: rigid bodies, rotational dynamics



Key ideas:Torque and CoupleMoment of inertia Rotational and translational equilibrium Angular acceleration Equations of rotational motion for uniform angular acceleration Newton’s second law applied to angular motion Conservation of angular momentum You have to be able to calculate torque for single forces and couples Solve problems involving moment of inertia, torque and angular acceleration Solve problems in which objects are in both rotational and translational equilibriumSolve problems using rotational quantities analogous to linear quantities – the rotational equivalent of SUVATSketch and interpret graphs of rotational motion Solve problems involving rolling without slipping (analysis will be limited to basic geometric shapes) The equation for the moment of inertia of a specific shape will be provided when necessary Graphs will be limited to angular displacement–time, angular velocity–time and torque–timeLINEAR ANGULARΓ=FrsinθI = ?mr2Γ=Iαω=2πfv = u + atωf= ωi+αtv2-u2=2as ωf2- ωi2=2αθs=ut+ 12at2θ=ωit+ 12αt2p = mvL=IωTranslational EK = 12mv2Rotational EK= 12Iω2Recall definitions for angular displacement θ=sr, angular velocity ω in rad s-1= ΔθΔt= vr where s=vΔtand angular acceleration ? (by analogy) = ΔωΔt (rad s-2)Torque Γ (Nm not J since F and r do not act along the same straight line)The turning effect of a force. Imagine a spanner or wrench tightening a bolt.If a force F of 120N is applied perpendicular to the lever arm at a distance r of 0.2m, then torque Γ = Fr (you learned this in MYP as the ‘moment’ or ‘turning effect’ of the force)More precisely, Γ = Frsin ???For max torque, ????900 so sin 900 ???In the second diagram, the torque is reduced because the effort force is no longer at 900 to the lever. Find the angle between the effort force and the lever.Solution: sin(angle required) =15/20 , so sin-1 0.75 = 48.60. Thus Γ= 120 x 0.15 = 18NmThe longer the lever, the greater the torque it can exert.The idea of a couple. The torque (or turning effect) due to a couple is the force x perpendicular distance between the two forces.1190928304700The idea of equilibrium. (exam: “what is meant by…”)An object in equilibrium has no net ‘influences’ to cause it to move in a straight line or to rotate. In other words, the sum of all forces acting is zero and also the sum of all torques about a point is also zero. IOW, it stays in a state of uniform motion in a straight line or is at rest (N1) and is not rotating about an axis.At an instant (i): if these two conditions are satisfied the object or system is in equilibrium.22218653111500Examples: force only: try these for revisionBalancing torquesRemember: it is the sum of all the ACW torques which equals the sum of the CW torques. Consider the example top left, sometimes called the Decorator Problem. Here it is, expanded a bit. A decorator stands on a plank held up by two ropes. In this case, the mass of the plank has been neglected. If it wasn’t, the weight force of the plank would act at its mid-point as the upper diagram shows and would itself produce a torque about A or B.Clearly T1> T2 because the decorator’s weight W is closer to T1The trick is to take moments about A, (or indeed B, it doesn’t matter), then solve as shown, as long as we know the numerical values of W, d1 and d2.5143510096500Example: decorator weight =800N, stands on a plank 4m long, ropes 1m from each end.Plank weight = 40N Find T1, T2 if d1 = 0.5m ................
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