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# Physics 1501: Lecture 21 Today ’ s Agenda

Physics 1501: Lecture 21 Today ’ s Agenda. Sec. 1-7. Announcements HW#8: due Oct. 28 Honors ’ students see me after class Midterm 1: average ~ 45 % … Topics Torque Rotational energy Rolling motion . 0 10 20 30 40 50 60 70 80 90 100. Sec. 21-28. Télécharger la présentation ## Physics 1501: Lecture 21 Today ’ s Agenda

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1. Physics 1501: Lecture 21Today’s Agenda Sec. 1-7 • Announcements • HW#8: due Oct. 28 • Honors’ students • see me after class • Midterm 1: average ~ 45 % … • Topics • Torque • Rotational energy • Rolling motion 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

2. Summary (with comparison to 1-D kinematics) Angular Linear And for a point at a distance R from the rotation axis: • x = Rv = R a = R

3. Rotation & Kinetic Energy... • 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. y x z y x z Direction of Rotation: • In general, the rotation variables are vectors (have a direction) • If the plane of rotation is in the x-y plane, then the convention is • CCW rotation is in the + zdirection • CW rotation is in the- zdirection

5. Rotational Dynamics:What makes it spin? TOT=I • This is the rotational version of FTOT = ma • Torque is the rotational cousin of force: • The amount of “twist” provided by a force. • Moment of inertia I is the rotational cousin of mass. • If I is big, more torque is required to achieve a given angular acceleration. • Torque has units of kg m2/s2 = (kg m/s2) m = Nm.

6. Comment on=I • When we write =I we are really talking about the z component of a more general vector equation. (we normally choose the z-axis to be the the rotation axis.) z=Izz • We usually omit the zsubscript for simplicity. z Iz z z

7. Example • To loosen a stuck nut, a man pulls at an angle of 45o on the end of a 50cm wrench with a force of 200 N. • What is the magnitude of the torque on the nut? • If the nut suddenly turns freely, what is the angular acceleration of the wrench? (The wrenchhas a mass of 3 kg, and its shape is that of a thin rod). 45o F=200N L=0.5m

8. So =/I = (70.7Nm) / (.25kgm2) = 283 rad/s2 Example: solution • Torque =Lfsin = (0.5m)(200N)(sin(45)) = 70.7 Nm • If the nut turns freely (no more friction), =I • We know  and we want , so we need to figure out I. 45o F=200N L=0.5m 

9. Torque and the Right Hand Rule: • The right hand rule can tell you the direction of torque: • Point your hand along the direction from the axis to the point where the force is applied. • Curl your fingers in the direction of the force. • Your thumb will point in the directionof the torque. F y r x  z

10. B  A C Review: The Cross Product • We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. • The cross product of two vectors is a third vector: AXB=C • The length of C is given by: C = ABsin  • The direction of C is perpendicular to the plane defined by A and B, and inthe direction defined by the right-handrule.

11. The Cross Product j • The cross product of unit vectors: ixi=0 ixj =k ix k = -j jxi = -k jx j =0 jx k =i kxi =j kx j =-i kx k =0 AXB = (AX i + AYj+ Azk) X (BX i + BYj+ Bzk) = (AX BX i x i +AX BY i xj+AX BZ i xk) + (AY BX jx i +AY BY jxj+ AY BZ jxk) + (AZ BX kx i +AZ BY kxj +AZ BZ kxk) i k

12. The Cross Product • Cartesian components of the cross product: C =AXB CX = AY BZ - BY AZ CY = AZ BX - BZ AX CZ = AX BY - BX AY B A C Note: B XA =- A X B

13. y x z Torque & the Cross Product: • So we can define torque as:  = r x F = r F sin  X = y FZ - z FY Y = z FX - x FZ Z = x FY - y FX F f  r

14. Rotational Work F • Work done by a constant torque in turning an object by an angle  is • Consider a wheel of radius R with a rope wrapped around it • You pull the rope with a constant tension T • W = F d cos  • Here d and F are parallel • W = F d • Total displacement is d = s =R  so • W = F R  • But FR =  since F and R are perpendicular • W =   s  F R

15. Work • 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: • 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 =  • Analogue of W= F•r • W will be negative if  and  have opposite sign !  F  R dr=Rd d axis

16. Work & Kinetic Energy: • Recall the Work Kinetic-Energy Theorem: K = WNET • This is true in general, and hence applies to rotational motion as well as linear motion. • So for an object that rotates about a fixed axis:

17. M R F Example: Disk & String • A massless string is wrapped 10 timesaround 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). • How fast is the disk spinning after the string has unwound?

