PHYSICS 149: Lecture 21 Chapter 8: Torque and Angular Momentum 8.2 Torque 8.4 Equilibrium Revisited 8.8 Angular Momentum Lecture 21 Purdue University, Physics 149 1
Midterm Exam 2 Wednesday, April 6, 6:30 PM 7:30 PM Place: PHY 333 Chapters 5-8 The exam is closed book. The exam is a multiple-choice test. There will be ~15 multiple-choice problems. Each problem is worth 10 points. Note that total possible score for the course is 1,000 points (see the course syllabus) The difficulty level is about the same as the level of textbook problems. You may make a single crib sheet you may write on both sides of an 8.5 11.0 sheet Lecture 20 Purdue University, Physics 149 2
Rotational Kinetic Energy Consider a mass M on the end of a string being spun around in a circle with radius r and angular frequency ω Mass has speed v = ω r Mass has kinetic energy K = ½ M v 2 = ½ M ω 2 r 2 Consider a disk with radius R and mass M, spinning with angular frequency ω Each piece of disk has speed v i = ωr i Each piece has kinetic energy K i = ½ m i v 2 = ½ m i ω 2 r 2 i Combine all the pieces ΣK i = Σ ½ m i ω 2 r 2 = ½ (Σ m i r i2 ) ω 2 = ½ I ω 2 Rotational Inertia r i I = Σ m i r 2 i (units kg m 2 ) M Lecture 21 Purdue University, Physics 149 3
Rotational Inertia Table For objects with finite number of masses, use I = Σ m r 2. For continuous objects, use table below. Lecture 20 Purdue University, Physics 149 4
Torque Rotational effect of force. Tells how effective force is at twisting or rotating an object. τ = ± r F perpendicular = r F sin θ τ = ±r perpendicular F r perpendicular = lever arm Units N m Sign, CCW rotation is positive F Lecture 20 Purdue University, Physics 149 5
Torque Torque measures the effectiveness of a force for twisting or turning an object. Magnitude: lever arm Sign: If a torque causes a CCW rotation, assign + sign. If a torque causes a CW rotation, assign - sign. Direction: If a torque causes a CCW rotation, the torque s direction is out of the plane and normal to the plane. If a torque causes a CW rotation, the torque s direction is into the plane and normal to the plane. Torque is a vector quantity. Units: N m But, torque is not a form of energy. Torque is denoted by τ beyond the scope of this class Lecture 20 Purdue University, Physics 149 6
Torque Rotational effect of force. Tells how effective force is at twisting or rotating an object. τ = ± r F perpendicular = r F sin θ Units N m Sign, CCW rotation is positive Work done by torque Recall W = F d cos θ For a wheel W = F tangential d = F tangential 2 π r θ / (2 π) (θ in radians) = F tangential r θ = τ θ P = W/t = τ θ/t = τ ω Lecture 21 Purdue University, Physics 149 7
The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case. In which of the cases is the torque on the nut the biggest? A) Case 1 B) Case 2 C) Case 3 ILQ: Torque In which of the cases is the torque on the nut the smallest? A) Case 1 B) Case 2 C) Case 3 Lecture 20 Purdue University, Physics 149 8
ILQ 1 Two forces produce the same torque. Does it follow that they have the same magnitude? A) Yes B) No Lecture 21 Purdue University, Physics 149 9
ILQ 2 Which of the forces in the figure produces a negative (clockwise) torque about the rotation axis indicated? A) 1 and 2 B) 1, 2, and 4 C) 3 only D) 4 only Lecture 21 Purdue University, Physics 149 10
ILQ The figure shows an overhead view of a meter stick that can pivot about the dot shown at the position marked 20 (for 20 cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque they produce, greatest first. a) 2 and 3 tie, 1, then 1 and 5 tie b) 1 and 3 tie, 4, then 2 and 5 tie c) All tie d) 1, 4, then 2 and 3 tie, 5 Lecture 21 Purdue University, Physics 149 11
