Chapter 1: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics
Translational vs Rotational / / 1/ m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv / / 1/ I d dt d dt I L I KE I Work P θ ω θ α ω τ α ω ω τθ τω c t s r v r a r a r Fr L pr θ ω ω α τ Connection
Center of Mass The geometric center or average location of the mass.
Rotational & Translational Motion Objects rotate about their Center of Mass. The Center of Mass Translates as if it were a point particle. v d r CM dt CM
Center of Mass The Center of Mass Translates as if it were a point particle and, if no external forces act on the system, momentum is then conserved. This means: EVEN if the bat EXPLODED into a thousand pieces, all the pieces would move so that the momentum of the CM is conserved that is, the CM continues in the parabolic trajectory!!!! THIS IS VERY VERY IMPORTANT!
System of Particles Center of Mass A projectile is fired into the air and suddenly explodes With no explosion, the projectile would follow the dotted line After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion! If no external forces act on the system, then the velocity of the CM doesn t change!!
Center of Mass: Stability If the Center of Mass is above the base of support the object will be stable. If not, it topples over.
Balance and Stability This dancer balances en pointe by having her center of mass directly over her toes, her base of support. Slide 1-88
Center of Mass The geometric center or average location of the mass. Extended Body: System of Particles: r CM 1 r M dm x CM i i mx M i
Center of Mass The geometric center or average location of the mass. Extended Body: r CM 1 r M dm
Center of Mass of a Solid Object Divide a solid object into many small cells of mass m. As m 0 and is replaced by dm, the sums become Before these can be integrated: dm must be replaced by expressions using dx and dy. Integration limits must be established. Slide 1-41
Example 1. The Center of Mass of a Rod Slide 1-4
Example 1. The Center of Mass of a Rod Slide 1-43
Example Extended Body: 1 rcm r dm M You must generate an expression for the density and the mass differential, dm, from geometry and by analyzing a strip of the sign. We assume the sign has uniform density. If M is the total mass then the total volume and density is given by: x CM V 1 M abt, ρ 1 abt M My dm ρ dv ρ ( ytdx) ytdx dx 1 abt ab 1 x dm M Where a, b and t are the width, height and thickness of the sign, respectively. Then the mass element for the strip shown is: a 3 a 1 My b x ( ) 3 3 0 0 x dx x x dx a M ab ab a a
Newton s nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. τ I α Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.
Torque: Causes Rotations τ Fr sinφ Fd lever arm: d rsinφ The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force The horizontal component of F (F cos φ) has no tendency to produce a rotation
Torque: Causes Rotations τ Fr sinφ Fd The direction convention is: Counterclockwise rotations are positive. Clockwise rotations are negative.
Newton s 1 st Law for Rotation If the sum of the torques is zero, the system is in rotational equilibrium. τ boy 500N 1.5m 750Nm τ 0 τ girl 50N 3m + 750Nm
Torque Is there a difference in torque? (Ignore the mass of the rope) NO! In either case, the lever arm is the same! What is it? 3m
Newton s nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. τ I α Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.
Torque is a Vector! τ r x F The direction is given by the right hand rule where the fingers extend along r and fold into F. The Thumb gives the direction of τ.
The Vector Product The magnitude of C is AB sin θ and is equal to the area of the parallelogram formed by A and B The direction of C is perpendicular to the plane formed by A and B The best way to determine this direction is to use the right-hand rule A B ˆ i ˆ j+ k ˆ ( AB AB ) ( AB AB ) ( AB AB ) y z z y x z z x x y y x
COMPARE! τ r F τ Fr sinφ Fd CROSS PRODUCT F and d must be mutually perpendicular! L r p L mvr sin φ CROSS PRODUCT L and p must be mutually perpendicular! DOT PRODUCT W F d Fd cos θ F and d must be mutually PARALLEL!
A B ˆ i ˆ j+ k ˆ ( AB AB ) ( AB AB ) ( AB AB ) y z z y x z z x x y y x Two vectors lying in the xy plane are given by the equations A 5i + j and B i 3j. The value of AxB is a. 19k b. 11k c. 19k d. 11k e. 10i j
Net external torques τ
Find the Net Torque τ Fr sinφ Fd τ Fd 1 1+ Fd ( 0 N)(.5 m) + (35 N)(1.10msin 60) OR ( 0 N)(.5 m) + (35Ncos30)(1.10 m) F and d must be mutually perpendicular! + 3.3 Nm CCW
Newton s nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. τ I α Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.
