Chapter 8: 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

§8.1 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.

is called rotational inertia or moment of inertia. The quantity Use the above expression when the number of masses that make up a body is small. Use the moments of inertia in table 8.1 for extended bodies.

Example: (a) Find the moment of inertia of the system below Example: (a) Find the moment of inertia of the system below. The masses are m1 and m2 and they are separated by a distance r. Assume the rod connecting the masses is massless.  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.

Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m , and r2 = 0.67 m. Example continued: 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?

Example (text problem 8.4): 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 table 8.1:

§8.2 Torque A rotating (spinning) body will continue to rotate unless it is acted upon by a torque. hinge Push Q: Where on a door do you push to open it? A: Far 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.

Torque method 1: F Hinge end  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).

The units of torque are 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.

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.

Line of action of the force Hinge end F Top view of door Line of action of the force  Hinge end F  r Lever arm The torque is: Same as before

Example (text problem 8.12): 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

Example: 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

X The lever arms are: Example continued: Lever arm for F2 F2=30 N 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:

The net torque is + 65.8 Nm and is the sum of the above results. Example continued: The torques are: The net torque is + 65.8 Nm and is the sum of the above results.

§8.3 Work done by a Torque The work done by a torque  is where  is the angle (in radians) 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.

Example (text problem 8.25): 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?

(b) How much work is done in this 30 sec period? Example continued: (b) How much work is done in this 30 sec period?

§8.4 Equilibrium The conditions for equilibrium are: 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.

Example (text problem 8.36): 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?

FBD for the bar: Example continued: X wbar Fy Fx  X Fsb x y FBD for the bar: 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.

Equation (3) can be solved for T: Example continued: Equation (3) can be solved for T: Equation (1) can be solved for Fx: Equation (2) can be solved for Fy:

§8.5 Equilibrium in the Human Body Example (text problem 8.43): 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.

Example continued: Fb “hinge” (elbow joint) Fca w 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), 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.

§8.6 Rotational Form of Newton’s 2nd Law 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.

Example (text problem 8. 57): A bicycle wheel (a hoop) of radius 0 Example (text problem 8.57): 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?

§8.7 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.

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  The object with the largest linear velocity (v) at the bottom of the ramp will win the race.

Apply conservation of mechanical energy: Example continued: Apply conservation of mechanical energy: Solving for v:

The moments of inertia are: Example continued: 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.

How do objects in the previous example roll? y FBD: x 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.

It is static friction that makes an object roll. 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 static friction that makes an object roll.

§8.8 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.

Example (text problem 8. 70): A turntable of mass 5 Example (text problem 8.70): A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.500 rev/sec. What is the angular momentum? Assume a uniform disk.

Example (text problem 8.79): 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? He/she is on ice, so we can ignore external torques.

§8.9 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.

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.

Consider a person holding a spinning wheel 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.

Initially, nothing happens 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.

Initially there is no angular momentum about the vertical axis 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?

It is not. The floor exerts a torque on the system (platform + person), thus angular momentum is not conserved here.

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