ILQ A heavy box is resting on the floor. You would like to push the box to tip it over on its side, using the minimum force possible. Which of the force vectors in the diagram shows the correct location and direction of the force? Assume enough friction so that the box does not slide; instead it rotates about point P. a) a b) b c) c r Lecture 21 Purdue University, Physics 149 12
Example The pull cord of a lawnmower engine is wound around a drum of radius of 6.00 cm. While the cord is pulled with a force of 75 N to start the engine, what magnitude torque does the cord apply to the drum? 75 N F = 75.0 N r = 6.00 cm = 0.0600 m θ = 90 6 cm τ = F r sinθ = 4.5 N m Lecture 21 Purdue University, Physics 149 13
Torque Example A person raises one leg to an angle of 30 degrees. An ankle weight (89 N) is attached a distance of 0.84 m from her hip. What is the torque due to this weight? 1) Draw Diagram 2) τ = F r sin θ = F r sin(90 30) 30 = 65 N m If she raises her leg higher, the torque due to the weight will A) Increase B) Same C) Decrease F=89 N Lecture 21 Purdue University, Physics 149 14
Equilibrium Conditions for Equilibrium Σ F = 0 Translational EQ (Center of Mass) Στ= 0 Rotational EQ Can choose any axis of rotation. Choose Wisely! A meter stick is suspended at the center. If a 1 kg weight is placed at x=0. Where do you need to place a 2 kg weight to balance it? A) x = 25 B) x=50 C) x=75 D) x=100 E) Impossible y 50 cm d 9.8 N pivot 19.6 N x Σ τ = 0 9.8 (0.5) (19.6)d = 0 d = 25 Lecture 21 Purdue University, Physics 149 15
Static Equilibrium and Center of Mass Gravitational Force Weight = mg Acts as force at center of mass Torque about pivot due to gravity τ = mgd Object not in static equilibrium pivot d W=mg Center of mass Lecture 21 Purdue University, Physics 149 16
Static Equilibrium Not in equilibrium Equilibrium pivot pivot d W=mg Center of mass Center of mass Torque about pivot 0 Torque about pivot = 0 A method to find center of mass of an irregular object Lecture 21 Purdue University, Physics 149 17
Equilibrium Acts A rod is lying on a table and has two equal but opposite forces acting on it. What is the net force on the rod? A) Up B) Down C) Zero y-direction: Σ F y = ma y +F F = 0 Will the rod move? A) Yes B) No y F Yes, it rotates! Lecture 21 Purdue University, Physics 149 18 F x
Example 8.6 The beam s weight is 425 N. For equilibrium, what should be the magnitudes of the forces? The entire gravitational force acts at the center of mass (recall in Ch 7). Choose a rotation axis. For Στ = 0 (rotational equilibrium), calculate and add all the torques. * torque due to gravity: τ g = (lever arm) (force) = +(2.44m / 2) (425N) = +518.5 N m (We assign + sign, because this torque causes a CCW rotation.) * torque due to F 1 : τ F1 = (lever arm) (force) = (2.44m 1.00m) (F 1 ) = (1.44m) (F 1 ) (We assign sign, because this torque causes a CW rotation.) Στ = τ g + τ F1 + τ F2 = +518.5 N m (1.44m) (F 1 ) + 0 = 0 F 1 = 360 N For ΣF = 0 (translational equilibrium), calculate and add all the forces. ΣF y = F 1 + F 2 mg = 360 N + F 2 425 N = 0 F 2 = 65 N Lecture 21 Purdue University, Physics 149 19
Example 8.6: Different Axis The beam s weight is 425 N. For equilibrium, what should be the magnitudes of the forces? The entire gravitational force acts at the center of mass (recall in Ch 7). Choose a rotation axis. For Στ = 0 (rotational equilibrium), calculate and add all the torques. * torque due to gravity: τ g = (lever arm) (force) = (2.44m / 2 1.00m) (425N) = 93.5 N m (We assign sign, because this torque causes a CW rotation.) * torque due to F 2 : τ F2 = (lever arm) (force) = +(2.44m 1.00m) (F 2 ) = +(1.44m) (F 2 ) (We assign + sign, because this torque causes a CCW rotation.) Στ = τ g + τ F1 + τ F2 = 93.5 N m + 0 + (1.44m) (F 2 )= 0 F 2 = 65 N For ΣF = 0 (translational equilibrium), calculate and add all the forces. Lecture 21 ΣF y = F 1 + F 2 mg = Purdue F 1 + 65 University, N 425 Physics N = 0 149 F 1 = 360 N 20
