chapter 11 n.
Skip this Video
Loading SlideShow in 5 Seconds..
CHAPTER-11 PowerPoint Presentation
Download Presentation

Loading in 2 Seconds...

play fullscreen
1 / 17

CHAPTER-11 - PowerPoint PPT Presentation

  • Uploaded on

CHAPTER-11. Rolling, Torque, and Angular Momentum. Ch 11-2 Rolling as Translational and Rotation Combined. Rolling Motion Rotation of a rigid body about an axis not fixed in space Smooth Rolling: Rolling motion without slipping Motion of com “O” and point “P”

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
    Presentation Transcript
    1. CHAPTER-11 Rolling, Torque, and Angular Momentum

    2. Ch 11-2 Rolling as Translational and Rotation Combined • Rolling Motion Rotation of a rigid body about an axis not fixed in space • Smooth Rolling: Rolling motion without slipping Motion of com “O” and point “P” • When the wheel rotates through angle , P moves through an arc length s given by s=R  Differentiating with respect to t We get ds/dt= R d/dt • vcom= R

    3. Ch 11-2 Rolling as Translational and Rotation Combined • Rolling motion of a rigid body: Purely rotational motion + Purely translational mption • Pure rotational motion:all points move with same angular velocity . • Points on the edge have velocity vcom= R with vtop= + vcom and vbot=- vcom • Pure translational motion:All points on the wheel move towards right with same velocity vcom

    4. The rear wheel on a clowns’ bicycle has twice the radius of the front wheel. (a) When the bicycle is moving , is the linear speed at the very top of the rear wheel greater than, less than, or the same as that of the very top of the front wheel? (b) Is the angular speed of the rear wheel greater than, less than, or the same as that of the front wheel? 1. (a) vtop-front=vtop-rear=2 vcom same; (b) vtop-front= vtop-rear = 2frontRfront= 2rearRrear rear/ front = Rfront /Rrear Rrear = 2 Rfront rear/ front = Rfront /Rrear= 1/2 rear < front less Ch-11 Check Point 1

    5. Ch 11-3 Kinetic Energy of Rolling • Rolling as a Pure Rotation about an axis through P Kinetic energy of rolling wheel rotating about an axis through P • K= (IP 2)/2 where IP= Icom+MR2 and R = vcom • K= (IP 2)/2= (Icom 2+MR2 2)/2 K= (Icom 2)/2 + (Mv2com)/2 K= KRot+KTrans

    6. Ch 11-4 The Forces of Rolling • In smooth rolling, static frictional force fs opposes the sliding force at point P Vcom=R; d/dt(Vcom)=d/dt(R) acom=R d/dt=R • Accelerating Torque acting clockwise; static frictional force fs tendency to rotate counter clockwise

    7. Ch 11-4-cont. Rolling Down a Ramp Rigid cylinder rolling down an incline plane,acom-x=? Components of force along the incline plane (upward) and perpendicular to plane Sliding force downward-static friction force upward; opposite trends fs-Mgsin=Macom-x; acom-x=(fs/M)-gsin To calculate fs apply Newtons Second Law for angular motion: Net torque= I Torque of fs about body com: fsR= I But =-acom-x/R; then fs =Icom/R=-Icomacom-x/R2 acom-x=(fs/M)-gsin =(-Icomacom-x/MR2)- gsin acom-x(1+Icom /MR2) = - gsin acom-x= - gsin/(1+Icom /MR2)

    8. Disk A and B are identical and rolls across a floor with equal speeds. The disk A rolls up an incline, reaching a maximum height h, and disk B moves up an incline that is identical except that is frictionless. Is the maximum height reached by disk B greater than, less than or equal to h? A is rolling and its kinetic energy before decent KA= Icom2 /2+ M(vcom)2/2 KB= M(vcom)2/2 vB<vA Height h of incline, given by conservation of mechanical energy K= - Ug; h=v2/2g hB<hA because vB<vA Ch-11 Check Point 2

