1 Two-Dimensional Rotational Dynamics 8.01 W09D2 W09D2 Reading Assignment: MIT 8.01 Course Notes: Chapter 17 Two Dimensional Rotational Dynamics Sections 17.1-17.5 Chapter 18 Static Equilibrium Sections 18.1-3 Announcements Problem Set 7 due Week 10 Tuesday at 9 pm in box outside 26-152 Math Review Week 10 Tuesday at 9 pm in 26-152 Exam 2 Regrade Policy If you would like any problem regraded on your exam, please write a note on the cover sheet and submit the exam no later then the following class. Exams have been photocopied so under no circumstances should you change any of your original answers. Any altered exams will automatically be graded zero.2 Rigid Bodies • Rigid body: An extended object in which the distance between any two points in the object is constant in time. • Effect of external forces on rigid body Main Idea: Rotational Motion about Center of Mass Torque about center of mass produces angular acceleration about center of mass is the moment of inertia about the center of mass is the angular acceleration about center of mass Analagous to Newton’s Second Law for Linear Motion Icm τcm= Icmαcm αcm Fext= macm τcmTorque as a Vector Force exerted at a point P on a rigid body. Vector from a point S to the point P. S S ,P Pτ= ×r FPFS ,PrTorque about point S due to the force exerted at point P:3 Concept Q.: Vector (cross) Product Consider a right-handed coordinate system with unit vectors . The vector product is equal to (ˆi,ˆj ,ˆk) ˆk ׈j 1.ˆi 2. −ˆi3.ˆj 4. −ˆj5.ˆk 6. −ˆk7. 0 8.09. 1Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule A ×B =AB sinθ=AB sinθ( )=A sinθ( )B (0 ≤θ≤π)Vector Product of Unit Vectors Unit vectors in Cartesian coordinates ( )ˆ ˆ ˆ ˆ| || | sin 2 1ˆ ˆ ˆ ˆ| || | sin(0) 0π× = =× = =i j i ji i i j ˆi ׈j =ˆkˆi ׈i =0ˆj ׈k =ˆiˆj ׈j =0ˆk ׈i =ˆjˆk ׈k =04 Components of Cross Product x y z x y zˆ ˆ ˆ ˆˆ ˆA A A , B B B= + + = + +A i j k B i j kˆ ˆˆ( ) ( ) ( )ˆ ˆˆy z z y z x x z x y y xx y zx y zA B A B A B A B A B A BA A AB B B× = − + − + −=A B i j ki j kConcept Question: Torque Consider two vectors with x > 0 and with Fx > 0 and Fz > 0 . The cross product points in the 1) + x-direction 2) -x-direction 3) +y-direction 4) -y-direction 5) +z-direction 6) -z-direction 7) None of the above directions r = xˆi r ×F F = Fxˆi + FzˆkTorque: Magnitude and Direction Magnitude of torque about a point S due to force acting at point P τS=rS ,P×FP⇒τS≡τS= rF⊥= rF sinθ Direction of torque: Perpendicular to the plane formed by and , and determined by the right-hand-rule. FP rS ,Pwhere is the magnitude of the force . FP F5 Concept Q.: Magnitude of Torque In the figure, a force of magnitude F is applied to one end of a lever of length L. What is the magnitude of the torque about the point S? 1. 2. 3. 4. 5. None of the above. LF sinφ LF cosφ LF tanφLFcotanφTorque Due to Uniform Gravitational Force The total torque on a rigid body due to the gravitational force can be determined by placing all the gravitational force at the center-of-mass of the object. τS,grav=rS ,i×Fgrav,ii=1N∑=rS ,i× migi=1N∑= mirS ,i×gi=1N∑=1mtotatmirS ,ii=1N∑⎛⎝⎜⎞⎠⎟× mtotatg=RS,cm× mtotatgCon. Q: Torque Due to Uniform Gravitational Force A box, with its center-of-mass off-center as indicated by the dot, is placed on an inclined plane. Which of the four orientations do you think will tip over?6 Fixed Axis Rotation Kinematics Angle variable Angular velocity Angular acceleration Mass element Radius of orbit Moment of inertia Parallel Axis Theorem θ ω = (dθ/ dt)ˆk α = (d2θ/ dt2)ˆk Δmi ri IS= Δmi(ri)2i=1i= N∑→ IS= dm(rdm)2body∫ IS= Md2+ IcmFixed Axis Rotational Dynamics z-Component of Torque about Point S Force: Torque about S: Tangential force on mass element produces z-component of torque Fi= Fθ,iˆθ + Fr ,iˆr + Fz ,iˆk rS ,i= riˆr + ziˆk τS ,i= (riˆr + ziˆk ) × (Fθ,iˆθ + Fr ,iˆr + Fz ,iˆk )= riFθ,iˆk + (ziFr ,i− riFz ,i)ˆθ − ziFθ,iˆr (τS ,i)z= riFθ,iˆk7 Newton’s Second Law and Torque Newton’s Second Law: z-Component of Torque: z-Component of the torque about an axis passing through S is the sum over all mass elements (τS ,i)z= riFθ,i= Δmiri2αz Fθ,i= Δmiriαz τS ,z= (τS ,1)z+ (τS ,2)z+ ⋅⋅⋅ = (τS ,i)zi=1i= N∑= Δmiri2i=1i= N∑⎛⎝⎜⎞⎠⎟αzCentral Forces: Internal Torques Cancel in Pairs If the internal forces between a pair of particles are directed along the line joining the two particles then the torque due to the internal forces cancel in pairs. This is a stronger version of Newton’s Third Law than we have so far used requiring that internal forces are central forces. With this assumption, the total torque is just due to the external forces However, so far no isolated system has been encountered such that the internal torques do not cancel in pairs. τS ,zext= (τS ,i)zi=1i= N∑ Fi, j (rS , i−rS , j)Torque, Moment of Inertia and Angular Acceleration z-Component of the total external torque about an axis passing through S is the sum over all elements Recall: Moment of Inertia about and axis passing through S : Summary: IS= Δmiri2i=1i= N∑ τS ,zext= ISαz τS ,zext= Δmiri2i=1i= N∑⎛⎝⎜⎞⎠⎟αz8 Concept Question: Chrome Inertial Wheel A fixed torque is applied to the shaft of the chrome inertial wheel. If the four weights on the arms are slid out, the component of the angular acceleration along the shaft direction will 1) increase. 2) decrease. 3) remain the same. Worked Example: Moment of Inertia Wheel An object of mass m is attached to a string which is wound around a disc of radius Rd and mass md. The object is released and takes a time t1 to fall a distance s. What is the moment of inertia of the disc? Analysis: Measuring Moment of Inertia Free body force diagrams and force equations: Constraint:
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