ILQ A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod? a) 0.25 kg b) 0.50 kg c) 1.0 kg d) 2.0 kg Lecture 21 Purdue University, Physics 149 21
ILQ Find the mass of the tomatoes which make this mobile a "balanced meal." Assume the rods are massless with the lengths indicated. a) 0.5 kg b) 1.0 kg c) 1.5 kg d) 2.0 kg Lecture 21 Purdue University, Physics 149 22
Equilibrium Example A 50 kg diver stands at the end of a 4.6 m diving board. Neglecting the weight of the board, what are the forces acted by the supports? Draw a FBD. Choose a rotation axis. Στ = 0 (rotational equilibrium) Στ = +(1.2m) (F 1 ) (4.6m) (50kg 9.8m/s 2 ) = 0 F 1 = 1880 N F 1 ΣF = 0 (translational equilibrium) ΣF y = F 1 F 2 mg = 1880 N F 2 (50kg 9.8m/s 2 )= 0 F 2 = 1390 N F 2 mg Lecture 21 Purdue University, Physics 149 23
ILQ A child's pull-toy has two wheels attached to an axle, which has a string wrapped around it. A cut-away view is shown. Assume that friction exists between the wheels and floor. When the string is pulled to the right, the wheels will roll to the a) right, winding up the string. b) left, unwinding the string. Lecture 21 Purdue University, Physics 149 24
Example: Dumbbell Consider a rotational motion around a point P. For normal force and F 3, θ = 0. So τ=0 no rotation For gravity, θ = 180. So τ=0 no rotation F 1 and F 2 cause CCW rotations. F 4 causes a CW rotation. Lecture 21 Purdue University, Physics 149 25
The Hammer You want to balance a hammer on the tip of your finger, which way is easier A) Head up B) Head down C) Same τ = I α m g R sin(θ) = mr 2 α Torque increases with R g sin(θ) / R = α Inertia increases as R 2 Lecture 21 Purdue University, Physics 149 26 R mg Angular acceleration decreases with R!, so large R is easier to balance.
Energy Conservation Friction causes an object to roll, but if it rolls w/o slipping friction does NO work! W = F d cos θ θ is zero for point in contact No dissipated work, energy is conserved Need to include both translation and rotation kinetic energy. K = ½ m v 2 + ½ I ω 2 Lecture 21 Purdue University, Physics 149 27
Translational + Rotational KE Consider a cylinder with radius R and mass M, rolling w/o slipping down a ramp. Determine the ratio of the translational to rotational KE. Translational: K T = ½ M v 2 Rotational: K R = ½ I ω 2 use and K R = ½ (½ M R 2 ) (V/R) 2 = ¼ M v 2 H = ½ K T Energy conservation: K i + U i = K f + U f Lecture 21 Purdue University, Physics 149 28
Rolling Act Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other. If both are placed at the top of the same ramp and released, which is moving faster at the bottom? A) bigger one B) smaller one C) same K i + U i = K f + U f Lecture 21 Purdue University, Physics 149 29
Define Angular Momentum Momentum Angular Momentum p = mv L = Iω conserved if ΣF ext = 0 conserved if Στ ext = 0 Vector Vector units: kg-m/s units: kg-m 2 /s For a particle: Lecture 21 Purdue University, Physics 149 30
Linear and Angular Linear Angular Displacement x θ Velocity v ω Acceleration a α Inertia m I Kinetic Energy ½ m v 2 ½ I ω 2 Newton s 2 nd Law F = ma τ = Iα Momentum p = mv L = Iω Lecture 21 Purdue University, Physics 149 31
Act: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity ω i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity ω f. A) ω f = ω i B) ω f = ½ ω i C) ω f = ¼ ω i z z ω i ω f Lecture 21 Purdue University, Physics 149 32
Act: Two Disks First realize that there are no external torques acting on the two-disk system. Angular momentum will be conserved! z 2 1 z ω 0 ω f Lecture 21 Purdue University, Physics 149 33
Demo You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body. Lecture 21 Purdue University, Physics 149 34