10/03/2013PHY 113 C Fall Lecture 111 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin 101 Plan for Lecture 11: Chapter 10 – rotational motion 1.Angular variables 2.Rotational energy 3.Moment of inertia 4.Torque

10/03/2013 PHY 113 C Fall Lecture 112

10/03/2013PHY 113 C Fall Lecture 113 Comments about Exam 1 Solutions posted onlineonline Please review the problems while the ideas are “fresh” in your mind iclicker question (serious) In future exams, would you like to have an additional “take” home component? Of course it would be assumed that all portions of the exam would be subject to strict honor code guidelines A.Yes B.No

10/03/2013PHY 113 C Fall Lecture 114 Review: iclicker question In which of the following cases does gravity do positive work? yfyf yiyi yiyi yfyf b. a.

10/03/2013PHY 113 C Fall Lecture 115 Review: iclicker question In which of the following cases does the work of gravity have a great magnitude? rfrf riri riri rfrf b. a.

10/03/2013PHY 113 C Fall Lecture 116 Review of center of mass: Example: 1 kg3 kg 1 m

10/03/2013PHY 113 C Fall Lecture 117 Center of mass example from Webassign Finding the center of mass For a solid object composed of constant density material, the center of mass is located at the center of the object.

10/03/2013PHY 113 C Fall Lecture 118 Another Webassign question from Assignment #9

10/03/2013PHY 113 C Fall Lecture 119 Another Webassign question from Assignment #9

10/03/2013PHY 113 C Fall Lecture 1110 Examples of two-dimensional collision; balls moving on a frictionless surface

10/03/2013PHY 113 C Fall Lecture 1111 Angular motion angular “displacement”   (t) angular “velocity”  angular “acceleration”  “natural” unit == 1 radian Relation to linear variables: s  = r (  f -  i ) v  = r  a  = r  s

10/03/2013PHY 113 C Fall Lecture 1112 Rotation at constant angular velocity   r vv v  = r 

10/03/2013PHY 113 C Fall Lecture 1113 Special case of constant angular acceleration:  =  0 :  t  =  i +   t  t  =  i +  i t +  ½   t 2 (  t    i    t  -  i   r1r1 v 1 =r 1  v 2 =r 2  r2r2

10/03/2013PHY 113 C Fall Lecture 1114 A wheel is initially rotating at a rate of f=30 rev/sec. R

10/03/2013PHY 113 C Fall Lecture 1115 A wheel is initially rotating at a rate of f=30 rev/sec. Because of a constant angular deceleration, the wheel comes to rest in 3 seconds. R

10/03/2013PHY 113 C Fall Lecture 1116 Example: Compact disc motion In a compact disk, each spot on the disk passes the laser-lens system at a constant linear speed of v  = 1.3 m/s.  1 =v  /r 1 =56.5 rad/s  2 =v  /r 2 =22.4 rad/s What is the average angular acceleration of the CD over the time interval  t=4473 s as the spot moves from the inner to outer radii?  = (  2 -  1 )/  t = rad/s 2 11 22

10/03/2013PHY 113 C Fall Lecture 1117 Object rotating with constant angular velocity (  = 0)  v=0 v=R  Kinetic energy associated with rotation: “moment of inertia” R

10/03/2013PHY 113 C Fall Lecture 1118 Moment of inertia: iclicker exercise: Which case has the larger I? A. a B. b

10/03/2013PHY 113 C Fall Lecture 1119 Moment of inertia:

10/03/2013PHY 113 C Fall Lecture 1120 Note that the moment of inertia depends on both a)The position of the rotational axis b)The direction of rotation mm dd I=2md 2 mm dd I=m(2d) 2 =4md 2

10/03/2013PHY 113 C Fall Lecture 1121 iclicker question: Suppose each of the following objects each has the same total mass M and outer radius R and each is rotating counter- clockwise at an constant angular velocity of  =3 rad/s. Which object has the greater kinetic energy? (a) (Solid disk) (b) (circular ring)

