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Honors Physics 1 Class 15 Fall 2013 Rolling motion Noninertial reference frames Fictitious forces 1 Rolling motion We will treat the case where 1) the axis of rotation passes through the center of mass. 2) the axis of rotation does not change direction. Example = Rolling wheel 2 11.2 Rolling 3 11.3 The Kinetic Energy of Rolling If we view the rolling as pure rotation about an axis through P, then (w is the angular speed of the wheel and IP is the rotational inertia of the wheel about the axis through P). Using the parallel-axis theorem (I P= Icom +Mh2), (M is the mass of the wheel, Icom is its rotational inertia about an axis through its center of mass, and R is the wheel’s radius, at a perpendicular distance h). Using the relation vcom =wR, we get: A rolling object, therefore has two types of kinetic energy: 1. a rotational kinetic energy due to its rotation about its center of mass (=½ Icomw2), 2. a translational kinetic energy due to translation of its center of mass (=½ Mv2com) 4 The Forces of Rolling: Friction and Rolling A wheel rolls horizontally without sliding while accelerating with linear acceleration acom. A static frictional force fs acts on the wheel at P, opposing its tendency to slide. The magnitudes of the linear acceleration acom, and the angular acceleration a can be related by: where R is the radius of the wheel. 5 The Forces of Rolling: Rolling Down a Ramp A round uniform body of radius R rolls down a ramp. The forces that act on it are the gravitational force Fg, a normal force FN, and a frictional force fs pointing up the ramp. 6 Sample problem: Rolling Down a Ramp b) What are the magnitude and direction of the frictional force on the ball as it rolls down the ramp? Calculations: First we need to determine the ball’s acceleration acom,x : A uniform ball, of mass M=6.00 kg and radius R, rolls smoothly from rest down a ramp at angle q=30.0°. a) The ball descends a vertical height h=1.20 m to reach the bottom of the ramp. What is its speed at the bottom? Calculations: Where Icom is the ball’s rotational inertia about an axis through its center of mass, vcom is the requested speed at the bottom, and w is the angular speed there. Substituting vcom/R for w, and 2/5 MR2 for Icom, We can now solve for the frictional force: 7 q A wheel and a cube, both of mass M are made to race down a slope. The wheel rolls without slipping. The cube slides without rolling. Which one gets to the bottom first? q Which one has greater KE at the bottom? 8 Energy analysis of rolling down a ramp q q a) A cubical block with mass M and side 2R is allowed to slide without rolling down a ramp/slide from height h to height 0. How fast does the center of mass travel at the bottom of the slope? b) A wheel with mass M and radius R is allowed to roll without slipping on a ramp of height h to the level ground at height 0. How fast does its center of mass travel at the bottom of the ramp? 9 The Yo-Yo 10 Galilean transformations Consider measurements of the motion of a mass m made from each of two coordinate systems. For simplicity let’s assume that corresponding axes are parallel and the scale units (time and distance) are the same. The origins of the two systems are displaced by S so r ra S If the two systems move at a fixed velocity V with respect to one another, then V dS dt Observer a sees the object moving with velocity va Observer sees the object moving with velocity v dra . dt dr dra V. dt dt 11 Galilean transformations 2 Observer a measures acceleration aa Observer measures acceleration a d 2 ra dt 2 d 2 r . d 2 r dV . 2 2 dt dt dt If V is constant then aa a and Fa maa ma F and Newton's second law is the same. 12 XKCD LINK 13 Linearly Accelerating Frame Fictitious forces How is Newton’s second law affected by measurement in accelerating frames? a aa A assuming constant A Fa maa F maa mA Fa mA Assuming the frame a is inertial, then the frame observer measures an additional force. F Fa F fict 14 Example: Hanging mass in a car A small weight of mass m hangs from a string in a car which accelerates at rate A=1/4g. a) What is the static angle of the string from the vertical? b) What is the tension in the string? q T T cosq mg 0 T sin q mA 0 tan q Tfict mg mA A mg g T m( g 2 A2 )1/2 15 Principle of Equivalence In the example problem, we treated acceleration A in the same way as we treated gravitational acceleration. The Principle of Equivalence states that there is no way to distinguish locally* between a gravitational acceleration and an acceleration of the coordinate system. *Locally means that we don’t look outside the system for the cause of an acceleration or gravity. 16 Rotating motion and fictitious forces Consider two reference frames, one rotating about the z axis, the other inertial. Origins coincide. Z axes coincide. What fictitious forces are observed in the rotating frame? We already have some experience with this from analyzing forces in the spinning terror ride and centripetal acceleration. 17 Rotating frames Let's consider the time derivatives of a changing particle position in inertial and rotating (primed parameters) frames. Let the x,y,z and x',y',z' coordinates of the particle coincide at t=0. At a later time r (t t ) r (t ) r so r r (t t ) r (t ). Since observer' and observer(inertial) started in the same position, r ' r (t t ) r '(t ) From the drawing we see that r r ' r '(t ) - r (t ) 18 Rotating frames We can easily calculate the difference between r (t ) and r '(t ) : r '(t ) r (t ) r t dr dr ' r or vinertial vrot r dt dt This approach was general and applies to the time derivative of so any vector, so we can apply it to v and a. dv dv a inertial in in vin dt in dt rot and using vin vrot r we have d ain vrot r vrot r dt rot arot ain 2 vrot r From which, F fict m 2 vrot m r 19 Fictitious forces in rotating frames F fict m2 vrot m r with vrot vin r m r centrifugal force m2 vrot Coriolis force 20