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**Unformatted text preview: **Chapter 2 Rolling Motion; Angular Momentum 2.1 The Important Stuff 2.1.1 Rolling Without Slipping When a round, symmetric rigid body (like a uniform cylinder or sphere) of radius R rolls
without slipping on a horizontal surface, the distance though which its center travels (when
the wheel turns by an angle 6) is the same as the arc length through which a point on the
edge moves: AxCM = s = R6 (2.1) These quantities are illustrated in Fig. 2.1.
The speed of the center of mass of the rolling object, vCM = ”61% and its angular speed
are related by ”CM 2 Raw (2.2) and the acceleration magnitude of the center of mass is related to the angular acceleration
by:
aCM 2 R06 (2.3) Figure 2.1: Illustration of the relation between A30, 8, R and 0 for a rolling object. 35 36 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM The kinetic energy of the object is:
Km“ : yaw? + éMUéM . (2.4) The ﬁrst term on the right side represents the rotational kinetic energy of the object about
its symmetry axis; the second term represents the kinetic energy the object would have if
it moved along with speed UCM without rotating (i.e. just translational motion). We can
remember this relation simply as: Kmu 2 Km + Ktrans. When a wheel rolls without slipping there may be a frictional force of the surface on
the wheel. If so, it is a force of static friction (which does no work) and depending on the
situation it could point in the same direction or opposite the motion of the center of mass;
in all cases it tends to oppose the tendency of the wheel to slide. 2.1.2 Torque as a Vector (A Cross Product) In the last chapter we gave a deﬁnition for the torque 7' acting on a rigid body rotating
around a ﬁxed axis. We now give a more general deﬁnition for “torque”; we deﬁne the
torque acting on a single particle (relative to some ﬁxed point O) when a force acts on it. Suppose the (instantaneous) position vector of a particle (relative to the origin 0) is r
and a single force F acts on it. Then the torque T acting on the particle is 7’2er (2.5) If gt is the angle between the position vector r and the force F then the torque T has
magnitude 7' = rFsinng 2.1.3 Angular Momentum of a Particle and of Systems of Particles There is yet more important quantity having to do with rotations that will be of help in
solving problems involving rotating objects; just as the linear momentum p was of importance
in problems with interacting particles, the angular momentum of objects which have motion
about a given axis will be useful when these objects interact with one another. Admittedly,
some of the ﬁrst deﬁnitions and theorems will be rather abstract! But we will soon apply
the ideas to simple objects which rotate around an axis and then the theorems and examples
will be quite downitoiearth. We start with a fundamental deﬁnition; if a particle has position vector r and linear
momentum p, both relative to some origin O, then the angular momentum of that particle
(relative to the origin) is deﬁned by: €=r><p=m(r><v) (2.6) . . 2
Angular momentum has units of kgsm . One can show that the net torque on a particle is equal to the time derivative of its
angular momentum: 2.1. THE IMPORTANT STUFF 37 Zr 22—? (2.7) This relation is analogous to the relation 2 F = ”fl—It) from linear motion.
For a set of mass points in motion, we deﬁne the total angular momentum as the (vector)
sum of the individual angular momenta: L=€1+£2+£3+... When we consider the total angular momentum, we can prove a theorem which is a bit
different in its content than Eq. 2.7. It’s a bit subtle; when the particles in a system all move
around they will be acted upon by forces from outside the system but also by the forces
they all exert on one another. What the theorem says is that the rate of change of the total
angular momentum just comes from the torques arising from forces exerted from outside the
system. This is useful because the external torques aren’t so hard to calculate. The theorem is: ZText : E (2.8) This tells us that when the sum of external torques is zero then L is constant (conserved).
