Page 1Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 1Lecture 17 (Catch up)Goals:Goals:••Chapter 12Chapter 12 Introduce and analyze torque Understand the equilibrium dynamics of an extended object in response to forces Employ “conservation of angular momentum” conceptAssignment: HW8 due March 17th Thursday, Exam ReviewPhysics 207: Lecture 17, Pg 2Rotational Motion: Statics and Dynamics Forces are still necessary but the outcome dependson the location from the axis of rotation This is in contrast to the translational motion and acceleration of the center of mass. Here the position of these forces doesn’t matter (doesn’t alter the physics we see) However: For rotational statics & dynamics: we must reference the specific position of the force relative to an axis of rotation. It may be necessary to consider more than one rotation axis! Vectors remain the key tool for visualizing Newton’s LawsPage 2Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 3Angular motion can be described by vectors With rotation the distribution of mass matters. Actual result depends on the distance from the axis of rotation. Hence, only the axis of rotation remains fixed in reference to rotation. We find that angular motions may be quantified by defining a vector along the axis of rotation. We can employ the right hand rule to find the vector directionPhysics 207: Lecture 17, Pg 4The Angular Velocity Vector• The magnitude of the angular velocity vector is .• The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule as illustrated above.• As ω increased the vector lengthensPage 3Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 5So: What makes it spin (causes ωωωω) ?τNET = |r| |FTang| |r| |F| sin θ If a force points at the axis of rotation the wheel won’t turn Thus, only the tangential component of the force matters With torque the position & angle of the force mattersFTangentialarFradialFA force applied at a distance from the rotation axis gives a torqueFradialFTangentialθr=|FTang| sin θPhysics 207: Lecture 17, Pg 6Rotational Dynamics: What makes it spin?τNET = |r| |FTang| |r| |F| sin θTorque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2= (kg m/s2) m = N mFTangentialaarFradialFFA force applied at a distance from the rotation axisτNET= r FTang= r m aTang= r m r α= (m r2) αFor every little part of the wheelPage 4Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 7For a point mass τNET= m r2α This is the rotational version of FNET= ma Moment of inertia, Moment of inertia, I ΣΣΣΣimiri2 ,is the rotational is the rotational equivalent of mass.equivalent of mass. If I is big, more torque is required to achieve a given angular acceleration.FTangentialaarFrandialFFThe further a mass is away from this axis the greater the inertia (resistance) to rotation (as wesaw on Thursday)τNET = I αPhysics 207: Lecture 17, Pg 8Rotational Dynamics: What makes it spin?τNET = |r| |FTang| |r| |F| sin θ A constant torque gives constant angular accelerationif and only if the mass distribution and the axis of rotation remain constant.FTangentialarFradialFA force applied at a distance from the rotation axis gives a torquePage 5Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 9Torque, like ω, is a vector quantity Magnitude is given by (1) |r| |F| sin θ(2) |Ftangential| |r|(3) |F| |rperpendicular to line of action | Direction is parallel to the axis of rotation with respect to the “right hand rule” And for a rigid object τ = I αrFF cos(90°−θ) = FTang. rFFradialFarFrθθθθ90°−θr sin θline of actionPhysics 207: Lecture 17, Pg 10Exercise Torque MagnitudeA. Case 1B. Case 2C. Same In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. Remember torque requires F, r and sin θor the tangential force component times perpendicular distanceLLFFaxiscase 1 case 2Page 6Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 11Example: Rotating Rod Again A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is the initial angular acceleration α?LmPhysics 207: Lecture 17, Pg 12Example: Rotating Rod A uniform rod of length L=0.5 m and mass m=1 kg is free …What is its initial angular acceleration ?1. For forces you need to locate the Center of MassCM is at L/2 ( halfway ) and put in the Force on a FBD2. The hinge changes everything!LmmgΣ F = 0 occurs only at the hingebutτz= I αz= - r F sin 90°at the center of mass andIEndαz= - (L/2) mgand solve forαzPage 7Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 13Work (in rotational motion) Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement dθ :where dr =R dθ dW = FTangentialdrdW = (FTangentialR) dθ dW = τ dθ (and with a constant torque) We can integrate this to find: W = τ θ = τ (θf−θi) Analog of W = F •∆r W will be negative if τ and θ have opposite sign !axis ofrotationRFFdr =RdθdθφPhysics 207: Lecture 17, Pg 14StaticsEquilibrium is established when0 motion nalTranslatioNet=ΣFr0 motion RotationalNet=ΣτrIn 3D this implies SIX expressions (x, y & z)Page 8Physics 207 – Lecture 17Physics 207: Lecture 17, Pg 15Example Two children (60 kg and 30 kg) sit on a horizontal teeter-totter. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless. The teeter-totter is a uniform bar of 30 kg its moment of inertia about the support point is 30 kg m2. Assuming you can treat both children as point like particles, what is the initial angular acceleration of the teeter-totter when the large child lifts up their legs off the ground (the smaller child can’t reach)? For the static case:0 motion RotationalNet=ΣτrPhysics 207: Lecture 17, Pg 16Example: Soln. Draw a Free Body diagram (assume g = 10 m/s2) 0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0 0= 2d + 1 – 4 d = 1.5
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