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DYNAMICAL TORQUE AND STATIC EQUILIBRIUM Learning Objectives After you complete the homework associated with this lecture, you should be able to: • Explain what torque is, and give examples that show it is related to force but differs from it. • Describe how torque changes rotational motion in a way analogous to how force changes linear motion. • Explain how the sign and magnitudes of torque components are determined in rotational situations. • Calculate changes in motion of objects due to torques. • Determine the forces that act on a static structure if it does NOT change its translational or rotational state. Cause of Rotation Consider two equal but opposite forces acting on a stick as shown Y 3P F ext = 0 We see C 3P F ext = m P a CM C but object will begin to rotate about its CM C This rotation is caused by TORQUE (symbol P τ) Y Pa CM = 0 in this case 1 [©2013 RJ Bieniek] 2 [©2013 RJ Bieniek] Angular Momentum Torque Causes Change in Rotational State C Linear momentum P P is the fundamental descriptor of translational motion. It is changed by forces. C Angular momentum P L is the fundamental descriptor of rotational motion. It is changed by torques. Angular Momentum of a Rigid Body The total angular momentum of a rigid body (all components P ) rotating about a symmetry have same angular velocity ω Changes in rotational motion are caused by the torque P τ vector. It is not a force, but is related to force. Torque is a vector, and thus has a direction. This comes from fundamental definition of torque as the cross product of position vector P r with force vector P F: Pτ = Pr × P F L axis is: PL = I axis ω P PL is in the same direction as angular velocity ωP . 3 [©2013 RJ Bieniek] We get its direction from the right-hand-thumb rule, and its magnitude from τ=|P r ×P F | = r F sin(θ) 4 [©2013 RJ Bieniek] TORQUE AND ANGULAR MOMENTUM C The time rate of change of a system’s total angular momentum VECTOR equals the net external torque VECTOR acting on it. P dL ' Pτnet ext ' j Pτext dt Torques produce change in angular momentum in a way analogous to how forces change linear momentum! 5 [©2013 RJ Bieniek] LINE OF ACTION AND MOMENT ARM The torque due to force P F is a function of P F and the position vector P r where is P F applied about point O. Important Important: rz is called the moment arm τ Angular Acceleration of Rigid Bodies Consider an rigid object that can rotate about an axis of symmetry, which we will label the z-axis. It has rotational inertia I about this axis. We have Pτnet ' P dL dt PL = I ωP and If Iz is constant, taking z-component of these gives us τnet , z ' d (Lz ) dt ' d (I ωz ) dt ' I d ωz dt 3τz = I αz <==> 3 Fx = M ax rotational linear 6 [©2013 RJ Bieniek] Choosing Positive Direction of Rotational Components 1) In your diagram, put a dot at the point about which you will calculate rotational quantities. If the object is rotating, pick the axis of rotation. 2) Draw a curved arrow (labelled z) around this point to show the sense of a positive rotation = |P τ | = magnitude of torque = r F sin(θ) = r [ F sin(θ) ] = r Fz = F [ r sin(θ) ] = F rz This defines direction of z-axis and you can use it to determine signs of z-components. DEMO : wrench turns bolt 7 [©2013 RJ Bieniek] 8 [©2013 RJ Bieniek] Survival Methods for Determining τz • • If a force tends to produce a positive rotation (i.e., in the direction of the curved z-axis arrow), then the corresponding torque’s z-component τz is positive. If a force tends to produce a negative rotation (i.e., in the direction of the opposite the curved arrow), then the corresponding torque’s z-component τz is negative. Example: A disk of mass M and radius R rotates on a frictionless axis through its center. A block of mass m is attached to a string that is wound around the disk’s inner core of radius r. What is the linear acceleration of the block? τ1z = + F1 r1z = + F1 d1 = + F1z r1 (positive) τ2z = S F2 r2z = S F2 d2 = S F2z r2 (negative) 9 [©2013 RJ Bieniek] 10 [©2013 RJ Bieniek] 3τz = I αz Form free-body of block where I = ½MR2 3Fx = (ST) + (+ mg) = m ax τT,z + V τ S,z + V τ Fg,z = I αz one equation with 2 unknowns (T and ax ) (+ r T) + 0 + 0 = I αz Get torques from extended free-body diagram of the disk Because the disk does not translate, we could get the magnitude of the support force P S from 3P F = MP a = 0. But why bother? F g are both zero, Since the moment arms of P S &P they produce zero torques and thus do not affect the rotational dynamics! 11 [©2013 RJ Bieniek] But ˆ r αz = ax (rolling without slipping) r T = Iz [ ax / r] r 2 T = Iz a x multiplying by r: r2 @ [force-eq on block]: (S r2 T) + r2 m g = r2 m ax adding to get rid of T: r2 m g = [r2 m + I ] ax Then solve: ax = m r2 g [mr2 + 12 MR2 ] 12 [©2013 RJ Bieniek] STATIC EQUILIBRIUM An object will not change its rotation if Y IP α = 3 Pτ o = 0 Y Pα = 0 Y Example of an odd-shaped beam 3P τo = 0 L no rotation change T? Thus, for motionless object to remain motionless: 3P F ext = 0 (no translation) odd beam and 3P τo = 0 Y 3 (τo)z = 0 (no rotation) M CAUTION : Remember to EXPLICITLY indicate a ref point and draw a labeled curved arrow around it for the choice of positive rotation. Demo Demo: m θ free pivot support D A ball of mass M is suspended from an oddly shaped beam of mass m. The beam can freely rotate about a pivot support. What is tension T in the cable? What (reactance) support force P S does the pivot supply? Static Equilibrium of a Simple Beam 13 [©2013 RJ Bieniek] 14 [©2013 RJ Bieniek] If we just want T, we could get it directly from torque eq because P S produces zero torque about O. But let us not see it for the moment, and do the "drill". Note the tension force P T lies along the strut, but support force P S has unknown direction. Vx + V w x + Sx = 0 3Fx = Tx + W = (+T cos θ) + Sx = 0 Hint Hint: If you choose an origin O that lies along the line of action of a force, that force’s moment arm is zero. Thus it will not contribute the sum of torques, which simplifies the algebra. 3 Fy = Ty + Wy + wy + Sy = 0 = +T sin θ + (SMg) + (Smg) + Sy = 0 2 equations in 3 Unknowns: T , Sx , Sy Y look for another equation τ Sz = (ST rz) + ( + MgD) + ( + mg[½L]) = 0 3τOz = τTz + τWz + τwz + V = S T (L sin θ) + M g D + ½ m g L = 0 T = [ M g D + ½ m g L ] / [L sin θ ] , then go above to get Sx and Sy note: τTz = ST rz = STz L = S (T sin θ) L (both the same) 15 [©2013 RJ Bieniek] 16 [©2013 RJ Bieniek]