ROLLING, TORQUE, and ANGULAR MOMENTUMANGULAR MOMENTUM

28 Pages·2009·0.49 MB·English

PH 221-3A Fall 2009
ROLLING, TORQUE, and
AANNGGUULLAARR MMOOMMEENNTTUUMM
Lectures 18-19
Chapter 11
((Hallidayy/Resnick/Walker,, Fundamentals of Phyysics 8th edition))
1
Chapter 11
Rolling, Torque, and Angular Momentum
In this chapter we will cover the following topics:
-Rollingg of circular objjects and its relationshipp with friction
-Redefinition of torque as a vector to describe rotational problems that
are more complicated than the rotation of a rigid body about a fixed
aaxxiiss
-Angular Momentum of single particles and systems or particles
-Newton’s second law for rotational motion
-CConservatiion off angullar MMomentum
-Applications of the conservation of angular momentum
2
Rolling as Translation and Rotation Combined
t = 0 t = t
1 2
Consider an object with circular cross section that rolls
along a surface without slipping. This motion, though
common, is complicated. We can simplify its study by
ttrreeaattiinngg iitt aass aa ccoommbbiinnaattiioonn ooff ttrraannssllaattiioonn ooff tthhee cceenntteerr ooff
mass and rotation of the object about the center of mass
Consider the two snappshots of a rollingg bicyycle wheel shown in the figgure.
An observer stationary with the ground will see the center of mass O of the wheel
move forward with a speed v . The point P at which the wheel makes contact
com
with the road also moves with the same speed. During the time interval t between
ds
the two snapshots both O and P cover a distance s. v = (eqs.1) During t
com
ddtt
the bicycle rider sees the wheel rotate by an angle θ about O so that
ds dθ
s = Rθ→ = R =Rω (eqs.2) If we cambine equation 1 with equation 2
ddtt ddtt
v = Rω
we get the condition for rolling without slipping.
com
3
v = Rω
com
We have seen that rolling is a combination of purely translational motion
with sppeed v and a ppurelyy rotaional motion about the center of mass
ccoomm
v
with angular velocity ω= com . The velocity of each point is the vector sum
R
off tthhe vellociittiies off tthhe ttwo mottiions. FFor tthhe ttransllattiionall mottiion tthhe
r
velocity vector is the same for every point (v ,see fig.b ). The rotational
com
velocityy varies from ppoint to ppoint. Its maggnitude is eqqual to ωr where r is
the distance of the point from O. Its direction is tangent to the circular orbit
(see fig.a). The net velocity is the vector sum of these two terms. For example
the velocity of point P is always zero. The velocity of the center of mass O is
r r
v (r = 0). Finally the velocity of the top point T is wqual to 2v .
4
com com
Problem 2. An automobile traveling at 80.0 km/h has tires of 75.0 cm diameter.
(a) What is the angular speed of the tires about their axes?
(b) If the car is brought to a stop uniformly in 30.0 complete turns of the tires
(without skidding), what is the magnitude of the angular acceleration of the wheels?
(c) How far does the car move during the braking?
TThhee iinniittiiaall ssppeeeedd ooff tthhee ccaarr iiss

v =(80 km/h)(1000 m/km)(1 h/3600 s) = 22.2 m/s.

The tire radius is R = 0.750/2 = 0.375 m.

(a) The initial speed of the car is the initial speed of the center of mass of the tire, so

v 22.2 m/s
com0
ω = = =59.3 rad/s.
0
R 0.375 m

((bb)) WWiitthh θθ == ((3300.00))((22ππ)) == 118888 rraadd aanndd ωω== 00, EEqq. 1100-1144 lleeaaddss ttoo

(59.3 rad/s)2
ω2 =ω2 +2αθ ⇒ α= =9.31 rad/s2.
0 2((188 rad))