We will encounter this theorem most often in problems where there is rotation about a ﬁxed
axis (and then once again we will only deal with the 2 components of 7' and L). 2.1.4 Angular Momentum for Rotation About a Fixed Axis An extended object is really a set of mass points, and it has a total angular momentum
(vector) about a given origin. We will keep things simple by considering only rotations
about an axis which is ﬁxed in direction (say, the z direction), and for that case we only
need to consider the component of L which lies along this axis, Lz. So, for rotation about a
ﬁxed axis the “angular momentum” of the rigid object is (for our purposes) just a number,
L. Furthermore, one can show that if the angular velocity of the object is w and its moment
of inertia about the given axis is I, then its angular momentum about the axis is L = [ca (2.9) Again, there is a correspondence with the equations for linear motion: px=muw (a) Lzlw 2.1.5 The Conservation of Angular Momentum In the chapter on Momentum (in Vol. 1) we used an important fact about systems for which
there is no (net) external force acting: The total momentum remains the same. One can
show a similar theorem which concerns net external torques and angular momenta. For a system on which there is no net external torque, the total angular momentum
remains constant: L,- = Lf. This principle is known as the Conservation of Angular
Momentum. 38 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM 2.2 Worked Examples 2.2.1 Rolling Without Slipping 1. An automobile traveling 80.0 km/hr has tires of 75.0 cm diameter. (a) What
is the angular speed of the tires about the axle? (b) If the car is brought to a
stop uniformly in 30.0 turns of the tires (without skidding), what is the angular
acceleration of the wheels? (0) How far does the car move during the braking?
[HRW5 12-31 (a) We know that the speed of the center of mass of each wheel is 80.0 km/hr. And the
radius of each wheel is R = (75.0 cm)/2 = 37.5 cm. Converting the speed to 3 we have: 1h 103m
km : km : m
80 h (80 h > (36008) <1km> 22.2 S From the relation between UCM and w for an object which rolls without slipping, we have: UCM = wR : w = 110%
and we get
(22.2 E) d
= —S = 59.3 i
w (0.375 m) s The angular speed of the wheel is 59.3 %.
(b) As the car comes to a halt, the tires go through 30.0 turns. Thus they have an angular
displacement of (with 60 = 0): 277 rad
1 rev 6 = (30.0 rev) < > = 188.5 rad . Also, when the wheel has come to a halt, its angular velocity is zero!
So we have the initial and ﬁnal angular velocities and the angular displacement. We can
get the angular acceleration of the wheel from Eq. 1.8. From that equation we get: 2— 2 oﬂ2—593ﬂi2
05:“ “0=( S) < S) = 9331““;d
2e 2(188.5rad) s The magnitude of the wheels’ angular acceleration is 9.33 :13. The minus sign in our result
indicates that a goes in the sense opposite to that of the initial angular velocity (and angular
displacement) of the wheel during the stopping. 2.2. WORKED EXAMPLES 39 (a) (b) Figure 2.2: (a) Constant horizontal applied to a rolling wheel in Example 3. (b) The forces acting on the
wheel, with the points of application as indicated. (c) As we saw, the angular displacement of any wheel during the stopping was 188.5 rad.
The radius of the wheel is R = 0.375 m, so from Eq. 2.1 the linear displacement of the wheel
(i.e. its center) is: 330M 2 R6 = (0.375 m)(188.5 rad) 2 70.7m so the car goes 70.7m before coming to a halt. 2. A bowling ball has a mass of 4.0 kg, a moment of inertia of 1.6 X 10’2 kg-m2 and
a radius of 0.10m. If it rolls down the lane without slipping at a linear speed of
4.0 ?, what is its total energy? [Ser4 11—5] The total (kinetic) energy of an object which rolls without slipping is given by Eq. 2.4.
To use this equation we have everything we need except the angular speed of the ball. From
Eq. 2.2 it is related to the linear velocity of the ball by UCM = Ru}, so the angular speed is ”CM (4.0 g) d
= — = —S = 40.0 i
w R (0.10m) s
and then the kinetic energy is
Kroll = %ICMW2 + %MUéM
2 are x 10-2 kg . m2)(40.0 %‘)2 + %(4.0kg)(4.0 g)?
