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

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## Definition of Torque

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**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.**Definition of Torque**Torque is defined as the tendency to produce a change in rotational motion. Examples:**Each of the 20-N forces has a different torque due to the**direction of force. Direction of Force 20 N q 20 N q 20 N Magnitude of force The 40-N force produces twice the torque as does the 20-N force. Location of force The forces nearer the end of the wrench have greater torques. 20 N 40 N 20 N 20 N 20 N Torque is Determined by Three Factors: • The magnitude of the applied force. • The direction of the applied force. • The location of the applied force.**6 cm**40 N 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: t = Fr Sin θ Units: Nm t = (40 N)(0.60 m) = 24.0 Nm, cw t = 24.0 Nm, cw**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.**cw**ccw Sign Convention for Torque By convention, counterclockwise torques are positive and clockwise torques are negative. Positive torque: Counter-clockwise, out of page Negative torque: clockwise, into page**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. r = 12cm sin 600 = 10.4 cm t = (80 N)(0.104 m) = 8.31 N m**Calculating Resultant Torque**• Read, draw, and label a rough figure. • Draw free-body diagram showing all forces, distances, and axis of rotation. • Extend lines of action for each force. • Calculate moment arms if necessary. • Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). • Resultant torque is sum of individual torques.**negative**20 N 30 N 300 300 2 m 6 m A 4 m 40 N Example 2:Find resultant torque about axis A for the arrangement shown below: Find t due to each force. Consider 20-N force first: r The torque about A is clockwise and negative. r = (4 m) sin 300 = 2.00 m t = Fr = (20 N)(2 m) = 40 N m, cw t20 = -40 N m**r**20 N 30 N negative 300 300 2 m 6 m A 4 m 40 N Example 2 (Cont.):Next we find torque due to 30-N force about same axis A. Find t due to each force. Consider 30-N force next. The torque about A is clockwise and negative. r = (8 m) sin 300 = 4.00 m t = Fr = (30 N)(4 m) = 120 N m, cw t30 = -120N m**positive**20 N 30 N r 300 300 2 m 6 m A 4 m 40 N Example 2 (Cont.):Finally, we consider the torque due to the 40-N force. Find t due to each force. Consider 40-N force next: The torque about A is CCW and positive. r = (2 m) sin 900 = 2.00 m t = Fr = (40 N)(2 m) = 80 N m, ccw t40 = +80N m**20 N**30 N 300 300 2 m 6 m A 4 m 40 N Example 2 (Conclusion):Find resultant torque about axis A for the arrangement shown below: Resultant torque is the sum of individual torques. tR = t20 + t20 + t20 = -40 N m -120 N m + 80 N m tR = - 80N m Clockwise**Summary: Resultant Torque**• Read, draw, and label a rough figure. • Draw free-body diagram showing all forces, distances, and axis of rotation. • Extend lines of action for each force. • Calculate moment arms if necessary. • Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (-). • Resultant torque is sum of individual torques.**Newtons 2nd law and rotation**• Define and calculate the moment of inertia for simple systems. • Define and apply the concepts of Newton’s second law.**Linear Inertia, m**F = 20 N 20 N 4 m/s2 a = 4 m/s2 m = = 5 kg F = 20 N Rotational Inertia, I R = 0.5 m t a (20 N)(0.5 m) 4 m/s2 a = 2 rad/s2 I = = = 2.5 kg m2 Inertia of Rotation Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation. Force does for translation what torque does for rotation:**v = wR**m m4 w m3 m1 m2 axis Object rotating at constant w. Rotational Inertia 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 = SmR2**L**L R R R I = mR2 I = ½mR2 Hoop Disk or cylinder Solid sphere Common Rotational Inertias**R**I = 0.120 kg m2 I = 0.0600 kg m2 I = mR2 Hoop R I = ½mR2 Disk 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.**x**f t A resultant torque tproduces angular acceleration a of disk with rotational inertia I. A resultant force F produces negative acceleration a for a mass m. w wo = 50 rad/s t = 40 N m R 4 kg Newton 2nd Law For many problems involving rotation, there is an analogy to be drawn from linear motion. m I**How many revolutions required to stop?**F wo = 50 rad/s R = 0.20 m F = 40 N w R t = Ia 4 kg 0 Newton’s 2nd Law for Rotation FR = (½mR2)a 2aq = wf2 - wo2 q = 12.5 rad = 1.99 rev a = 100rad/s2

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