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Chapter 8

Chapter 8. Torque and Angular Momentum. Torque and Angular Momentum. Rotational Kinetic Energy Rotational Inertia Torque Work Done by a Torque Equilibrium (revisited) Rotational Form of Newton’s 2 nd Law Rolling Objects Angular Momentum. Rotational KE and Inertia.

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Chapter 8

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  1. Chapter 8 Torque and Angular Momentum

  2. Torque and Angular Momentum • Rotational Kinetic Energy • Rotational Inertia • Torque • Work Done by a Torque • Equilibrium (revisited) • Rotational Form of Newton’s 2nd Law • Rolling Objects • Angular Momentum Chap08-Torque-Revised 3/6/2010

  3. Rotational KE and Inertia For a rotating solid body: For a rotating body vi = ri where ri is the distance from the rotation axis to the mass mi. Chap08-Torque-Revised 3/6/2010

  4. Moment of Inertia The quantity is called rotational inertia or moment of inertia. Use the above expression when the number of masses that make up a body is small. Use the moments of inertia in the table in the textbook for extended bodies. Chap08-Torque-Revised 3/6/2010

  5. Moments of Inertia Chap08-Torque-Revised 3/6/2010

  6. Chap08-Torque-Revised 3/6/2010

  7. r1 r2 m1 m2 Moment of Inertia Example: The masses are m1 and m2 and they are separated by a distance r. Assume the rod connecting the masses is massless. Q: (a) Find the moment of inertia of the system below. r1 and r2 are the distances between mass 1 and the rotation axis and mass 2 and the rotation axis (the dashed, vertical line) respectively. Chap08-Torque-Revised 3/6/2010

  8. Moment of Inertia Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m , and r2 = 0.67 m. (b) What is the moment of inertia if the axis is moved so that is passes through m1? Chap08-Torque-Revised 3/6/2010

  9. Moment of Inertia What is the rotational inertia of a solid iron disk of mass 49.0 kg with a thickness of 5.00 cm and a radius of 20.0 cm, about an axis through its center and perpendicular to it? From the table: Chap08-Torque-Revised 3/6/2010

  10. hinge Push Torque A torque is caused by the application of a force, on an object, at a point other than its center of mass or its pivot point. Q: Where on a door do you normally push to open it? A: Away from the hinge. A rotating (spinning) body will continue to rotate unless it is acted upon by a torque. Chap08-Torque-Revised 3/6/2010

  11. F Hinge end  Torque Torque method 1: Top view of door r = the distance from the rotation axis (hinge) to the point where the force F is applied. F is the component of the force F that is perpendicular to the door (here it is Fsin). Chap08-Torque-Revised 3/6/2010

  12. Torque The units of torque are Newton-meters (Nm) (not joules!). By convention: • When the applied force causes the object to rotate counterclockwise (CCW) then  is positive. • When the applied force causes the object to rotate clockwise (CW) then  is negative. Chap08-Torque-Revised 3/6/2010

  13. Torque Torque method 2: r is called the lever arm and F is the magnitude of the applied force. Lever arm is the perpendicular distance to the line of action of the force. Chap08-Torque-Revised 3/6/2010

  14. F r  Hinge end  Line of action of the force Lever arm The torque is: Same as before Torque Top view of door Chap08-Torque-Revised 3/6/2010

  15. Torque Problem The pull cord of a lawnmower engine is wound around a drum of radius 6.00 cm, while the cord is pulled with a force of 75.0 N to start the engine. What magnitude torque does the cord apply to the drum? F=75 N R=6.00 cm Chap08-Torque-Revised 3/6/2010

  16. F2=30 N 30 F3=20 N 10 X 45 F1=25 N Torque Problem Calculate the torque due to the three forces shown about the left end of the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the bar. Chap08-Torque-Revised 3/6/2010

  17. Lever arm for F2 F2=30 N 30 F3=20 N 10 X 45 Lever arm for F3 F1=25 N Torque Problem The lever arms are: Chap08-Torque-Revised 3/6/2010

  18. Torque Problem The torques are: The net torque is +65.8 Nm and is the sum of the above results. Chap08-Torque-Revised 3/6/2010

  19. Work done by the Torque The work done by a torque  is where  is the angle (in radians) that the object turns through. Following the analogy between linear and rotational motion: Linear Work is Force x displacement. In the rotational picture force becomes torque and displacement becomes the angle Chap08-Torque-Revised 3/6/2010

  20. Work done by the Torque A flywheel of mass 182 kg has a radius of 0.62 m (assume the flywheel is a hoop). (a) What is the torque required to bring the flywheel from rest to a speed of 120 rpm in an interval of 30 sec? Chap08-Torque-Revised 3/6/2010

  21. Work done by the Torque (b) How much work is done in this 30 sec period? Chap08-Torque-Revised 3/6/2010

  22. Equilibrium The conditions for equilibrium are: Linear motion Rotational motion For motion in a plane we now have three equations to satisfy. Chap08-Torque-Revised 3/6/2010

  23. Using Torque A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.0 N. A cable, inclined at a 35 angle with the boom, is attached at a distance of 2.38 m from the hinge at the wall. The weight of the sign is 120.0 N. What is the tension in the cable and what are the horizontal and vertical forces exerted on the boom by the hinge? Chap08-Torque-Revised 3/6/2010

  24. y Fy T  X Fx x wbar Fsb Using Torque This is important! You need two components for F, not just the expected perpendicualr normal force. FBD for the bar: Apply the conditions for equilibrium to the bar: Chap08-Torque-Revised 3/6/2010