= 44.8J The total kinetic energy of the ball is 44.8 J. 3. A constant horizontal force of 10N is applied to a wheel of mass 10kg and
radius 0.30m as shown in Fig. 2.2. The wheel rolls without slipping on the
horizontal surface, and the acceleration of its center of mass is 0.60 8%. (a) What
are the magnitude and direction of the frictional force on the Wheel? (b) What
is the rotational inertia of the wheel about an axis through its center of mass
and perpendicular to the plane of the wheel? [HRW5 12—9] 40 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM (a) The forces which act on the wheel along with where these forces are applied are shown
in Fig 2.2 (b). In addition to the applied force of 10N which points to the right, there is
a force of static friction between the surface and the wheel (of magnitude f3), which for
now we draw pointing to the left (we can ask: Does it really point that way?). There are
vertical forces acting on the wheel (from gravity and the normal force of the surface) but
these clearly cancel out and for now we don’t need to worry about them. Even though the wheel will be rolling during its motion, Newton’s 2nd law still holds,
and the sum of the horizontal forces gives may, Here the wheel is clearly accelerating to the
right and so with the choice of directions given in the ﬁgure, we ﬁnd: 2ng = 10.0N — f5 = max = (10kg)(0.60 5%) = 6.0N so that
f, = 10.0N — 6.0N : 4.0N and since this is positive, the frictional force does indeed point to the left, as we guessed.
Actually, it wasn’t so hard to guess that, since only a leftward frictional force could make
the wheel rotate clockwise ias we know it must herei but for some problems in rolling
motion, the direction of the static friction force may not be so evident. (b) Rotational inertia is related to net torque and angular acceleration by way of 7' 2 la.
It is true that in this problem the rotating object is also accelerating but it turns out this
relation still holds as long as we choose the center of mass to be the rotation axis. Considering the four forces in Fig. 2.2 (b), the applied force and the (effective) force of
gravity are applied at the center of the wheel, so they give no torque about its center. The
normal force of the surface is applied at the rim, but its direction is parallel to the line which
joins the application point to the center so it too gives no torque. All that remains is the
friction force, applied at a distance R from the center and perpendicular to the line joining
this point and the axis; this force gives a clockwise rotation, so if we take the clockwise
direction as the positive sense for rotations, then the net torque on the wheel is 7 = +f57“ = (4.0N)(0.30m) = 1.2N . m From Eq. 2.3 we know the angular acceleration of the wheel; it is _ CLCM (0.60 8%) — — = — = 2.0 ﬂ .
0‘ R (0.30m) 52
and then from 7' 2 la we get the rotational inertia:
1.2N -=:=(—dm)=0.60kg-m2
oz (2.0 2%) 4. A round, symmetrical object of mass M, radius R and moment of inertia I
rolls without slipping down a ramp inclined at an angle 6. Find the acceleration
of its center of mass. 2.2. WORKED EXAMPLES 41 Figure 2.3: Round, symmetrical object with mass M, radius R and moment of inertia I rolls down a ramp
sloped at angle 0 from the horizontal. The problem is diagrammed in Fig. 2.3. We show the forces acting on the object and
where they are applied. The force of gravity, Mg is (effectively) applied at the center
of the object. As usual we decompose this force into its components down the slope and
perpendicular to the slope. The slope exerts a normal force N at the point of contact. Finally
there is a force of static friction f5 from the surface; this force points along the surface and
