Math Review Night Vector Product, Angular Momentum, and TorqueSummary: Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule ( )( )sin sin sin (0 )θ θ θ θ π× = = = ≤ ≤A B A B A B A B Properties of Cross Products ( )( )c c c× = − ×× = × = ×+ × = × + ×A B B AA B A B A BA B C A C B C Cross 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 jˆ ˆ ˆ ˆˆˆ ˆ ˆ ˆˆˆ ˆˆ ˆ ˆ× = × =× = × =× = × =i j k i i 0j k i j j 0k i j k k 0Components 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 kCheckpoint Problem: Vector Product Find a unit vector perpendicular to and . A =ˆi +ˆj −ˆk B = −2ˆi −ˆj + 3ˆkCross Product of Unit Vectors Unit vectors in cylindrical coordinates ˆr ׈θ =ˆkˆθ ׈k =ˆrˆk ׈r =ˆθ A ×B = −B ×A ˆr ׈r =ˆθ ׈θ =ˆk ׈k =0Angular Momentum of a Point Particle Point particle of mass m moving with a velocity Momentum Fix a point S Vector from the point S to the location of the object Angular momentum about the point S SI Unit vm=p vSrS S= ×L r p [kg ⋅ m2⋅ s-1]Cross Product: Angular Momentum of a Point Particle Magnitude: a) moment arm b) Perpendicular momentum S S= ×L r p sinS Sθ=L r p ,sinS Srθ⊥= r,S Sr⊥=L p,sinSpθ⊥= pS Sp⊥=L rAngular Momentum of a Point Particle: Direction Direction: Right Hand RuleWorked Example: Angular Momentum and Cross Product A particle of mass m = 2 kg moves with a uniform velocity At time t, the position vector of the particle with respect ot the point S is Find the direction and the magnitude of the angular momentum about the origin, (the point S) at time t. v = 3.0 m ⋅ s-1ˆi + 3.0 m ⋅ s-1ˆj rS= 2.0 m ˆi + 3.0 mˆjSolution: Angular Momentum and Cross Product The angular momentum vector of the particle about the point S is given by: The direction is in the negative direction, and the magnitude is LS=rS×p =rS× mv= (2.0m ˆi + 3.0m ˆj) × (2kg)(3.0m ⋅ s−1ˆi + 3.0m ⋅ s−1ˆj)= 12kg ⋅ m2⋅ s−1ˆk + 18kg ⋅ m2⋅ s−1(−ˆk)= − 6.0kg ⋅ m2⋅ s−1ˆk.,,× =× = −× = × =i j kj i ki i j j 0 2 16.0 kg m s .S−= ⋅ ⋅LˆkAngular Momentum and Circular Motion of a Point Particle: Fixed axis of rotation: z-axis Angular velocity Velocity Angular momentum about the point S ˆˆˆR Rω ω= × = × =v r k r θω2ˆ ˆ ˆS S Sm Rmv RmR mRω ω= × = × = = =L r p r v k k k ˆω≡ kωCheckpoint Problem: angular momentum of dumbbell A dumbbell is rotating at a constant angular speed about its center (point A). How does the angular momentum about the point B compared to the angular momentum about point A, (as shown in the figure)?Checkpoint Problem: angular momentum of a single particle! A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane. a) Find the magnitude and the direction of the angular momentum relative to the origin. b) Is this angular momentum relative to the origin constant? If yes, why? If no, why is it not constant? 0LTorque 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 ,PrP θ PF,rS PS Torque about point S due to the force exerted at point P:Torque: Magnitude and Direction Magnitude of torque about a point S: τS= rF⊥= rF sinθ Direction of torque: Perpendicular to the plane formed by and . Determined by the Right-Hand-rule. FP rS ,Pwhere is the magnitude of the force . FP FPθ FP rS , PSPθ FPS rS ,PτSCheckpoint Problem: Torque Consider two vectors with x > 0 and with Fx > 0 and Fz > 0 . What is the direction of the cross product ? ˆx=r i×r F F = Fxˆi + FzˆkConditions for Static Equilibrium (1) Translational equilibrium: the sum of the forces acting on the rigid body is zero. (2) Rotational Equilibrium: the vector sum of the torques about any point S in a rigid body is zero. total 1 2...= + + =F F F 0 ,1 ,2...S Sτ τ τ= + + =totalS0 Worked Example: Lever Law Pivoted Lever at Center of Mass in Equilibrium Show that (Lever Law): d1m1g = d2m2gWorked Example: Pivoted Lever Apply Newton’s 2nd law to each body: Third Law: Pivot force: Torque about pivot: Lever Law: N1, B− m1g = 0 N2,B− m2g = 0 N1, B= NB,1N2, B= NB,2 Fpivot= (mB+ m1+ m2)g Fpivot− mBg − NB,1− NB,2= 0 d1N1ˆk − d2N2ˆk =0 d1m1g = d2m2gGeneralized Lever Law F1=F1 ,+F1,⊥F2=F2,+F2,⊥ F1,⊥= F1sinθ1F2,⊥= F2sinθ2 F1 , F1,⊥ F2 , F2,⊥ τcmtotal=τcm,1+τcm,2=0 → (d1F1,⊥− d2F2,⊥)ˆk = 0 d1F1,⊥= d2F2,⊥Problem Solving Strategy Force: 1. Identify System and draw all forces and where they act on Free Body Force Diagram 2. Write down equations for static equilibrium of the forces: sum of forces is zero Torque: 1. Choose point to analyze the torque about. 2. Choose sign convention for torque 3. Calculate torque about that point for each force. (Note sign of torque.) 4. Write down equation corresponding to condition for static equilibrium: sum of torques is zeroCheckpoint Problem: Lever Law Suppose a beam of length s = 1.0 m and mass m = 2.0 kg is balanced on a pivot point that is placed directly beneath the center of the beam. Suppose a mass m1 = 0.3 kg is placed a distance d1 = 0.4 m to the right of the pivot point. A second mass m2 = 0.6 kg is placed a distance d2 to the left of the pivot point to keep the beam static. (1) What is the force that the pivot exerts on the beam? (2) What is the distance d2 that maintains static
View Full Document