  25. Using Torque Equation (3) can be solved for T: Equation (1) can be solved for Fx: Equation (2) can be solved for Fy: Chap08-Torque-Revised 3/6/2010

  26. Equilibrium in the Human Body Find the force exerted by the biceps muscle in holding a one liter milk carton with the forearm parallel to the floor. Assume that the hand is 35.0 cm from the elbow and that the upper arm is 30.0 cm long. The elbow is bent at a right angle and one tendon of the biceps is attached at a position 5.00 cm from the elbow and the other is attached 30.0 cm from the elbow. The weight of the forearm and empty hand is 18.0 N and the center of gravity is at a distance of 16.5 cm from the elbow. Chap08-Torque-Revised 3/6/2010

  27. y Fb “hinge” (elbow joint) x Fca w MCAT type problem Chap08-Torque-Revised 3/6/2010

  28. Newton’s 2nd Law in Rotational Form Compare to Chap08-Torque-Revised 3/6/2010

  29. Rolling Object A bicycle wheel (a hoop) of radius 0.3 m and mass 2 kg is rotating at 4.00 rev/sec. After 50 sec the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces? Chap08-Torque-Revised 3/6/2010

  30. Rolling Objects An object that is rolling combines translational motion (its center of mass moves) and rotational motion (points in the body rotate around the center of mass). For a rolling object: If the object rolls without slipping then vcm = R. Note the similarity in the form of the two kinetic energies. Chap08-Torque-Revised 3/6/2010

  31. h  Rolling Example Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height. Which object reaches the bottom of the ramp first? This we know - The object with the largest linear velocity (v) at the bottom of the ramp will win the race. Chap08-Torque-Revised 3/6/2010

  32. Rolling Example Apply conservation of mechanical energy: Solving for v: Chap08-Torque-Revised 3/6/2010

  33. Rolling Example Example continued: Note that the mass and radius are the same. The moments of inertia are: For the disk: Since Vsphere> Vdisk the sphere wins the race. For the sphere: Compare these to a box sliding down the ramp. Chap08-Torque-Revised 3/6/2010

  34. The Disk or the Ring? Chap08-Torque-Revised 3/6/2010

  35. y N w How do objects in the previous example roll? FBD: Both the normal force and the weight act through the center of mass so  = 0. This means that the object cannot rotate when only these two forces are applied. x The round object won’t rotate, but most students have difficulty imagining a sphere that doesn’t rotate when moving down hill. Chap08-Torque-Revised 3/6/2010

  36. y N Fs w  x Add Friction FBD: Also need acm = R and The above system of equations can be solved for v at the bottom of the ramp. The result is the same as when using energy methods. (See text example 8.13.) It is the addition of static friction that makes an object roll. Chap08-Torque-Revised 3/6/2010

  37. Angular Momentum Units of p are kg m/s Units of L are kg m2/s When no net external forces act, the momentum of a system remains constant (pi = pf) When no net external torques act, the angular momentum of a system remains constant (Li = Lf). Chap08-Torque-Revised 3/6/2010

  38. Angular Momentum Units of p are kg m/s Units of L are kg m2/s When no net external forces act, the momentum of a system remains constant (pi = pf) When no net external torques act, the angular momentum of a system remains constant (Li = Lf). Chap08-Torque-Revised 3/6/2010

  39. Angular Momentum Example A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.500 rev/sec. Assume a uniform disk. What is the angular momentum? Chap08-Torque-Revised 3/6/2010

  40. Angular Momentum Example A skater is initially spinning at a rate of 10.0 rad/sec with I=2.50 kg m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces I to 1.60 kg m2? The skater is on ice, so we can ignore external torques. Chap08-Torque-Revised 3/6/2010

  41. The Vector Nature of Angular Momentum Angular momentum is a vector. Its direction is defined with a right-hand rule. Chap08-Torque-Revised 3/6/2010

  42. The Right-Hand Rule Curl the fingers of your right hand so that they curl in the direction a point on the object moves, and your thumb will point in the direction of the angular momentum. Angular Momentum is also an example of a vector cross product Chap08-Torque-Revised 3/6/2010

  43. The Vector Cross Product The magnitude of C C = ABsin(Φ) The direction of C is perpendicular to the plane of A and B. Physically it means the product of A and the portion of B that is perpendicular to A. Chap08-Torque-Revised 3/6/2010

  44. The Cross Product by Components Since A and B are in the x-y plane A x B is along the z-axis. Chap08-Torque-Revised 3/6/2010

  45. Memorizing the Cross Product Chap08-Torque-Revised 3/6/2010

  46. The Gyroscope Demo Chap08-Torque-Revised 3/6/2010

  47. Angular Momentum Demo Consider a person holding a spinning wheel. When viewed from the front, the wheel spins CCW. Holding the wheel horizontal, they step on to a platform that is free to rotate about a vertical axis. Chap08-Torque-Revised 3/6/2010

  48. Angular Momentum Demo Initially, nothing happens. They then move the wheel so that it is over their head. As a result, the platform turns CW (when viewed from above). This is a result of conserving angular momentum. Chap08-Torque-Revised 3/6/2010

  49. Angular Momentum Demo Initially there is no angular momentum about the vertical axis. When the wheel is moved so that it has angular momentum about this axis, the platform must spin in the opposite direction so that the net angular momentum stays zero. Is angular momentum conserved about the direction of the wheel’s initial, horizontal axis? Chap08-Torque-Revised 3/6/2010

  50. It is not. The floor exerts a torque on the system (platform + person), thus angular momentum is not conserved here. Chap08-Torque-Revised 3/6/2010

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