(c) Rθ = 70.7 m for the distance traveled.
5
v Rolling as Pure Rotation
T
A
Another way of looking at rolling is shown in the figure
v
A
v WWe consiidder rolllliing as a pure rottattiion abboutt an axiis
O B
of rotation that passes through the contact point P
v
B
between the wheel and the road. The anggular
v
velocity of the rotation is ω= com
R
In order to define the velocity vector for each point we must know its magnitude
as well as its direction. The direction for each point on the wheel points along the
r
ttaannggeenntt ttoo iittss cciirrccuullaarr oorrbbiitt. FFoorr eexxaammppllee aatt ppooiinntt AA tthhee vveelloocciittyy vveeccttoorr vv iiss
A
perpendicular to the dotted line that connects pont A with point P. The speed
of each point is given by: v =ωr. Here r is the distance between a particular
point and the contact point P. For example at point T r = 2R.
Thus v = 2Rω= 2v . For point O r = R thus v =ωR = v
T com O com
For poiint P r = 0 thhus v = 0
P
6
The Kinetic Energy of Rolling
Consider the rolling object shown in the figure
It is easier to calculate the kinetic energy of the rolling
body by considering the motion as pure rotation
aabboouutt tthhee ccoonnttaacctt ppooiinntt PP. TThhee rroolllliinngg oobbjjeecctt hhaass mmaassss MM
and radius R.
1
The kinetic energy K is then given by the equation: K = I ω2. Here I is the
PP PP
22
rotational inertia of the rolling body about point P. We can determine I using
P
1
(( ))
tthhee ppaarraalllleell aaxxiiss tthheeoorreemm. II == II ++ MMRR2 →→ KK == II ++ MMRR2 ωω2
P com com
2
1 1 1 1 1
( )
K = I + MR2 ω2 = I ω2 + MR2ω2 K = I ω2 + Mv2
com com com com
22 22 22 22 22
The expression for the kinetic energy consists of two terms. The first term
corresponds to the rotation about the center of mass O with angular velocity ω.
The second term is associated with the kinetic energy due to the translational
motion of evey point with speed v 7
com
Problem 9. A solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without
slipping a distance L=6.0 m down a roof that is inclined at the angle θ=30o.
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
((bb)) TThhee rrooooff'ss eeddggee iiss aatt hheeiigghhtt HH=55..00 mm.. HHooww ffaarr hhoorriizzoonnttaallllyy ffrroomm tthhee rrooooff'ss eeddggee
does the cylinder hit the level ground?
(a) We find its angular speed as it leaves the roof using conservation of energy. Its initial
kkiinneettiicc eenneerrggyy iiss KK = 00 aanndd iittss iinniittiiaall ppootteennttiiaall eenneerrggyy iiss UU = MMgghh wwhheerree
ii ii
h = 6.0sin30° = 3.0 m (we are using the edge of the roof as our reference level for
computing U). Its final kinetic energy (as it leaves the roof) is

K = 1 Mv2 + 1 Iω2.
ff 2 2

Here we use v to denote the speed of its center of mass and ω is its angular speed — at
the moment it leaves the roof. Since (up to that moment) the ball rolls without sliding we
can set v = Rω = v where R = 0.10 m. Using I = 1 MR2 (Table 10-2(c)), conservation of
2
energy leads to

1 1 1 1 3
Mgh= Mv2 + Iω2 = MR2ω2 + MR2ω2 = MR2ω2.
2 2 2 4 4

The mass M cancels from the equation, and we obtain

1 4 1 4 c hb g
ω= gh = 9.8 m s2 3.0 m = 63 rad s.
RR 33 00.1100 mm 33
8
((b)) Now this becomes a pprojjectile motion of the tyyppe examined in Chappter 4. We pput the
origin at the position of the center of mass when the ball leaves the track (the “initial”
position for this part of the problem) and take +x leftward and +y downward. The result
of part (a) implies v = Rω = 6.3 m/s, and we see from the figure that (with these positive
0
direction choices) its components are

v =v cos30°=5.4 m s
0x 0

v =v sin30°=3.1m s.
0y 0

The projectile motion equations become

1
x = v t and y = v t + gt2 .
0x 0y
22

We first find the time when y = H = 5.0 m from the second equation (using the quadratic
formula, choosing the positive root):

−v + v2 +2gH
0y 0y
t = =0.74s.
g

b gb g
Then we substitute this into the x equation and obtain x = 5.4 m s 0.74 s = 4.0 m.
9
Friction and Rolling
When an object rolls with constant speed (see top figure)
it has no tendency to slide at the contact point P and thus
no frictional force acts there. If a net force acts on the
r r
aa == 00 rroolllliinngg bbooddyy iitt rreessuullttss iinn aa nnoonn-zzeerroo aacccceelleerraattiioonn aa
com com
for the center of mass (see lower figure). If the rolling
object accelerates to the right it has the tendency to slide
r
at point P to the left. Thus a static frictional force f
s
opposes the tendency to slide. The motion is smooth
rolling as long as f < f
s s,max
The rolling condition results in a connection beteen the magnitude of the
acceleration a of the center of mass and its angular acceleration α
com
dv dω
v =ωR We take time derivatives of both sides → a = com = R = Rα
com com
ddtt ddtt
a = Rα
com 10