rotação corpo rigido

8/10/2019 rotao corpo rigido

1/17

CHAPTER-11

Rolling, Torque, and Angular Momentum


2/17

Ch 11-2 Rolling as Translational andRotation Combined

Rolling MotionRotation of a rigid body aboutan axis not fixed in spaceSmooth Rolling:

Rolling motion without slippingMotion of com O and point P

When the wheel rotates throughangle , P moves through an arc

length s given bys=R Differentiating with respect to tWe get ds/dt= R d /dt v com= R


3/17

Ch 11-2 Rolling as Translational andRotation 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 v com= R

with v top= + v com and v bot= - v com

Pure translational motion : All points on the wheel move towards rightwith same velocity v com


4/17

Ch-11 Check Point 1

The rear wheel on a clownsbicycle has twice the radiusof the front wheel.(a) When the bicycle ismoving , is the linear speedat the very top of the rearwheel greater than, lessthan, or the same as that ofthe very top of the frontwheel?(b) Is the angular speed ofthe rear wheel greater than,less than, or the same asthat of the front wheel?

1. (a) v top-front =v top-rear =2 v comsame;

(b) v top-front = v top-rear= 2 front Rfront = 2 rear Rrear rear / front = Rfront /R rear

Rrear = 2 R frontrear / front = Rfront /R rear = 1/2rear < front

less


5/17

Ch 11-3 Kinetic Energy of Rolling

Rolling as a Pure Rotation about anaxis through P

Kinetic energy of rolling wheel rotatingabout an axis through P

K= (IP2)/2where I P= I com+MR2 and R = v com

K= (IP 2)/2= (I com 2 +MR2 2)/2K= (Icom 2)/2 + (Mv 2com)/2

K= KRot+KTrans


6/17

Ch 11-4 The Forces of Rolling

In smooth rolling, staticfrictional force f s opposesthe sliding force at point P

Vcom=R ;

d/dt(V com)=d/dt(R )acom=R d /dt=R Accelerating Torque actingclockwise; static frictionalforce f s tendency to rotate

counter clockwise


7/17

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 planeSliding force downward-static friction force

upward; opposite trends

f s-Mgsin =Macom-x ; a com-x = (f s/M)-gsin To calculate f s apply Newtons Second Lawfor angular motion: Net torque= I

Torque of f s about body com: f sR= I But =-a com-x /R; thenf s =I com /R=-I com a com-x /R 2

acom-x =(f s/M)-gsin =(-I com a com-x /MR 2 )- gsin acom-x (1+I com /MR 2 ) = - gsin

acom-x = - gsin /(1+ I com /MR 2 )


8/17

Ch-11 Check Point 2

Disk A and B areidentical and rollsacross a floor with equalspeeds. The disk A rollsup an incline, reaching amaximum height h, anddisk B moves up anincline that is identicalexcept that isfrictionless. Is the

maximum heightreached by disk Bgreater than, less thanor equal to h?

A is rolling and itskinetic energy beforedecent

KA= I com 2 /2+ M(vcom)2/2 K

B= M(v

com)2/2vB


9/17

Ch 11-5 The Yo-Yo

Yo-Yo is Physics teaching Lab. Yo-Yo rolls down its stringfor a distance h and then climbsback up.During rolling down yo-yo losespotential energy (mgh) and gainstranslational kinetic energy(mv 2com/2) and rotational kineticenergy ( I com 2/2).As it climbs up it loses

translational kinetic energy andgains potential energy .For yo-yo, equations of inclineplane modify to =90acom=- g/(1+ I com /MR 0 2 )


10/17

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 F y F z


11/17

The position vector r of aparticle points along thepositive direction of a z-axis.If the torque on the particleis (a) zero(b) in the negative direction ofx and(c) in the negative direction ofy, in what direction is theforce producing the torque

Ch-11 Check Point 3

=rxF =rfsin (a) =rfsin =0 ( =0, 180)(b) i = k x F, i .e. F alon g j

(c) j=k x F i .e . F alo n g -i


12/17

Ch-11 Check Point 4In part a of the figure,

particles 1 and 2 move aroundpoint O in opposite directions,in circles with radii 2m and 4m. In part b, particles 3 and 4travel in the same directionalong straight lines atperpendicular distance of 4mand 2m from O. Particle 5move directly away from O.

All five particles have thesame mass and same constantspeed.

(a) Rank the particlesaccording to magnitude oftheir angular ,momentum aboutpoint O, greatest first

(b) which particles havenegative angular momentumabout point O.

= r mv

r = 4m for 1 and 3

=2m for 2 and 4=0 for 5

Ans: (a) 1 and 3 tie, then 2and 4 tie, then 5 (zero); (b) 2 and 3


13/17

Ch 11-7,8,9 Angular Momentum

l =r x p =rp sin = r p = r p Newtons Second Law:

F net = dp/dt; net = dl/dtFor system of particlesL= l i ; net = dL/dt


14/17

Ch-11 Check Point 5The figure shows theposition vector r of aparticle at a certaininstant, and four choicesfor the directions of forcethat is to accelerate theparticle. All four choice liein the xy plane.(a) Rank the choicesaccording to the magnitudeof the time rate of change(dl/dt) they produce in theangular momentum f theparticle about point O,greatest first(b) Which choice results ina negative rate of changeabout O?

=(dl/dt)=rxF

1 =

3 = |rxF

1|= |rxF

3|

and 2 = 4 =0


15/17

Ch 11-7 Angular Momentum of a RigidBody Rotating about a Fixed Axis

Magnitude of angular momentumof mass m i l i = r i x p i =r i p i sin90= r i m i v il i ( r i and p i )

Component of l i along Z-axisl iZ = l i sin =r i sin90 m i v i =r i m i v i

v i = r i l iZ =r i m i v i =r i m i (r i )=r i 2 m i

L z = l iZ = ( r i 2 m i ) =I (rigid body fixed axis)


16/17

Ch-11 Check Point 6In the figure, a disk, a hoop

and a solid sphere are made tospin about fixed central axis(like a top) by means of stringswrapped around them, with thestring producing the sameconstant tangential force F onall three objects. The threeobjects have the same massand radius, and they areinitially stationary. Rank theobjects according to

(a) angular momentum abouttheir central axis(b) their angular speed,greatest first, when the stringhas been pulled for a certaintime t.

net =dl/dt=FR; l= net x tSince net =FR for all three objects,lhoop=ldisk=lsphere

f= i+ t; net=I =FR; =FR/Ii=0; f= i+ t= t=FRt/If= t=FRt/I

I hoop=MR2 ; I Disk=MR2/2;I sphere = 2/5 MR 2f-hoop =FRt/I hoop =FRt/MR2 f-Disk =FRt/I Disk =2(FRt/MR2)f-Sphere =FRt/I Sphere =5(FRt/MR2)/2

Sphere, Disk and hoop angular speed


17/17

Ch 11-11: Conservation of Angular momentum

Newtons Second Law inangular form:

net = dL/dtIf net = 0 then

L = a constant (isolatedsystem)Law of conservation ofangular momentum:

L i = L I i i = I f f

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