we can pretty quickly see that it must point up the slope because it is the friction force
which gives the object an angular acceleration, which (here) is in the clockwise sense. Apply Newton’s 2nd law ﬁrst: The forces perpendicular to the slope cancel, so that
N 2 Mg cos6 (but we won’t need this fact). If aCM is the acceleration of the GM of the
object down the slope, then adding the forces down the slope gives Mgsind — f3 = MaCM (2.10) Now we look at the net torque on the rolling object about its center of mass. The force
of gravity acts at the center, so it gives no torque. The normal force of the surface acts at
the point of contact, but since it is parallel to the line joining the pivot and the point of
application, it also gives no torque. The force of friction is applied at a distance R from the
pivot and it is perpendicular to the line joining the pivot and point of application. So the
friction force gives a torque of magnitude 7' = Rfs sin90° = Rfs . and if we take the clockwise sense to be positive for rotations, then the net torque on the
object about its CM is Tnet : Rfs From the relation 7' = 105 we then have
T=RfS=Ioz (2.11) but we can also use the fact that for rolling motion (without slipping) the linear acceleration
of the CM and the angular acceleration are related by: CLCMZTOK :> 04:— 42 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM and using this in Eq. 2.11 gives laCM l(ICM
R 8 _ j 8 — 2.12 where we choose to isolate f, (the magnitude of the friction force) so that we can put the
result into Eq. 2.10. When we do that, we get: [aCM MgsinQ— R2 = MaCM Now solve for aCM: I I
Mgsind =MaCM—l— 23M 2 (M+—> aCM Our result is sensible in that if I is very small then (LCM is nearly equal to gsinQ, the
result for a mass sliding with no rolling motion. 5. A uniform sphere rolls down an incline. (a) What must be the incline angle
if the linear acceleration of the center of the sphere is to be 0.10 g? (b) For this
angle, what would be the acceleration of a frictionless block sliding down the
incline? [HRW5 12—7] (a) We will use the formula for CLCM (for rolling without slipping down a slope) in a previous
problem. Note that we are not given the mass of the sphere! But it turns out that we don’t
need it, because for a uniform sphere, we have I _ 2
MR2 _ 5
and as we can see from our earlier result,
_ gsin6
aoM — —<1 + M11?) 7 (LCM just depends on the combination [/ (MR2). Solving this equation for sin 6 and plugging
in the given numbers, we get: l | sin9=aC—M<1+ I ) <0'109)(1+§)=(0.10) = 0.14
g MR2 9 OWIKI 2.2. WORKED EXAMPLES 43 Which gives us
6 = sin—1(0.14) = 8.00 (b) As we saw in the Chapter on forces (Volume 1) when a mass slides down a frictionless
incline its linear acceleration is given by a = gsinQ. For the slope angle found in part (a),
this is a = gsinQ = (9.80 8%) sin 8.00 = 1.4 8%
We can also calculate a/g:
= Sing 2 Sin8.0O = 0.14 ‘QIQ so for the frictionless case we have a = 0149. 2.2.2 Torque as a Vector (A Cross Product) 6. What are the magnitude and direction of the torque about the origin on
a plum (!) located at coordinates (—2.0m,0,4.0m) due to force F whose only component is (a) F36 : 6.0N, (b) F93 : —6.0 N, (0) F2 : 6.0N, and (d) FZ : —6.0N ?
[HRW5 12—21] (a) The (vector) torque on a point particle is given by
'r = r x F I ﬁnd it easiest to set up the cross product in determinant notation, discussed in Chapter
1 of Volume 1. We note that the units of the result must be N - m; then the cross product
of r and F is i j k
er= —2.0 0.0 4.0 N-m=(+(4.0)(6.0)N-m)j=(+24.0N-m)j
6.0 0.0 0.0 The torque 7' has magnitude 24.0N ' m and points in the +y direction.
