Two-Dimensional Rotational Dynamics 8.01 W09D2

Today’s Reading Assignment: W09D2 Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3;

Rigid Bodies Rigid body: An extended object in which the distance between any two points in the object is constant in time. Examples: sphere, disk … Effect of external forces (the solid arrows represent forces): Both translate and rotate Translate Rotate

Main Idea: Fixed Axis Rotation of Rigid Body Torque produces angular acceleration about center of mass is the moment of inertial about the center of mass is the angular acceleration about center of mass

Recall: Fixed Axis Rotation Kinematics Angle variable Angular velocity Angular acceleration Mass element Radius of orbit Moment of inertia Parallel Axis Theorem

Torque as a Vector Force exerted at a point P on a rigid body. Vector from a point S to the point P. Torque about point S due to the force exerted at point P:

Summary: Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule

Torque: Magnitude and Direction Magnitude of torque about a point S due to force acting at point P where is the magnitude of the force . Direction of torque: Perpendicular to the plane formed by and .Determined by the Right-Hand-rule.

Properties of Cross Products

Cross Product of Unit Vectors Unit vectors in Cartesian coordinates

Components of Cross Product

Concept Question: Torque Consider two vectors with x > 0 and with Fx > 0 and Fz > 0 . The cross product points in the + x-direction -x-direction +y-direction -y-direction +z-direction -z-direction None of the above directions

Concept Question 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? FL sinθ FL cosθ FL tanθ None of the above

Recall: Rotational Kinematics Individual element of mass Radius of orbit Tangential velocity Tangential acceleration Radial Acceleration

Dynamics: Newton’s Second Law and Torque about S Tangential force on mass element produces torque Newton’s Second Law Torque about S z-component of torque about S

Moment of Inertia and Torque Component of the total torque about an axis passing through S is the sum over all elements Recall: Moment of Inertia about and axis passing through S : Summary:

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. Grammar, punctuation corrected

Torque and Static Equilibrium

Conditions for Static Equilibrium 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.

Concept Question A 1 kg rock is suspended by a massless string from one end of a 1 m measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the 0.25 m from the left end? 0.25 kg. 0.5 kg. 1.0 kg. 2.0 kg. 4.0 kg. 6. Impossible to determine.

Concept Question: Tipping A box, with its center-of-mass off-center as indicated by the dot, is placed on an inclined plane. In which of the four orientations shown, if any, does the box tip over?

Problem Solving Strategy: Static Equilibrium Force: Identify System and draw all forces and where they act on Free Body Force Diagram Write down equations for static equilibrium of the forces: sum of forces is zero Torque: Choose point to analyze the torque about. Choose sign convention for torque Calculate torque about that point for each force. (Note sign of torque.) Write down equation corresponding to condition for static equilibrium: sum of torques is zero

Table Problem: Standing on a Hill A person is standing on a hill that is sloped at an angle α with respect to the horizontal. The person’s legs are separated by a distance d, with one foot uphill and one downhill. The center of mass of the person is at a distance h above the ground, perpendicular to the hillside, midway between the person’s feet. Assume that the coefficient of static friction between the person’s feet and the hill is sufficiently large that the person will not slip. a) What is the magnitude of the normal force on each foot? b) How far must the feet be apart so that the normal force on the upper foot is just zero? (This is the instant when the person starts to rotate and fall over.)

Fixed Axis Rotational Dynamics

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. The object is released and takes a time t 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: Rotational equation: Constraint: Solve for moment of inertia: Time to travel distance s:

Demo: Moment of Inertia Wheel Measuring the moment of inertia.

Properties of Wheel Radius of disc: Mass of disc: Mass of weight holder: Theoretical result:

Rotational Work Tangential force Displacement vector work for a small displacement

Rotational Work Newton’s Second Law Tangential acceleration Work for small displacement Summation becomes integration for continuous body

Rotational Work Rotational work for a small displacement Torque about S Infinitesimal rotational work Integrate total work

Rotational Work-Kinetic Energy Theorem Infinitesimal rotational work Integrate rotational work Kinetic energy of rotation about S

Rotational Power Rotational power is the time rate of doing rotational work Product of the applied torque with the angular velocity

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 increase. decrease. remain the same.

Problem Solving Strategy: Two Dimensional 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. Indicate positive direction for increasing rotational angle. 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

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. As the mass falls, the rotor undergoes an angular acceleration of magnitude a1. After the string detaches from the rotor, the rotor coasts to a stop with an angular acceleration of magnitude a2. Let g denote the gravitational constant. What is the moment of inertia of the rotor assembly (including the washer) about the rotation axis?

Solution: Moment of Inertia of Rotor Force and rotational equations while weight is descending: Constraint: Rotational equation while slowing down Solve for moment of inertia: Speeding up Slowing down

Worked Example: Change in Rotational Energy and Work While the rotor is slowing down, use work-energy techniques to find frictional torque on the rotor.

Next Reading Assignment: W09D3 Young and Freedman: 1.10 (Vector Product) 10.1-10.2, 10.5-10.6 ; 11.1-11.3