18. Recall thatIfor a disk about its central axis is given by: So = 792.5 rad/s Disk & String... WNET = W = 62.8 J = K W = t.q = F x r.q = (10 N)(0.10 m)(10*2p) = 62.8 J M R 

19. Lecture 21, ACT 1Work & 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 r = 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 ? w2 w1 (a)disk 1 (b)disk 2 (c)same F F

20. Rolling Motion • Now consider a cylinder rolling at a constant speed. VCM CM The cylinder is rotating about CM and its CM is moving at constant speed (vCM). Thus its total kinetic energy is given by :

21. Rolling Motion • Consider again a cylinder rolling at a constant speed. Q VCM CM  P At any instant the cylinder is rotating about point P. Its kinetic energy is given by its rotational energy about that point. KTOT = 1/2 IPw2

22. Rolling Motion • We can find IP using the parallel axis theorem V = 2VCM Q V = r w V = VCM CM IP = ICM + MR2  V = 0 P KTOT = 1/2 (ICM + MR2 ) w2 KTOT = 1/2 ICM w2 + 1/2 M (R2w2 ) = 1/2 ICM w2 + 1/2 M VCM2 !

23. Connection with CM motion • For a system of particles, the kinetic energy is : = KCM = KREL • For a solid object rotating about it’s center of mass, we now see that the first term becomes: Substituting but

24. Connection with CM motion... • So for a solid object which rotates about its center of mass and whose CM is moving: VCM 

25. Rolling Motion • Cylinders of different I rolling down an inclined plane: K = - U = Mgh v = 0 = 0 K = 0 R M h v = R

26. Rolling... • If there is no slipping (due to friction): v  v 2v v Where v = R In the lab reference frame In the CM reference frame

27. The rolling speed is always lower than in the case of simple sliding since the kinetic energy is shared between CM motion and rotation. Rolling... hoop: c=1 disk: c=1/2 sphere: c=2/5 etc... Use v= R and I = cMR2. c c So: c c

28. Lecture 21, ACT 2Rolling Motion • A race !! Two cylinders are rolled down a ramp. They have the same radius but different masses, M1 > M2. Which wins the race to the bottom ? A) Cylinder 1 B) Cylinder 2 C) It will be a tie M1 M2 Active Figure h M? q

29. v ? M Example :Rolling Motion • A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? Ball has radius R M M M M M h M q

30. Example :Rolling Motion • Use conservation of energy. Ei = Ui + 0 = Mgh Ef = 0 + Kf = 1/2 Mv2 + 1/2 I w2 = 1/2 Mv2 + 1/2 (1/2MR2)(v/R)2 Mgh = 1/2 Mv2 + 1/4 Mv2 v2 = 4/3 g h v = ( 4/3 g h ) 1/2

31. Consider a roller coaster. We can get the ball to go around the circle without leaving the loop. Note: Radius of loop = R Radius of ball = r

32. 1 2 How high do we have to start the ball ? Use conservation of energy. Also, we must remember that the minimum speed at the top is vtop = (gR)1/2 E1 = mgh + 0 + 0 E2 = mg2R + 1/2 mv2 + 1/2 Iw2 = 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2)(v/r)2 = 2mgR + 1/2 mgR + (2/10)m (gR) = 2.7 mgR

33. 1 2 How high do we have to start the ball ? E1 = mgh + 0 + 0 E2 = 2.7 mgR mgh = 2.7 mgR h = 2.7 R h = 1.35 D (The rolling motion added an extra 2/10 R to the height: without it, h = 2.5 R)

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