2. Torque is a twist or turn that tends to produce rotation. * * * Applications are found in many common tools around the home or industry where it is necessary to turn, tighten or loosen devices.

3. Definition of Torque Torque is defined as the tendency to produce a change in rotational motion. Examples:

4. Torque is Determined by Three Factors: The magnitude of the applied force.The magnitude of the applied force. The direction of the applied force.The direction of the applied force. The location of the applied force.The location of the applied force. The magnitude of the applied force.The magnitude of the applied force. The direction of the applied force.The direction of the applied force. The location of the applied force.The location of the applied force. 20 N Magnitude of force 40 N The 40-N force produces twice the torque as does the 20-N force. Each of the 20-N forces has a different torque due to the direction of force. 20 N Direction of Force 20 N   Location of force The forces nearer the end of the wrench have greater torques. 20 N

5. Units for Torque Torque is proportional to the magnitude of F and to the distance r from the axis. Thus, a tentative formula might be:  = Fr Sin θ Units: N  m 6 cm 40 N  = (40 N)(0.60 m) = 24.0 N  m, cw  = 24.0 N  m, cw

6. Direction of Torque Torque is a vector quantity that has direction as well as magnitude. Turning the handle of a screwdriver clockwise and then counterclockwise will advance the screw first inward and then outward.

7. Sign Convention for Torque By convention, counterclockwise torques are positive and clockwise torques are negative. Positive torque: Counter-clockwise, out of page cw ccw Negative torque: clockwise, into page

8. Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate r.  = (80 N)(0.104 m) = 8.31 N m r = 12 cm sin 60 0 = 10.4 cm

9. Calculating Resultant Torque Read, draw, and label a rough figure.Read, draw, and label a rough figure. Draw free-body diagram showing all forces, distances, and axis of rotation.Draw free-body diagram showing all forces, distances, and axis of rotation. Extend lines of action for each force.Extend lines of action for each force. Calculate moment arms if necessary.Calculate moment arms if necessary. Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-).Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). Resultant torque is sum of individual torques.Resultant torque is sum of individual torques. Read, draw, and label a rough figure.Read, draw, and label a rough figure. Draw free-body diagram showing all forces, distances, and axis of rotation.Draw free-body diagram showing all forces, distances, and axis of rotation. Extend lines of action for each force.Extend lines of action for each force. Calculate moment arms if necessary.Calculate moment arms if necessary. Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-).Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). Resultant torque is sum of individual torques.Resultant torque is sum of individual torques.

10. Example 2: Find resultant torque about axis A for the arrangement shown below: 30 0 6 m 2 m 4 m 20 N 30 N 40 N A Find  due to each force. Consider 20-N force first: r = (4 m) sin 30 0 = 2.00 m  = Fr = (20 N)(2 m) = 40 N m, cw The torque about A is clockwise and negative.  20 = -40 N m r negative

11. Example 2 (Cont.): Next we find torque due to 30-N force about same axis A. 30 0 6 m 2 m 4 m 20 N 30 N 40 N A Find  due to each force. Consider 30-N force next. r = (8 m) sin 30 0 = 4.00 m  = Fr = (30 N)(4 m) = 120 N m, cw The torque about A is clockwise and negative.  30 = -120 N m r negative

12. Example 2 (Cont.): Finally, we consider the torque due to the 40-N force. Find  due to each force. Consider 40-N force next: r = (2 m) sin 90 0 = 2.00 m  = Fr = (40 N)(2 m) = 80 N m, ccw The torque about A is CCW and positive.  40 = +80 N m 30 0 6 m 2 m 4 m 20 N 30 N 40 N A r positive

13. Example 2 (Conclusion): Find resultant torque about axis A for the arrangement shown below: 30 0 6 m 2 m 4 m 20 N 30 N 40 N A Resultant torque is the sum of individual torques.  R = - 80 N m Clockwise  R =  20 +  20 +  20 = -40 N m -120 N m + 80 N m

14. Summary: Resultant Torque Read, draw, and label a rough figure.Read, draw, and label a rough figure. Draw free-body diagram showing all forces, distances, and axis of rotation.Draw free-body diagram showing all forces, distances, and axis of rotation. Extend lines of action for each force.Extend lines of action for each force. Calculate moment arms if necessary.Calculate moment arms if necessary. Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-).Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). Resultant torque is sum of individual torques.Resultant torque is sum of individual torques. Read, draw, and label a rough figure.Read, draw, and label a rough figure. Draw free-body diagram showing all forces, distances, and axis of rotation.Draw free-body diagram showing all forces, distances, and axis of rotation. Extend lines of action for each force.Extend lines of action for each force. Calculate moment arms if necessary.Calculate moment arms if necessary. Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-).Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). Resultant torque is sum of individual torques.Resultant torque is sum of individual torques.

15. Newtons 2 nd law and rotation Define and calculate the moment of inertia for simple systems.Define and calculate the moment of inertia for simple systems. Define and apply the concepts of Newton’s second law.Define and apply the concepts of Newton’s second law. Define and calculate the moment of inertia for simple systems.Define and calculate the moment of inertia for simple systems. Define and apply the concepts of Newton’s second law.Define and apply the concepts of Newton’s second law.

16. Inertia of Rotation Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation. F = 20 N a = 4 m/s 2 Linear Inertia, m m = = 5 kg 20 N 4 m/s 2 F = 20 N R = 0.5 m  = 2 rad/s 2 Rotational Inertia, I I = = = 2.5 kg m 2 (20 N)(0.5 m) 4 m/s 2  Forcetorque Force does for translation what torque does for rotation:

17. Rotational Inertia m2m2 m3m3 m4m4 m m1m1 axis  v =  R Object rotating at constant  Rotational Inertia is how difficult it is to spin an object. It depends on the mass of the object and how far away the object if from the axis of rotation (pivot point). Rotational Inertia Defined: I =  mR 2

18. Common Rotational Inertias L L R R R I = mR 2 I = ½mR 2 Hoop Disk or cylinder Solid sphere

19. Example 1: A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias. R I = mR 2 Hoop R I = ½mR 2 Disk I = 0.120 kg m 2 I = 0.0600 kg m 2

20. Newton 2 nd Law For many problems involving rotation, there is an analogy to be drawn from linear motion. x f R 4 kg     50 rad/s  = 40 N m A resultant force F produces negative acceleration a for a mass m. I m A resultant torque  produces angular acceleration  of disk with rotational inertia I.

21. Newton’s 2nd Law for Rotation R 4 kg  F    50 rad/s R = 0.20 m F = 40 N  = I  How many revolutions required to stop? FR = (½mR 2 )   = 100  rad/s 2 2  f 2 -  o 2 0  = 12.5 rad = 1.99 rev

22. Summary – Rotational Analogies QuantityLinearRotational DisplacementDisplacement x Radians  InertiaMass (kg) I (kg  m 2 ) ForceNewtons NTorque N·m Velocity v “ m/s ”  Rad/s Acceleration a “ m/s 2 ”  Rad/s 2

23. CONCLUSION: Chapter 5A Torque