Chapter 8 Rotational Motion I.Rotational Motion A.How is rotational motion different from linear motion? B.Rotational Displacement (theta =  ) 1)How far the object rotates, similar to distance in linear motion 2)Measured in degrees or radians or revolutions 3)1 radian has an arc s = r. s(m)/r(m) =1 a)Revolutions can be converted to degrees or radians b)1 revolution = 360 degrees c)1 revolution = 2  Radians d)1 Radian = 57.3 degrees

C.Rotational Velocity (omega =  ) 1)Velocity is a rate of distance/time 2)Rotational Velocity is the rate of rotational displacement/time 3)RPM = revolutions per minute is an  4)Revolutions/second = Similar to meters/second in linear motion 5)Radians/second is another common unit for  D.Rotational Acceleration (alpha =  ) 1)Change in  = rotational acceleration 2)Units = revolutions/s 2 or radians/s 2

Linear motionRotational motion

D.Uniform Rotational Acceleration 1)Rotational Acceleration at a constant rate is like Uniform Linear Acceleration 2)Example calculation:  = rev/s 2, t = 60 s, r = 1 m Linear MotionRotational Motion

E.Relationship between Linear and Rotational Motions 1)Who is going faster? 2)Linear velocity depends on how far from the center an object is (r) 3)v = r  (for  in rad/s only) 4)Example: r = 1 m,  = 0.6  radians v = r  = (1 m)(0.6  rad/s) = 1.88 m/s If r = 2 m v = r  = (2 m)(0.6  rad/s) = 3.76 m/s II.Torque and Balance A.How is something balanced? 1)F 1 = F 2 and d 1 = d 2 (l 1 = l 2 ) 2)Law of the lever: F 1 d 1 = F 2 d 2 3)Torque = tau =  = F x l 4)Torque is the ability to cause rotation about a pivot point l1l1 l2l2

5)l must be measured perpendicular to the force vector 6)Increase the length to increase torque C.Adding Torques 1)If F causes counter-clockwise rotation we call  = (+) 2)If F causes clockwise rotation we call  = (-) 3)Total torque is found by vector addition 4)System is balanced if Total Torque = 0 (F 1 = -F 2 ) +  1 -2-2 -  T l1l1 l2l2

5)Example Calculation: F 1 = 5 N, l 1 = 0.2 m, F 2 = 3 N, l 2 = ? to balance  1 = (F 1 )(l 1 ) = (5 N)(0.2 m) = 1 Nm  2 = -  1 = -1 Nm = (F 2 )(l 2 ) D.Center of Gravity 1)How far can the child “walk the plank”? 2)Center of Gravity = point at which an object’s weight causes no net torque (balanced) 3)All of the plank’s weight is “centered” at its C.O.G. 4)Child won’t tip the plank until torque he causes is larger than that of the plank. The pivot point is the edge of the dock. plankchild

5)Finding C.O.G. for a complex shape a)Suspend the object from 2 different points b)Draw lines extending from the suspending line c)Intersection of the points gives center of gravity 6)Center of Gravity can be below pivot point a)The torque will always try to bring C.O.G. back into position b)The object is automatically balanced by gravity,  = 0 at that point 7)Try touching your toes with your heels and back against a wall. Where is your center of gravity?