1. Physics 1501: Lecture 20, Pg 1 Physics 1501: Lecture 20 Today’s Agenda l Announcements çHW#7: due Oct. 21 l Midterm 1: average ~ 45 % … l Topics çMoments of Inertia çTorque 0 10 20 30 40 50 60 70 80 90 100 Sec. 21-28 0 10 20 30 40 50 60 70 80 90 100 Sec. 1-7

2. Physics 1501: Lecture 20, Pg 2 Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R

3. Physics 1501: Lecture 20, Pg 3 Rotation & Kinetic Energy... Point Particle Rotating System l The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

4. Physics 1501: Lecture 20, 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 ! l So where

5. Physics 1501: Lecture 20, Pg 5 Calculating Moment of Inertia... l For a discrete collection of point masses we found: l For a continuous solid object we have to add up the mr 2 contribution for every infinitesimal mass element dm.  We have to do an integral to find I : r dm

6. Physics 1501: Lecture 20, Pg 6 Lecture 20: Act 1 Moment of Inertia l Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold. çWhich one has the biggest moment of inertia about an axis through its center? same mass & radius solid hollow (a) solid aluminum(b) hollow gold(c) same

7. Physics 1501: Lecture 20, Pg 7 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.

8. Physics 1501: Lecture 20, Pg 8 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 I CM I END Let’s first find I CM :

9. Physics 1501: Lecture 20, Pg 9 Parallel Axis Theorem: Example L D=L/2 M x CM I CM I END So I PARALLEL = I CM + MD 2 We know We get the same result if evaluating:

10. Physics 1501: Lecture 20, Pg 10 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

11. Physics 1501: Lecture 20, Pg 11 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

12. Physics 1501: Lecture 20, Pg 12 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  = m a  = m  r r F  = m r 2  r aaaa  F m r^ ^^ ^ FF l Multiply by r :

13. Physics 1501: Lecture 20, Pg 13 Rotational Dynamics: What makes it spin? r F  = mr 2  use Define torque:  = r F .   is the tangential force F  times the lever arm r. 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. r aaaa  F m r^ ^ FF

14. Physics 1501: Lecture 20, Pg 14 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 .

15. Physics 1501: Lecture 20, Pg 15 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.

16. Physics 1501: Lecture 20, Pg 16 Lecture 20, Example 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. a)How long will it take for the object to hit the ground ? b)What will be the velocity of the object when it hits the ground ? c)What is the tension on the cord ?  M m h =10 m T

17. Physics 1501: Lecture 20, Pg 17 Lecture 20, Example Solution  M m h =10 m T v 2 - v o 2 = 2 a h v = (2 a h) 1/2 v = a / t t = v / a mg - T = a m a = R  I  =  =T R T = I  / R = (1/2) MR 2 a / R 2 T = Ma/2 mg - Ma/2 = a m a = mg / (m+M/2) = 2g/3 a = 2g/3 = 2 x 9.8 m/s 2 / 3 = 6.5 m/s 2 v = (2 a h) 1/2 =(2 x 6.5m/s 2 10m) 1/2 =11m/s t = v / a = 11 m/s / 6.5 m/s 2 = 1.7 s T = Ma/2 = 10kg 6.5m/s 2 / 2 = 32 N

18. Physics 1501: Lecture 20, Pg 18 Torque  = r F   = r Fsin   = r sin  F l Recall the definition of torque: r  r F  FF FrFr  r = “distance of closest approach”

19. Physics 1501: Lecture 20, Pg 19 Torque  = r Fsin  So if  = 0 o, then  = 0 And if  = 90 o, then  = maximum r F r F

20. Physics 1501: Lecture 20, Pg 20 Lecture 20, Act 2 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

21. Physics 1501: Lecture 20, Pg 21 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 

22. Physics 1501: Lecture 20, Pg 22 Review: The Cross Product l We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. çThe cross product of two vectors is a third vector: A B C A X B = C C l The length of C is given by: C = ABsin  C AB l The direction of C is perpendicular to the plane defined by A and B, and in the direction defined by the right-hand rule. ABC 

23. Physics 1501: Lecture 20, Pg 23 The Cross Product l The cross product of unit vectors: i i = 0 i j = k i k = -j i x i = 0 i x j = k i x k = -j j i = -k j j = 0 j k = i j x i = -k j x j = 0 j x k = i k i = j k j = -i k k = 0 k x i = j k x j = -i k x k = 0 i + jk) i + jk) A X B = (A X i + A Y j + A z k) X (B X i + B Y j + B z k) i x i + i x j + i x k) = (A X B X i x i + A X B Y i x j + A X B Z i x k) j x i + j x j + j x k) + (A Y B X j x i + A Y B Y j x j + A Y B Z j x k) k x i + k x j + k x k) + (A Z B X k x i + A Z B Y k x j + A Z B Z k x k) i j k

24. Physics 1501: Lecture 20, Pg 24 The Cross Product l Cartesian components of the cross product: C A B C = A X B C X = A Y B Z - B Y A Z C Y = A Z B X - B Z A X C Z = A X B Y - B X A Y A B C Note: B X A = - A X B

25. Physics 1501: Lecture 20, Pg 25 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 

26. Physics 1501: Lecture 20, Pg 26 Work 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 = F R d  cos(  ) = F R d  cos(90-  ) = F R d  sin(  ) = F R sin(  ) 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  

27. Physics 1501: Lecture 20, Pg 27 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:

28. Physics 1501: Lecture 20, Pg 28 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 through 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? F R M

29. Physics 1501: Lecture 20, Pg 29 Disk & String... W NET = W = 62.8 J =  K Recall that  I  for a disk about its central axis is given by: So  = 792.5 rad/s  R M W =    = F x r.  = (10 N)(0.10 m)(10*2  ) = 62.8 J

30. Physics 1501: Lecture 20, Pg 30 Lecture 20, ACT 3 Work & Energy 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. çWhich 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