Torque and Angular Momentum 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 2nd Law Rolling Objects Angular Momentum MFMcGraw Chap08-Torque-Revised 3/6/2010

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. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Questions 1, 11, and 13. MFMcGraw Chap08-Torque-Revised 3/6/2010

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

Chap08-Torque-Revised 3/6/2010 Moments of Inertia MFMcGraw Chap08-Torque-Revised 3/6/2010

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

Chap08-Torque-Revised 3/6/2010 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 r2 m1 m2 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. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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? MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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: MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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. hinge Push Q: Where on a door do you normally push to open it? A: Away from the hinge. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Question 16. A rotating (spinning) body will continue to rotate unless it is acted upon by a torque. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 Torque method 1: Hinge end F  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). MFMcGraw Chap08-Torque-Revised 3/6/2010

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. CW is clockwise and CCW is counterclockwise. MFMcGraw Chap08-Torque-Revised 3/6/2010

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

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

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

Chap08-Torque-Revised 3/6/2010 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. 45 10 30 F1=25 N F3=20 N F2=30 N X MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 Torque Problem Lever arm for F3 Lever arm for F2 45 10 30 F1=25 N F3=20 N F2=30 N X This is a fairly complex figure. It will probably be useful if you walk your students through its construction. The lever arms are: MFMcGraw Chap08-Torque-Revised 3/6/2010

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

Chap08-Torque-Revised 3/6/2010 Work done by the Torque The work done by a torque  is where  is the angle (in radians) that the object turns through. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Questions 17 and 20. 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 MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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? MFMcGraw Chap08-Torque-Revised 3/6/2010

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

Chap08-Torque-Revised 3/6/2010 Equilibrium The conditions for equilibrium are: Linear motion Rotational motion Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Questions 8, 9, 12, and 14. For motion in a plane we now have three equations to satisfy. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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? MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 Using Torque This is important! You need two components for F, not just the expected perpendicualr normal force. FBD for the bar: T wbar Fy Fx  X Fsb x y Apply the conditions for equilibrium to the bar: Fsb is the force that the sign exerts on the boom. Drawing a free body diagram for the sign and applying Newton’s second and third laws gives the magnitude of this force as the weight of the sign. L is the length of the boom (given as 3.00 m) and x is the distance from the hinge to the location of the cable (given as 2.38 m). Note: the lines of action and the lever arms are omitted from the figure for clarity. MFMcGraw Chap08-Torque-Revised 3/6/2010

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

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

Chap08-Torque-Revised 3/6/2010 MCAT type problem Fb w Fca “hinge” (elbow joint) x y Fca is the force that the carton exerts on the arm. Drawing a free body diagram for the carton and applying Newton’s second and third laws gives the magnitude of this force as the weight of the carton. x1 is the lever arm for the “biceps force” (given as 5.00 cm; assumes the force Fb is nearly perpendicular to the forearm since no angle is given), x2 is the lever arm for the “weight of the forearm” (given as 16.5 cm), and x3 is the lever arm for the “force of the carton on the arm” (given as 35.0 cm). Note: the lines of action and the lever arms are omitted for clarity. MFMcGraw Chap08-Torque-Revised 3/6/2010

Newton’s 2nd Law in Rotational Form Compare to Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Question 18. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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? MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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: Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Questions 2, 3, 4, and 5. If the object rolls without slipping then vcm = R. Note the similarity in the form of the two kinetic energies. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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? h  This we know - The object with the largest linear velocity (v) at the bottom of the ramp will win the race. MFMcGraw Chap08-Torque-Revised 3/6/2010

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

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

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

How do objects in the previous example roll? y 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. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 Add Friction y x N w Fs  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. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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). Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Questions 6, 7, and 19. MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 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). Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Questions 6, 7, and 19. MFMcGraw Chap08-Torque-Revised 3/6/2010

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? MFMcGraw Chap08-Torque-Revised 3/6/2010

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

The Vector Nature of Angular Momentum Angular momentum is a vector. Its direction is defined with a right-hand rule. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 8, Questions 10 and 15. MFMcGraw Chap08-Torque-Revised 3/6/2010

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

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

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

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

Chap08-Torque-Revised 3/6/2010 The Gyroscope Demo MFMcGraw Chap08-Torque-Revised 3/6/2010

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

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

Chap08-Torque-Revised 3/6/2010 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? MFMcGraw Chap08-Torque-Revised 3/6/2010

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

Chap08-Torque-Revised 3/6/2010 Summary Rotational Kinetic Energy Moment of Inertia Torque (two methods) Conditions for Equilibrium Newton’s 2nd Law in Rotational Form Angular Momentum Conservation of Angular Momentum MFMcGraw Chap08-Torque-Revised 3/6/2010

Chap08-Torque-Revised 3/6/2010 X Z Y MFMcGraw Chap08-Torque-Revised 3/6/2010