Transcript Chapter 11

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”
 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
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
Ch-11 Check Point 1
 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 11-3 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
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
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)
Ch-11 Check Point 2
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 11-5 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)
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
Ch-11 Check Point 3
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
(a)=rfsin =0 (=0, 180)
(b)–i = k x F, i.e. F along j
(c) –j=k x F i.e. F along -i
Ch-11 Check Point 4
 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.
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
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
Ch-11 Check Point 5
 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 11-7 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)
Ch-11 Check Point 6
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 11-11: 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