10/03/2013PHY 113 C Fall Lecture 1122 Various moments of inertia: solid cylinder: I=1/2 MR 2 solid sphere: I=2/5 MR 2 solid rod: I=1/3 MR 2 R R R

10/03/2013PHY 113 C Fall Lecture 1123 Calculation of moment of inertia: Example -- moment of inertia of solid rod through an axis perpendicular rod and passing through center: R

10/03/2013PHY 113 C Fall Lecture 1124 iclicker exercise: Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first? A B C

10/03/2013PHY 113 C Fall Lecture 1125

10/03/2013PHY 113 C Fall Lecture 1126 iclicker exercise: Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first? A B C

10/03/2013PHY 113 C Fall Lecture 1127 Review of rotational energy associated with a rigid body

10/03/2013PHY 113 C Fall Lecture 1128 Note that for a given center of rotation, any solid object has 3 moments of inertia; some times two or more can be equal j i k iclicker exercise: Which moment of inertia is the smallest? (A) i (B) j (C) k d d mm I B =2md 2 I C =2md 2 I A =0

10/03/2013PHY 113 C Fall Lecture 1129 From Webassign:

10/03/2013PHY 113 C Fall Lecture 1130 CM

10/03/2013PHY 113 C Fall Lecture 1131 iclicker exercise: Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first? A B C

10/03/2013PHY 113 C Fall Lecture 1132 How to make objects rotate. Define torque:  = r x F  = rF sin  r F   F sin  Note: We will define and use the “vector cross product” next time. For now, we focus on the fact that the direction of the torque determines the direction of rotation.

10/03/2013PHY 113 C Fall Lecture 1133 Another example of torque:

10/03/2013PHY 113 C Fall Lecture 1134

10/03/2013PHY 113 C Fall Lecture 1135 Newton’s second law applied to rotational motion Newton’s second law applied to center-of-mass motion mimi FiFi riri didi

10/03/2013PHY 113 C Fall Lecture 1136 An example: A horizontal 800 N merry-go-round is a solid disc of radius 1.50 m and is started from rest by a constant horizontal force of 50 N applied tangentially to the cylinder. Find the kinetic energy of solid cylinder after 3 s. K = ½ I  2  i  t =  t In this case I = ½ m R 2 and  = FR R F

10/03/2013PHY 113 C Fall Lecture 1137 Re-examination of “Atwood’s” machine T1T1 T2T2 T2T2 T1T1 I T 1 -m 1 g = m 1 a T 2 -m 2 g = -m 2 a  =T 2 R – T 1 R = I  = I a/R R

10/03/2013PHY 113 C Fall Lecture 1138 Another example: Two masses connect by a frictionless pulley having moment of inertia I and radius R, are initially separated by h=3m. What is the velocity v=v 2 = -v 1 when the masses are at the same height? m 1 =2kg; m 2 =1kg; I=1kg m 2 ; R=0.2m. m1m1 m2m2 v1v1 v2v2 h h/2

10/03/2013PHY 113 C Fall Lecture 1139 Kinetic energy associated with rotation: Distance to axis of rotation Rolling: Rolling motion reconsidered:

10/03/2013PHY 113 C Fall Lecture 1140 Kinetic energy associated with rotation: Distance to axis of rotation Rolling: Rolling motion reconsidered:

10/03/2013PHY 113 C Fall Lecture 1141 Newton’s law for torque: F fsfs Note that rolling motion is caused by the torque of friction:

10/03/2013PHY 113 C Fall Lecture 1142 Bicycle or automobile wheel:  fsfs

10/03/2013PHY 113 C Fall Lecture 1143 iclicker exercise: What happens when the bicycle skids? A.Too much torque is applied B.Too little torque is applied C.The coefficient of kinetic friction is too small D.The coefficient of static friction is too small E.More than one of these