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Physics 211 Lecture 15 Today’s Concepts: a) Parallel Axis Theorem b) Torque & Angular Acceleration Mechanics Lecture 15, Slide 1 Parallel Axis Theorem Smallest when D = 0 Mechanics Lecture 15, Slide 2 ACT 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 3 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 Hooray! 3M M L Demo: rod w/spheres 0 L/4 L/2 Mechanics Lecture 15, Slide 4 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 Xcm = L/4, and that is at axis B Iaxis = Icm + MtotD2 Mtot = 4 M IB = Icm IA = Icm + MtotD2 = cm + (4 M)(L/4) 2 IC = Icm + MtotD2 = cm + (4 M)(L/4) 2 3M M L 0 L/4 L/2 Mechanics Lecture 15, Slide 5 Your Comments Yes, D is the distance from CM to the new axis of rotation. Good, the moment of inertia has the smallest value for rotation about the center of mass! Any other axis will give a larger moment of inertia. When we say “moment of inertia” we’re not finished. We always have to say — or think — “moment of inertia about some particular axis of rotation”. Be careful. Moment of inertia is about an axis, not with respect to a point. The CM is on axis B. Great! Right Hand Rule for finding Directions Demo: bike wheel I didn't really get how to figure out the direction of angular acceleration by the right hand rule. Why is it perpendicular to the direction of rotation? So confusing!!!!!! Please help me!!!!!!!!!!! Mechanics Lecture 15, Slide 7 ACT 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? Demo: roll ball up board A) Into the page B) Out of the page C) Up D) Down Mechanics Lecture 15, Slide 8 ACT 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 ACT 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 t = rF sin(q ) Demo: wrench & bolts Mechanics Lecture 15, Slide 11 Torque and F=ma I am still not comfortable with the fact that the torque equation is somehow the same as the net force equation. Can we go over how that is possible in lecture? r r F = ma Fq = maq Fq = mr rFq r= mr r r F = mr 2 r t = I v F q r 2 Mechanics Lecture 15, Slide 12 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 bigger? 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 13 CheckPoint In which case is the torque due to the force about the rotation axis bigger? axis A) Case 1 B) Case 2 C) Same 90o L/2 F A) Perpendicular force means more torque. B) F*L = torque. L is bigger in Case 2 and the force is the same. C) Fsin30 is F/2 and its radius is L so it is FL/2 which is the same as the other one as it is FL/2. Case 1 axis L 30o F Case 2 Mechanics Lecture 15, Slide 14 CheckPoint In which case is the torque due to the force about the rotation axis bigger? axis A) Case 1 B) Case 2 C) Same Be careful; both expressions have a factor of 0.5 in them but for different reasons. C) Fsin30 is F/2 and its radius is L so it is FL/2 which is the same as the other one as it is FL/2. 90o L/2 F Case 1 axis L 30o F Case 2 Mechanics Lecture 15, Slide 15 Your Comments Great! Great! And it is — for different reasons! And it is — for different reasons! Great! Great! I got carried away with “Great!” but I’m happy. It seems like “torque” may be well understood. Similarity to 1D motion Mechanics Lecture 15, Slide 17 Mechanics Lecture 15, Slide 18 ACT 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 bigger 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? A) F2 = F1 F2 F1 B) F2 = 2F1 M, R C) F2 = 4F1 Case 1 M, 2R Case 2 A) If the angular acceleration is the same, they must have the same tangential force. B) Twice the radius=twice the force. C) twice the radius means 4 times the moment of inertia, thus 4 times the torque, thus 4 times the force Mechanics Lecture 15, Slide 21 CheckPoint How are the magnitudes of the two forces related if the angular acceleration of the two hoops is the same? A) F2 = F1 F2 F1 B) F2 = 2F1 C) F2 = 4F1 M, R Case 1 M, 2R Case 2 I2 = 4 I1 because of the radius. Torque1 = R F1 = I1 alpha Torque2 = 2R F2 = I2 alpha F2 = I2 alpha/2R F1 = I1 alpha/R F2/F1 = (I2/2)/(I1) = (1/2)(I2/I1) = (1/2)(4) F2/F1 = 2 or F2 = 2 Fa Mechanics Lecture 15, Slide 22 Wait! We can’t have F = R; they have different units! But why? ??? What is “it”? Is the “bigger” force 2x or 4x as big? Tell me more. Why angular velocity instead of angular accelertion. I don’t understand the significance of “torque * I “. Tell me more. Tell me more. Mechanics Lecture 15, Slide 23 t = R F sin q q q = 90 0 q = 00 q = 90 36 = 54 Mechanics Lecture 15, Slide 24 t = R F sin 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 25 (i) 1 I DISK = MR2 2 (ii) t = I 1 2 (iii) K = Iw 2 Use (i) & (ii) Use (iii) Mechanics Lecture 15, Slide 26