Presentation is loading. Please wait.

Presentation is loading. Please wait.

Two-Dimensional Rotational Dynamics 8.01 W09D2 Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3;

Similar presentations


Presentation on theme: "Two-Dimensional Rotational Dynamics 8.01 W09D2 Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3;"— Presentation transcript:

1 Two-Dimensional Rotational Dynamics 8.01 W09D2 Young and Freedman: 1.10 (Vector Product), , 10.4, ;

2 Announcements Next Reading Assignment Young and Freedman: 1.10 (Vector Product) , ;

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): Translate Rotate Both translate and rotate

4 Table Problem: Pulley System and Energy Using energy techniques, calculate the speed of block 2 as a function of distance that it moves down the inclined plane using energy techniques. Let I P denote the moment of inertia of the pulley about its center of mass. Assume there are no energy losses due to friction and that the rope does not slip around the pulley.

5 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

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

7 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:

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

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

10 Properties of Cross Products

11 Cross Product of Unit Vectors Unit vectors in Cartesian coordinates

12 Components of Cross Product

13 Concept Question: Torque Consider two vectors with x > 0 and with F x > 0 and F z > 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

14 Concept Question: Torque Answer Answer 4. We calculate the cross product noting that in a right handed choice of unit vectors, and Since and, the direction of the cross product is in the negative y-direction.

15 Concept Question: 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.FL sin  2.FL cos  3.FL tan  4.None of the above

16 Concept Question: Magnitude of Torque Answer Answer 2

17 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.

18 Fixed Axis Rotational Dynamics

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

20 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

21 Torque, Moment of Inertia and Angular Acceleration 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:

22 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.

23 Concept Question: Chrome Inertial Wheel Answer Answer 2. Torque about the central axis is proportional to the component of the angular acceleration along that axis. The proportionality constant is the moment of inertia about that axis. By pushing the weights out, the moment of inertia has increased. If the applied torque is constant then the component of the angular acceleration must decrease.

24 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. The object is released and takes a time t to fall a distance s. What is the moment of inertia of the disc?

25 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 :

26 Demo: Moment of Inertia Wheel Measuring the moment of inertia. Radius of disc: Mass of disc: Mass of weight holder: Theoretical result:

27 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

28 Rotor Moment of Inertia

29 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?

30 Torque and Static Equilibrium

31 Conditions 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.

32 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?

33 Concept Question: Tipping Answer Answer 3. In order to tip over, the box must pivot about its bottom left corner. Only in case 3 does the torque about this pivot (due to gravity) rotate the box in such a way that it tips over.

34 Problem Solving Strategy: Static Equilibrium 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 zero

35 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.)

36 Rotational Work Tangential force Displacement vector work for a small displacement

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

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

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

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

41 Table Problem: Change in Rotational Energy and Work Suppose that a rotor of moment of inertia I r is slowing down during the interval [t 1, t 2 ] according to ω(t)= ω(t 1 )-α t, where α = ω(t 1 )/(t 2 ). Use work energy techniques to find the frictional torque acting on the rotor.

42 Table Problem Solution: Change in Rotational Energy and Work Suppose that a rotor of moment of inertia I r is slowing down during the interval [t 1, t 2 ] according to ω(t)= ω(t 1 )-α t, where α = ω(t 1 )/(t 2 ). Use work energy techniques to find the frictional torque acting on the rotor.


Download ppt "Two-Dimensional Rotational Dynamics 8.01 W09D2 Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3;"

Similar presentations


Ads by Google