Physics 151: Lecture 21, Pg 1 Physics 151: Lecture 21 Today’s Agenda l Topics çMoments of InertiaCh çTorqueCh. 10.6, 10.7

Physics 151: Lecture 21, Pg 2 Lecture 22, ACT 1 Rotational Definitions Your goofy friend likes to talk in physics speak. She sees a disk spinning and says “ooh, look! There’s a wheel with a negative  and with antiparallel  and  !!” Which of the following is a true statement ? (a) (a) The wheel is spinning counter-clockwise and slowing down. (b) (b) The wheel is spinning counter-clockwise and speeding up. (c) (c) The wheel is spinning clockwise and slowing down. (d) The wheel is spinning clockwise and speeding up 

Physics 151: Lecture 21, Pg 3 Example: l A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s 2. The diameter of the wheel is 40 cm. What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s 2 ? a. 39 cm/s b. 42 cm/s c. 45 cm/s d. 35 cm/s e. 53 cm/s See text: 10.1

Physics 151: Lecture 21, Pg 4 Moment of Inertia Notice that the moment of inertia I depends on the distribution of mass in the system. çThe further the mass is from the rotation axis, the bigger the moment of inertia. l For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics ! See text: 10.4 l So where

Physics 151: Lecture 21, Pg 5 Parallel Axis Theorem Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass is known, = I CM l The moment of inertia about an axis parallel to this axis but a distance R away is given by: I PARALLEL = I CM + MR 2 So if we know I CM, it is easy to calculate the moment of inertia about a parallel axis. See text: 10.5

Physics 151: Lecture 21, Pg 6 Parallel Axis Theorem: Example l Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod. I PARALLEL = I CM + MD 2 L D=L/2 M x CM We know So which agrees with the result from the board. I CM I END See text: 10.5

Physics 151: Lecture 21, Pg 7 Direction of Rotation: l In general, the rotation variables are vectors (have a direction) l If the plane of rotation is in the x-y plane, then the convention is çCCW rotation is in the + z direction çCW rotation is in the - z direction x y z x y z

Physics 151: Lecture 21, Pg 8 Direction of Rotation: The Right Hand Rule l To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector ! l We normally pick the z-axis to be the rotation axis as shown.   =  z   =  z   =  z l For simplicity we omit the subscripts unless explicitly needed. x y z x y z See text: 10.1

Physics 151: Lecture 21, Pg 9 Rotational Dynamics: What makes it spin? l Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the direction at some instant:  a  =  r Now use Newton’s 2nd Law in the  direction:  F  = ma  = m  r r aaaa  F m r^ ^^ See text: 10.6 and 10.7 ^ FF l Multiply by r : rF  = mr 2  video

Physics 151: Lecture 21, Pg 10 Rotational Dynamics: What makes it spin? rF  = mr 2  use Define torque:  = rF .   is the tangential force F  times the lever arm r. r aaaa  F m r^ ^ FF See text: 10.6 and 10.7 l Torque has a direction: ç+ z if it tries to make the system spin CCW. ç- z if it tries to make the system spin CW.

Physics 151: Lecture 21, Pg 11 Rotational Dynamics: What makes it spin? l So for a collection of many particles arranged in a rigid configuration: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3  FF4FF4 FF1FF1 FF3FF3 FF2FF2 ii I Since the particles are connected rigidly, they all have the same . See text: 10.6 and 10.7

Physics 151: Lecture 21, Pg 12 Rotational Dynamics: What makes it spin?   TOT = I  l This is the rotational version of F TOT = ma l Torque is the rotational cousin of force: ç The amount of “twist” provided by a force. Moment of inertia I is the rotational cousin of mass. Moment of inertia I is the rotational cousin of mass.  If I is big, more torque is required to achieve a given angular acceleration. l Torque has units of kg m 2 /s 2 = (kg m/s 2 ) m = Nm. See text: 10.6 and 10.7

Physics 151: Lecture 21, Pg 13 Torque See Figure  = rF   = r Fsin    r sin  F l Recall the definition of torque: r  r F  FF FrFr  r = “distance of closest approach” See text: 10.6, 10.7, 11.2

Physics 151: Lecture 21, Pg 14 Example: l You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is : a. 0. b. mgr. c. 2mgr. d. a function of the angular velocity. e. small at first, then increasing as the Frisbee loses the torque given it by your hand. See text: 10.1

Physics 151: Lecture 21, Pg 15 Torque  = r Fsin  So if  = 0 o, then  = 0 And if  = 90 o, then  = maximum r F r F See Figure See text: 10.6 and 10.7