    9. Ch 11-5 The Yo-Yo • Yo-Yo is Physics teaching Lab. • Yo-Yo rolls down its string for a distance h and then climbs back up. • During rolling down yo-yo loses potential energy (mgh) and gains translational kinetic energy (mv2com/2) and rotational kinetic energy ( Icom2/2). • As it climbs up it loses translational kinetic energy and gains potential energy . • For yo-yo, equations of incline plane modify to =90 • acom=- g/(1+Icom /MR02)

    10. Ch 11-6 The Torque Revisited • =r xF =r Fsin  = r F= r F • Vector product =r x F • =i j k x y z  Fx Fy Fz

    11. The position vector r of a particle points along the positive direction of a z-axis. If the torque on the particle is (a) zero (b) in the negative direction of x and (c) in the negative direction of y, in what direction is the force producing the torque =rxF=rfsin =rfsin =0 (=0, 180) –i = k x F, i.e. F along j (c) –j=k x F i.e. F along -i Ch-11 Check Point 3

    12. In part a of the figure, particles 1 and 2 move around point O in opposite directions, in circles with radii 2m and 4m . In part b, particles 3 and 4 travel in the same direction along straight lines at perpendicular distance of 4m and 2m from O. Particle 5 move directly away from O. All five particles have the same mass and same constant speed. (a) Rank the particles according to magnitude of their angular ,momentum about point O, greatest first (b) which particles have negative angular momentum about point O. Ch-11 Check Point 4 L = rmv r= 4m for 1 and 3 =2m for 2 and 4 =0 for 5 Ans: (a) 1 and 3 tie, then 2 and 4 tie, then 5 (zero); (b) 2 and 3

    13. Ch 11-7,8,9 Angular Momentum • l =r x p =rp sin = r p= r p • Newtons Second Law: Fnet= dp/dt; net= dl/dt For system of particles L=li ; net= dL/dt

    14. The figure shows the position vector r of a particle at a certain instant, and four choices for the directions of force that is to accelerate the particle. All four choice lie in the xy plane. (a) Rank the choices according to the magnitude of the time rate of change (dl/dt) they produce in the angular momentum f the particle about point O, greatest first (b) Which choice results in a negative rate of change about O?  =(dl/dt)=rxF 1 = 3 = |rxF1|= |rxF3| and 2 = 4 =0 Ch-11 Check Point 5

    15. Ch 11-7 Angular Momentum of a Rigid Body Rotating about a Fixed Axis • Magnitude of angular momentum of mass mi li= ri x pi =ri pi sin90= ri mivi li  ( ri and pi) • Component of li along Z-axis liZ= li sin  =ri sin90mivi=ri mivi vi = ri  liZ=ri mivi=ri mi (ri )=ri 2mi  • Lz =  liZ= (ri 2mi ) =I  • (rigid body fixed axis)

    16. In the figure, a disk, a hoop and a solid sphere are made to spin about fixed central axis (like a top) by means of strings wrapped around them, with the string producing the same constant tangential force F on all three objects. The three objects have the same mass and radius, and they are initially stationary. Rank the objects according to (a) angular momentum about their central axis (b) their angular speed, greatest first, when the string has been pulled for a certain time t. net =dl/dt=FR; l=net x t Since net =FR for all three objects, lhoop=ldisk=lsphere f=i+t; net=I=FR; =FR/I i=0; f=i+t= t=FRt/I f=t=FRt/I Ihoop=MR2 ; IDisk=MR2/2; Isphere = 2/5 MR2 f-hoop =FRt/Ihoop =FRt/MR2 f-Disk =FRt/IDisk =2(FRt/MR2) f-Sphere =FRt/ISphere =5(FRt/MR2)/2 Sphere, Disk and hoop angular speed Ch-11 Check Point 6

    17. Ch 11-11: Conservation of Angular momentum • Newtons Second Law in angular form: net= dL/dt If net= 0 then L = a constant (isolated system) • Law of conservation of angular momentum: Li = L Ii i = If f