REVIEW #2B Circular Dynamics Statics and Torque Work and Energy Harmonic MotionCircular DynamicsCircular Motion: Position • The position vector of an object moving in a circular orbit of radius R • Components: • The angle θ(t) is a function of time! • Uniform Circular motion:Angular velocity and velocity • ω ≡ dθ dt ⋅ − 1⎡⎣ ⎤⎦ • Not degrees/sec! • Angular velocity rad secUnits Magnitude and direction of velocity • Period (T) and frequency (f):Direction of the velocity direction of the velocity at time t is perpendicular to the position vector i.e. tangent to the circular orbit! ∆r t∆ • directions as ∆t approaches zero. •The Sequence of chordTangential Velocity • Direction of velocity is constantly changing Gives Radial Acceleration • An object traveling with uniform circular motion (ω(t)=ω) is always accelerating towards the center!Uniform Circular Motion: ω(t)=ω 0aθ = • circle at uniform angular velocity. • remains constant. Object is constrained to move in a Magnitude of the velocity (speed) • The tangential acceleration is zero so by Newton’s Second Note: ω(t)=ωLaw, total tangential force acting on the object is zero. • Object is always accelerated towards center!Non-uniform Circular Motion: Acceleration • circular orbit, the acceleration has two components, the tangential component, and the . Radial component Tangential component When an object moves non-uniformly in a radial componentComparison of linear and circular motion at constant acceleration • Linear motion • Circular motionProblem solving strategy involving circular motion • Always has a component of acceleration pointing radially inward. • May or may not have tangential component of acceleration. • Draw a free body diagram for all forces. • Note: mv2/R is not a force but mass times acceleration and does not appear on force diagram. • Choose one unit vectors to point in the radial direction.PRS Question A puck of inertia M is moving in a circle at uniform speed on a frictionless table as shown above. It is held by a string which holds a suspended bob, also of inertia M, at rest below the table. Half of the length of the string is above the tabletop and half below. What is the centripetal acceleration of the moving puck? 1. less than g 2. g 3. greater than g 4. zeroCross Product, Torque, and Static EquilibriumPoint Masses Rigid Bodies Generalization of Newton’s Laws: • External forces accelerate the center of mass • External torques cause angular accelerationTorque Quantitative measure of the tendency of a force to cause or change the rotational motion of a body! Sign convention: • (1) Magnitude of the Counterclockwise always torque about S • (2) Direction S P positive!Moment Arm of the Force • Moment Arm: • Torque:Torque from Tangential Force Break the force into components parallel and perpendicular to the displacement of the force from the axis of the torque: s θP,1,2SSStatic Equilibrium (1) The sum of the forces acting on the rigid body is zero: (2) The vector sum of the torques about any point S in a rigid body is zero: SPRS Question You are using a wrench to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut?Work, Kinetic Energy, and PowerWork-energy Definitions Now we give each term a descriptive name: Kinetic Energy Work Restating the Work-Kinetic Energy relationship conventionally: Kinetic energy is a physical quantity that changes due to work The work Wf,i depends on the path to f from i and the force, Fx and the relative direction of the force and the path The expression for work is only valid for constant forceWork Done Along an Arbitrary Path Key Idea: Add work along incremental path, proportional to For straight path: scalar product of force and increment. Result = line integral r⋅∆ri r Fi ∆Wi = For Curved Path: Add straight increments: i = N rFi = rF i∫ rfpath ( rr ) rdr⋅r⋅∆riW = lim ∑f,o N →∞ →0i =1r∆ri r0PRS Question on Work Consider the work associated with three actions that a person might take: A. Lift a 4 kg mass from the floor to h=1m B. Hold a 5 kg mass 1 m above the floor C. Lower a 3 kg mass from h=2m to the floor Order these actions from greatest to least by the work done: 1. A,B,C 2. C,B,A 3. C,A,B 4. A,C,B 5. None of abovePRS: Work Due to Friction A force Fxa is applied to a block and causes it to slide from x = 0m to x = 5m and back to x = 3m. The block experiences a friction force, Ff = 7N. At the start and finish the block is not moving. The total work done by the friction force on the block is: 1. 21j 2. 49j 3. 56j 4. -21j 5. -49j 6. -56j 7. Need more information about motionInstantaneous Power • limit of the average power ∆W ∆xP = lim = F lim = F vapplied , x ∆t →0 ∆t applied , x x∆t →0 ∆tPotential Energy Total Mechanical EnergyMechanical Energy - Model E f = Ei + W f ,i noncons Similar to Work-Kinetic Energy Only non-conservative Work changes Mechanical Energy Ef ≡ Kf +U (xf ) Definition of total Mechanical Energy W noncons rf noncons ()Usual Definition but only f,o =i∫ path r0 For non-conservatives rdr⋅ Wcons Replaced by Potential Energy (difference) rrrFcons W f ,i ≡−U (xf ) +U (xi )Model: Potential Energy and Force The potential energy must represent the work: cons W f ,0 ≡−U (xf ) +U (x0) i.e. Work depends on endpointsThis is possible ONLY if cons rfW =i∫r0path rF (DEFINITION of CONSERVATIVE) is independent of the path cons ()rdr⋅ rr f,o rF ()f r0 ()any convenient path =− Fcons ()⋅ dx X0 defines zero of U rr()Like an indefinite integralUsually Uxx ∫ rdr⋅x x0 F ()=−dU( x )xx dx rr rfDefine i∫U =− cons Path independence allows this Get Force from U:Potential of Spring x U 2 x F 1 x U4 x U 3 Which graph is potential for spring? x ()=−∫Fcons ()⋅ dxHint: Uxx x0PRS: Energy Curve - Emin 10 6 0 -8 -4 Consider the following sketch of potential energy for a particle as a function of position. There are no dissipative forces or internal sources of energy. What is the minimum total mechanical energy that the particle can have if you know that it has travelled over the entire region of X shown? 1. -8 2. 6 3. 10 4. It depends on direction of travel 5. Can’t say - Potential Energy uncertain by a constantPotential Energy of Spring r •Force: F = Föi =−kxöi x x = xf • Work done: spring 2−kx 1 (W =∫()dx
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