Physics 151: Lecture 21, Pg 16 Lecture 21, Act 1 Torque l In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. (a) (a) case 1 (b) (b) case 2 (c) (c) same L L FF axis case 1case 2

Physics 151: Lecture 21, Pg 17 Lecture 21, ACT 2 A uniform rod of mass M = 1.2kg and length L = 0.80 m, lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. If a force (F = 5.0 N,  = 40°) acts as shown, what is the resulting angular acceleration about the pivot point ? a. 16 rad/s 2 b. 12 rad/s 2 c. 14 rad/s 2 d. 10 rad/s 2 e. 33 rad/s 2

Physics 151: Lecture 21, Pg 18 Torque and the Right Hand Rule: l The right hand rule can tell you the direction of torque: çPoint your hand along the direction from the axis to the point where the force is applied. çCurl your fingers in the direction of the force. çYour thumb will point in the direction of the torque. r F x y z  See text: 11.2  r F DIRECTION:  =  r X F   r F sin MAGNITUDE:  =  r F sin 

Physics 151: Lecture 21, Pg 19 Torque & the Cross Product: r F x y z  l So we can define torque as:  r F  = r x F = r F sin   X = y F Z - z F Y  Y = z F X - x F Z  Z = x F Y - y F X  See text: 11.2

Physics 151: Lecture 21, Pg 20 How Much WORK is Done ? F Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d  : Fdr  dW = F. dr = FRd  cos(  ) = FRd  cos(90-  ) = FRd  sin(  ) = FRsin(  ) d   dW =  d  We can integrate this to find: W =  l Analogue of W = F  r W will be negative if  and  have opposite sign ! R F dr=Rd  dd axis   See text: 10.8

Physics 151: Lecture 21, Pg 21 Work & Kinetic Energy: Recall the Work Kinetic-Energy Theorem:  K = W NET l This is true in general, and hence applies to rotational motion as well as linear motion. l So for an object that rotates about a fixed axis: See text: 10.8

Physics 151: Lecture 21, Pg 22 Example: Disk & String l A massless string is wrapped 10 times around a disk of mass M=40 g and radius R=10cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F=10N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning). çHow fast is the disk spinning after the string has unwound? See example F R M See text: 10.8  = rad/s

Physics 151: Lecture 21, Pg 23 Lecture 21, ACT 2 Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both are made of identical material (i.e. their density  = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the pull ? (a) (a) disk 1 (b) (b) disk 2 (c) (c) same FF 11 22

Physics 151: Lecture 21, Pg 24 Example 2 l A rope is wrapped around the circumference of a solid disk (R=0.2m) of mass M=10kg and an object of mass m=10 kg is attached to the end of the rope 10m above the ground, as shown in the figure.  M m h =10 m T a)How long will it take for the object to hit the ground ? a)What will be the velocity of the object when it hits the ground ? a)What is the tension on the cord ? 1.7 s 11m/s 32 N

Physics 151: Lecture 21, Pg 25 Example: Rotating Road l A uniform rod of length L=0.5m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is a) angular speed when it reaches the lowest point ? b) initial angular acceleration ? c) initial linear acceleration of its free end ? See example See text: 10.8 L m  = 7.67 rad/s a)  = 30 rad/s 2 b) c) a = 15 m/s 2

Physics 151: Lecture 21, Pg 26 Lecture 22, ACT 2 A campus bird spots a member of an opposing football team in an amusement park. The football player is on a ride where he goes around at angular velocity  at distance R from the center. The bird flies in a horizontal circle above him. Will a dropping the bird releases while flying directly above the person’s head hit him? a. Yes, because it falls straight down. b. Yes, because it maintains the acceleration of the bird as it falls. c. No, because it falls straight down and will land behind the person. d. Yes, because it mainatins the angular velocity of the bird as it falls. e. No, because it maintains the tangential velocity the bird had at the instant it started falling.

Physics 151: Lecture 21, Pg 27 Example: l A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a moment of inertia I = kg m 2. The cord does not slip on the pulley. What is the magnitude of the acceleration of m? a. 2.4 m/s 2 b. 2.8 m/s 2 c.3.2 m/s 2 d. 4.2 m/s 2 e. 1.7 m/s 2 See text: 10.1

Physics 151: Lecture 21, Pg 28 Recap of today’s lecture l Chapter 10, çCalculating moments of inertia çTourque çRight Hand Rule