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
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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 τ cm about Center of Mass Torque about center of mass produces angular acceleration about center of mass τ cm = I cm αcm I cm α cm 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 F ext = m a cm
Torque as a Vector Force F P exerted at a point P on a rigid body. Vector r S,P from a point S to the point P. Torque about point S due to the force exerted at point P: τ = r F S S,P P
Concept Q.: Vector (cross) Product Consider a right-handed coordinate system with unit vectors (î, ĵ, ˆk) ˆk ĵ. The vector product is equal to 1. î 2. î 3. ĵ 4. ĵ 5. ˆk 6. ˆk 7. 0 8. 0 9. 1
Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors A B = A B sinθ = A ( B sinθ ) = ( A sinθ ) B (0 θ π ) Direction: determined by the Right-Hand-Rule
Vector Product of Unit Vectors Unit vectors in Cartesian coordinates ˆi ˆj = ˆi ˆj sin π 2 = 1 ( ) ˆi ˆi = ˆi ˆj sin(0) = 0 î ĵ = ˆk î î = 0 ĵ ˆk = î ĵ ĵ = 0 ˆk î = ĵ ˆk ˆk = 0
Components of Cross Product A = A ˆi + A ˆj + A kˆ, B = B ˆi + B ˆj + B kˆ x y z x y z A B = ( A B A B ) ˆi + ( A B A B ) ˆj + ( A B A B ) kˆ = y z z y z x x z x y y x ˆi ˆj kˆ A A A x y z B B B x y z
Concept Question: Torque Consider two vectors with x > 0 and with F x > 0 and F z > 0. The cross product points in the r = xî F = r F F x î + F zˆk 1) + x-direction 2) -x-direction 3) +y-direction 4) -y-direction 5) +z-direction 6) -z-direction 7) None of the above directions
Torque: Magnitude and Direction Magnitude of torque about a point S due to force acting at point P τ S = r S,P F P τ S F τ S = rf = rf sinθ where is the magnitude of the force. F P Direction of torque: Perpendicular to the plane formed by and, and determined by the right-hand-rule. F P rs,p
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. LF sinφ LF cosφ LF tanφ 4. LFcotanφ 5. None of the above.
Torque 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. τ S,grav = = N r S,i F grav,i = i=1 1 N m totat m i rs,i i=1 = R S,cm m totat g N N r S,i m ig = m rs i,i g i=1 mtotat g i=1
Con. Q: Torque Due to Uniform 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? Gravitational Force
Fixed Axis Rotation Kinematics Angle variable Angular velocity θ ω = (dθ / dt) ˆk Angular acceleration α = (d 2 θ / dt 2 ) ˆk Mass element Radius of orbit Moment of inertia Δm i r i i=n I S = Δm i (r i ) 2 I S = dm(r dm ) 2 i=1 body Parallel Axis Theorem I S = Md 2 + I cm
Fixed Axis Rotational Dynamics
z-component of Torque about Point S Force: Torque about S: F i = F θ,i ˆθ + Fr,iˆr + F z,i ˆk r S,i = r iˆr + z i ˆk τ S,i = (r iˆr + z i ˆk) (F θ,i ˆθ + Fr,iˆr + F z,i ˆk) = r i F θ,i ˆk + (z i F r,i r i F z,i )ˆθ z i F θ,iˆr Tangential force on mass element produces z-component of torque ( τ S,i ) z = r i F θ,i ˆk
Newton s Second Law and Torque Newton s Second Law: F θ,i = Δm i r i α z z-component of Torque: (τ S,i ) z = r i F θ,i = Δm i r i 2 α z z-component of the torque about an axis passing through S is the sum over all mass elements τ S,z = (τ S,1 ) z + (τ S,2 ) z + = i=n i=n 2 (τ S,i ) z = Δm i r i α z i=1 i=1
Central 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 ext τ S,z i=n = (τ S,i ) z i=1 However, so far no isolated system has been encountered such that the internal torques do not cancel in pairs. F i, j ( r S,i r S, 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 ext τ S,z i=n 2 = Δm i r i α z Recall: Moment of Inertia about and axis passing through S : i=1 i=n i=1 I S = Δm i r i 2 Summary: ext τ S,z = I S α z
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 R d and mass m d. The object is released and takes a time t 1 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: F T m d g = 0 mg T = ma T = m(g a) Constraint: a = R d α α = a / R d Rotational equation: R d T = I cm α R d T = I cm α R d m(g a) = I cm a / R d Solve for moment of inertia: I cm = mr d 2 ( g a 1) Time to travel distance s: t 1 = 2s a 1 a = t 2 1 2s I cm = mr 2 d ( gt 2 1 2s 1)
Demo: Moment of Inertia Wheel Measuring the moment of inertia. Radius of disc: Mass of disc: R d = 0.50 m m d = 5.223 kg Mass of weight holder: m = 0.150 kg Theoretical result: I cm = 1 2 m R 2 d d
Problem Solving Strategy: Fixed Axis Rotation Step 1: Draw free body force diagrams for each object and indicate the point of application of each force Step 2: Select point to compute torque about (generally select center of mass) Step 3: Choose coordinate system. Use right-hand rule for consistency between choice of positive direction for increasing rotational angle and positive direction for angular acceleration. Step 4: Apply Newton s Second Law and Torque Law to obtain equations Step 5: Look for constraint condition between rotational acceleration and any linear accelerations. Step 6: Design algebraic strategy to find quantities of interest
Rotor Moment of Inertia Experiment 3 and 4 set-up: string wrapped around rotor and pulley and connected to falling weight
Table Problem: Moment of Inertia Wheel A steel washer is mounted on a cylindrical rotor. The inner radius of the washer is R. A massless string, with an object of mass m attached to the other end, is wrapped around the side of the rotor and passes over a massless pulley. Assume that there is a constant frictional torque about the axis of the rotor. The object is released and falls.we choose coordinates such that as the mass falls, the rotor undergoes an angular acceleration with a positive component α 1 > 0. After the string detaches from the rotor, the rotor coasts to a stop with a component of angular acceleration α 2 < 0. Let g denote the gravitational constant. What is the moment of inertia of the rotor assembly (including the washer) about the rotation axis?
Table Problem: Atwood s Machine A pulley of mass m p, radius R, and moment of inertia I cm about the center of mass, is suspended from a ceiling. An inextensible string of negligible mass is wrapped around the pulley and attached on one end to an object of mass m 1 and on the other end to an object of mass m 2, with m 1 > m 2. At time t = 0, the objects are released from rest. Find the magnitude of the acceleration of the objects.