Rotation Kinematics Rigid Bodies Kinetic Energy featuring moments of Inertia Torque Rolling
Angular Motion We think about rotation in the same basic way we do about linear motion How far does it go? How fast does it get there? In other words, displacement, speed and acceleration (somewhat modified since displacement is ultimately zero)
Angular Motion Consider the reference point P in the figure. How far does it move when the disc is rotated? s=arclength=r How fast does it move through that arc? ds d =r dt dt r is constant for a rigid body
Angular Motion ds d =r dt dt What is ds? dt It has units of distance over time so it must be a speed. At any given moment, what is the speed of the point P? v P =v R v T =0 v T v P =v T So the speed with which the point P moves along the arc s is just the tangential velocity
Angular Motion d v T =r dt What is d? dt Don't have a nice analog for it. It has units of rad/s (or degrees/s). We give it a special name omega vt d = =angular velocity= dt r If ω is not constant then d dv t a T = = angular acceleration dt r dt r
Angular Motion Both α and ω are vectors because tangential velocity & acceleration are vectors. So we have to define the positive and negative directions for them in a consistent fashion. y The convention: A counter-clockwise rotation is a positive angular velocity Δθ A clockwise rotation is a negative angular velocity x This gives a positive angular velocity α is positive if ω is increasing in magnitude, and negative if it's decreasing
Right-Hand Rule Choose a rotation axis (may be obvious from the problem, may not be). Curl the fingers of your right hand around the axis in the direction of rotation. If your thumb points away from the is positive. If it points object toward the object then is negative
Kinematics Linear Translational Rotational (rigid body) x f =x 0 v 0 t a t f = 0 0 t t v f =v 0 a t f f = 0 t 0 f v =v a x 0 = for constant accelerations a & α Relations for a point on a rigid body under rotation v=r at =r v ac = =r r
Kinetic Energy Any point on a rigid body with K = m v mass mi has a kinetic energy i i i The entire body has kinetic energy K R = K i = mi v i = mi r i since all points on the body rotate with the same rotational speed moment of inertia K R = I i i I = m r
Moments of Inertia In the usual fashion, for continuous rather than discrete distributions of mass... I = r dm If the density of something is known, then it might be easier to use I = r dv
Moments of Inertia Does this make sense? Consider the following... m Our old friend the ball on a string R K R = v ω m R I K T = m v
Moments of Inertia Let's consider the moment (in time) before the string is cut K R = I K R = mr i i What is I? I = m r =m R What is ω? =v / R v = m v =K T R The kinetic energy just after the string is cut
Moments of Inertia v R v ω m m v m m ω m m R R K = I I K = I m v K T = m v m v Which tells us that moments of inertia are additive
Moments of Inertia What about this? Same general problem, but now the mass of the rod is not negligible ω m mr R What's the kinetic energy? K R = I R I m = 3 m R m R =3mR
Parallel Axis Theorem I =I CM md where D is the distance from the center of mass to the axis of rotation From the table 0. I Rod = ml What is the moment of inertia for the same rod rotating about one end? I =I CM m D = m L m L = ml 4 =3mL which is also in Table 0.
Torque Torque describes the tendency of a thing to rotate due to a force applied some distance from a rotation axis F = r F r the magnitude of the torque is =r F sin r is called the moment arm of the force If the torque tends to produce a CCW rotation it's positive, if it tends to produce a CW rotation it's negative
Torque 0 F r 0 F r Right-hand rule for torque follows from the rule for ω Lay your right hand along r, and curl your fingers in the direction of F. If your thumb points away from the object τ is positive. If it points toward the object τ is negative
Net Torque F T =m at = F T r= mat r recall that at =r = mr r= mr =I =I This generalizes to ext =I Newtons nd law for rotation
Work & Power Work s dw = F d s= F sin r d W =F dw = d dw d Power P= = = dt dt d d d d ext =I =I =I =I dt d dt d ext d =dw =I d Tiny rotation small enough that F is constant and s is straight f W = i I d = Iw f i I w = K R Work-Kinetic Energy Theorem for rotation
Example Pulley A pulley of radius R, and mass M is mounted on a frictionless horizontal axle as shown. A light cord is wrapped around the pulley and connected to an object of mass m. When the wheel is released the object falls, the cord unwraps, and the wheel rotates. Find the angular acceleration of the wheel, the acceleration of the object, and the tension in the cord. =I =TR =TR / I The magnitude of the torque on the pulley is What is the net torque? The object is falling with an acceleration F y =mg T =ma mg T TR mg a= =R α= T = m I +mr / I the moment of a wheel is ½ MR =a/ R= T = 3 mg a= g g g =3 = R I /mr R 3R g = g 3 I /mr
Example Atwood Machine How fast are the blocks falling when the pulley is not longer of negligible mass? Assume it is released from rest and the string doesn't slip on the pulley We still have conservation of energy... E f =E i K f U f =K i U i m v f m v f I f m gh m gh=0 m v f m v f I v f / R = m gh m gh v f m m I / R =m gh m gh gh m m v= = m m I / R m m M P f gh m m
Summary