NEW: Rotational Inertia The resistance of an object to rotate. The further away the mass is from the axis of rotation, the greater the rotational inertia.
Rotational Inertia The resistance of an object to rotate. Extended Body: System of Particles: I r dm I i i mr
Moments of Inertia of Various Rigid Objects
Rotational Inertia Depends on the axis.
Calculate Rotational Inertia For a system of point particles: I mr I i i i Where r is the distance to the axis of rotation. SI units are kg. m
Moment of Inertia of a Rigid Object: The Rod I r dm
Calculate: Moment of Inertia of a Rigid Object I r dm R dm The trick is to write dm in terms of the density: dm ρ dv Divide the cylinder into concentric shells with radius r, thickness dr and length L: R 1 4 I r dm r ( πρlr dr) πρlr 0 But ρ M/ π RL dv π rl dr I z 1 MR
Newton s nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. τ I α Force thing Inertia thing Acceleration thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.
Tangential and Angular a t dv dt Acceleration ω α d dt ( ω ) d r dt d r dt ω at αr
A 50 N m torque acts on a wheel of moment of inertia 150 kg m. If the wheel starts from rest, how long will it take the wheel to make a quarter turn (90 degrees)? 1 Use : θ ω0t + αt τ Iα α τ 1 θ θ θ ω0t+ αt t α τ / I I t π /rad 50 N m /150kg m 3.1s
a) τ? b) α? m R 4.3kg.314m τ Fr+ Fr 1 ( + 90N 15 N).314m 11Nm τ τ α I 1 mr 11Nm 1/ 4.3kg(.314m) 9. rad / s
Superposition of Inertia The Parallel Axis Th m Superposition: Inertia ADD I I i The theorem states I I CM + MD I is about any axis parallel to the axis through the center of mass of the object I CM is about the axis through the center of mass D is the distance from the center of mass axis to the arbitrary axis
Moment of Inertia for a Rod Rotating The moment of inertia of the rod about its center is The position of the CM is D½ L Therefore, Around One End CM ICM I I + MD 1 1 ML L 1 1 I ML + M ML 1 3 It is easier to rotate a rod about its center than about an end.
Rotational Inertia: Parallel Axis Theorem A uniform rod (mass.0 kg, length 0.60 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 below the horizontal? a. 15 rad/s b. 1 rad/s c. 18 rad/s d. 9 rad/s e. 3 rad/s
Rotational Energy A rotating object has kinetic energy because all particles in the object are in motion. The kinetic energy due to rotation is called rotational kinetic energy. Adding up the individual kinetic energies, and using v i r i ω : 013 Pearson Education, Inc. Slide 1-47
Rotational Energy Define the object s moment of inertia: Then the rotational kinetic energy is simply The units of moment of inertia are kg m. Moment of inertia depends on the axis of rotation. Mass farther from the rotation axis contributes more to the moment of inertia than mass nearer the axis. This is not a new form of energy, merely the familiar kinetic energy of motion written in a new way. 013 Pearson Education, Inc. Slide 1-48
Rotational Inertia Which reaches the bottom first? (Same mass and radius)
Why Solid Cylinder? PROVE IT! ICM mr ICM 1 mr
Rolling The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid The green line shows the path of the center of mass of the object which moves in linear motion.
Rolling without Slipping All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity v CM, and the point P moves with a velocity v CM. (Why?) Fig. 10.8, p.317
For pure rolling motion, (no slipping) as the cylinder rotates through an angle θ, its center moves a linear distance s Rθ with speed v CM. At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs.
The motion of a rolling object can be modeled as a combination of pure translation and pure rotation. Translation: CM moves with v CM. Rotation: All points rotate about P with angular speed ω. Fig. 10.9, p.318
Rolling Without Slipping Friction If an object rolls without slipping then the frictional force that causes the rotation is a static force. If no slipping occurs, the point of contact is momentarily at rest and thus friction is static and does no work on the object. No energy is dissipated and Mechanical Energy is Conserved. If the object slips, then friction is not static, does work on the object and dissipates energy.
Rolling Without Slipping K 1 P I ω Kinetic Energy Use Parallel-axis Th : I I + MR m P CM 1 K ( ICM + MR ) ω 1 1 K I ω + mv CM CM The total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass.