Torque and Equilibrium Torque is a twist or

Torque and Equilibrium

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 is defined as the tendency to produce a change in rotational motion. Examples:

Torque is Determined by Three Factors: • The magnitude of the applied force. • The direction of the applied force. • The location of the applied force. Each The 40 -N of the force 20 -Nthe The forces nearer forces produces different the end ofhas theatwice wrench torque as due doestorques. tothe 20 have greater direction -N force. of force. Magnitude Locationofofof force Direction Force 20 N q 2020 N 20 NN q 20 40 NN 20 N

Units for 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: N m t = (40 N)(0. 60 m) = 24. 0 N m, cw t = 24. 0 N m, cw 6 cm 40 N

Direction of 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.

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

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 = 12 cm 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.

Example 2: Find resultant torque about axis A for the arrangement shown below: Find t due to each force. Consider 20 -N force first: negative 30 N r 300 2 m 6 m 40 N 20 N 300 A 4 m r = (4 m) sin 300 The torque about A is clockwise and negative. t = Fr = (20 N)(2 m) t 20 = -40 N m = 2. 00 m = 40 N m, cw

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. r negative 30 N 300 2 m 6 m 40 N A 4 m r = (8 m) sin 300 The torque about A is clockwise and negative. t = Fr = (30 N)(4 m) t 30 = -120 N m = 4. 00 m = 120 N m, cw

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: r = (2 m) sin 900 = 2. 00 m t = Fr = (40 N)(2 m) = 80 N m, ccw positive 30 N r 300 2 m 6 m 40 N 20 N 300 A 4 m The torque about A is CCW and positive. t 40 = +80 N m

Example 2 (Conclusion): Find resultant torque about axis A for the arrangement shown below: Resultant torque is the sum of individual torques. 20 N 300 300 2 m 6 m 40 N A 4 m t. R = t 20 + t 20 = -40 N m -120 N m + 80 N m t. R = - 80 N 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 2 nd law and rotation • Define and calculate the moment of inertia for simple systems. • Define and apply the concepts of Newton’s second law.

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 F = 20 N R = 0. 5 m a = 2 rad/s 2 Linear Inertia, m 20 N 4 m = m/s 2 = 5 kg Rotational Inertia, I t (20 N)(0. 5 m) 2 I =a = = 2. 5 kg m 4 m/s 2 Force does for translation what torque does for rotation:

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). v = w. R m w m 4 m 1 axis m 3 m 2 Object rotating at constant w. Rotational Inertia Defined: I = Sm. R 2

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

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 = 0. 120 kg m 2 I = m. R 2 Hoop R I = ½m. R 2 Disk I = 0. 0600 kg m 2

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

Newton’s 2 nd Law for Rotation How many revolutions required to stop? t = Ia FR = (½m. R 2)a a = 100 rad/s 2 F R 4 kg w wo = 50 rad/s R = 0. 20 m F = 40 N 0 2 aq = wf 2 - wo 2 q = 12. 5 rad = 1. 99 rev

Summary – Rotational Analogies Quantity Linear Rotational Displacement x Radians Inertia Mass (kg) I (kg m 2) Force Newtons N Torque N·m Velocity v “ m/s ” Rad/s Acceleration a “ m/s 2 ” Rad/s 2

CONCLUSION: Chapter 5 A Torque