefore the amount of energy that becomes translational is less, meaning it has a lower linear velocity than that of the solid cylinder. C) Since they both have the same mass, and start at the same height, they will have the same potential energy, and will have the same kinetic energy at the bottom. Mechanics Lecture 16, Slide 12 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 Mechanics Lecture 16, Slide 13 A) Small cylinder B) Large cylinder C) Both reach the bottom at the same time A) The small cylinder has less inertia than the larger cylinder. B) larger cylinder has more potential energy to begin C) The mass and radius of each cancel out when combining the moment of inertia and kinetic energy equations, which means velocity only depends on the moment of inertia. Mechanics Lecture 16, Slide 14 Acceleration Mechanics Lecture 16, Slide 15 Acceleration Mechanics Lecture 16, Slide 16 Acceleration Suppose a ball rolls down a ramp with a coefficient of friction just big enough to keep the ball from slipping. An identical ball rolls down an identical ramp with a coefficient of friction of 1. Do both balls reach the bottom at the same time? (y/n) Acceleration depends only on the shape, not on mass or radius. Mechanics Lecture 16, Slide 17 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 a R T2 T1 a m2 m1 a Mechanics Lecture 16, Slide 18 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: M a A) T1 > T2 B) T1 = T2 C) T1 < T2 R T2 T1 a2 m2 m1 a1 Mechanics Lecture 16, Slide 19 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 hanging masses use F = ma m1g - T1 = m1a M a R T2 - m2g = m2a T2 T1 m2 m1 a m1g a m2g Mechanics Lecture 16, Slide 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. a For the pulley use = Ia = I R a 1 T1R - T2R = I = MRa R 2 1 (Since I = MR 2 for a disk) 2 M a R T1 T2 a m2 m1 a m1g m2g Mechanics Lecture 16, Slide 21 Atwood's Machine with Massive Pulley: We have three equations and three unknowns (T1, T2, a). Solve for a. m1g - T1 = m1a (1) T2 - m2g = m2a (2) 1 T1 - T2 = Ma (3) 2 m1 - m2 g a = m1 m2 M 2 M a R T2 T1 m2 m1 a m1g a m2g Mechanics Lecture 16, Slide 22 a f v M R = Ia a= I 5 g fR MgR = = = I 2 2R 2 5 MR Mechanics Lecture 16, Slide 23 a f v M R F = Ma F Mg = g a= = M M Mechanics Lecture 16, Slide 24 a = g 5 g a= 2R v M R a v = R v v0 v = v0 - at t = at t Once v=R it rolls without slipping v0 v = v0 - at v = R R = a Rt t v0 - at = a tR 5 g v0 - gt = t 2 7 g v0 = t 2 t= 2 7 g v0 Mechanics Lecture 16, Slide 25 a M R v 1 2 x = v0t - at 2 v = v0 - at Plug in a and t found in parts 2) & 3) a = g t= 2 7 g v0 Mechanics Lecture 16, Slide 26 a M R a = g v Interesting aside: how v is related to v0 : v = v0 - at t= 2 7 g v0 2v0 v = v0 - ( g ) 7 g 2 v = v0 - v0 7 5 v = v0 7 Doesn’t depend on v = 0.714 v0 We can try this… Mechanics Lecture 16, Slide 27