(b) Is is fairly clear that if we had had Fm = —6.0 N in part (a), we would have gotten ’7': (—24.0N-m)j so the torque would have magnitude 24.0 N - m and point in the —y direction. (c) When the only component of F is FZ = 6.0 N, then we have i j k
er= —2.0 0.0 4.0 N-m=(—(—2.0)(6.0)N-n1)j=(+12.0N-m)j
0.0 0.0 6.0 so the torque would have magnitude 12.0 N - m and point in the +y direction. ((1) If instead we have only Fz = —6.0 N then the sign of the result in part (c) changes, and
the torque would have magnitude 24.0 N - m and point in the —y direction. 44 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM 6'5 kg 2.2 m/s if
1.5 In
all 2.8 m 0 Figure 2.4: Two masses and their motion relative to the point 07 as in Example 8. 2.2.3 Angular Momentum of a Particle and of Systems of Particles 7. The position vector of a particle of mass 2.0 kg is given as a function of time by
r = (6.0i+5.0tj) m when t is given in seconds). Determine the angular momentum
of the particle as a function of time. [Ser4 11—17] The angular momentum of a point mass is given by 2 = r X p. The velocity of our particle is given by d d . . - m
V : ﬁr : $63.01 + 5.0m) m — (5.0J) g and its momentum is
p = mv = (2.0 kg)(5.0j) § = (10.0j) 1% Now take the cross product to get 3: i j k
€=r><p= 6.0 5.015 0.0 5%:(50010hssg
0.0 10.0 0.0 8. Two objects are moving as shown in Fig. 2.4. What is their total angular
momentum about point 0? [HRVV5 12—27] In this problem we must use the deﬁnition of the angular momentum of a particle (with
respect to some origin 0):
K = r x p First consider the 3.1 kg mass. Its momentum vector has magnitude P1 = 772101 2 (3.1kg)(3.6 §) 2 11.2 kgT-m 2.2. WORKED EXAMPLES 45 and is directed upward in this picture. The vector r which goes from point 0 to the particle
has magnitude 2.8m and points to the right. By the rightihand rule for cross products, the
vector r X p points up out of the page, which is along the +2 axis; and since the two vectors
are perpendicular, the magnitude of r X p is 2 |r1 >< p1| 2 mm sin 90° = (2.8m)(11.2 1%) = 31.2 kg:
and so the angular momentum of this particle about 0 is
31 2 (+312 ﬂyim
The 6.5 kg mass has a momentum vector of magnitude
p2 = mm = (6.5kg)(2.2 g) = 14.3 kg?“ and is directed to the right. The vector r which goes from the point 0 to the particle has
magnitude 1.5m and points straight up. By the rightihand rule for cross products, the
vector r X p points into the page, which is along the —z axis; and since the two vectors are
perpendicular, the magnitude of r X p is |r2 >< p2l = r2192 sin 90° 2 (1.5m)(14.3 1%) = 21.5 k331i
and so the angular momentum of this particle about 0 is 32 = (—21.5 kg‘m2)k S The total angular momentum of the system is 3 = 31% = (31.2 — mm W = (+9.8 Wm. S S 2.2.4 Angular Momentum for Rotation About a Fixed Axis 9. A uniform rod rotates in a horizontal plane about a vertical axis through one
end. The rod is 6.00m long, weighs 10.0 N, and rotates at 240 rev/min clockwise
when seen from above. Calculate (a) the rotational inertia of the rod about the
axis of rotation and (b) the angular momentum of the rod about that axis. [HRW5
12—45] (a) The mass of the rod is
M = W/g = (10.0 N)/(9.80 8%) = 10.2 kg and its angular velocity in units of % is . > = 25.1 @
min 608 S w_240rev 27rrad (1min
_ 1rev 46 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM We know the formula for the moment of inertia of a uniform rod rotating about an axis at
one of its ends (see Chapter 1, Fig. 1) so we calculate I as: 2 2 2
1: gML = (10.2kg)(6.00m) = 12.2kg-m l
3 The rotational inertia of the rod (about the given axis) is 12.2 kg - m2. (b) The angular momentum of the rotating rod will be given by L = I to. We ﬁnd that the
magnitude of the angular momentum is: L = 1w = (12.2 kg - m2)(25.1 %) = 309 kg“? S The vector L would point upward (along the +2 axis if the rotation were counterclockwise
as seen from above. That is not the case (it is clockwise) so the direction of L is downward. 2.2.5 The Conservation of Angular Momentum 10. Suppose that the Sun runs out of nuclear fuel and suddenly collapses to form
a white dwarf star, with a diameter equal to that of the Earth. Assuming no
mass loss, what would then be the Sun’s new rotation period, which currently
is about 25 days? Assume that the Sun and the white dwarf are uniform, solid
spheres; the present radius of the Sun is 6.96 X 108 m. [HRW5 12-55] In this ...

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