Download Chapter 11 Questions

Chapter 11 Questions
1. A cannonball rolls down an incline without sliding. If the roll is now repeated with an incline that is less
steep but of the same height as the first incline, are (a) the ball’s time to reach the bottom and (b) its
translational kinetic energy at the bottom greater than, less than, or the same as previously?
l
pp pppp pppppppppppppppp p pp p pp p ppp p p p p p p p pl
p
XXX
PP
θ1 > θ2
6
XXX
P
PP
A
A
h
X
XXX
PP
B
B
PP
X X
P
p
p
p
p
p
p
p
p
?
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p p θ2 X
p pppX
p p pXp p p p
θ1
PP
X
The cannonball rolls without sliding (vcom = ωr) starting from rest and from the same height h (point A).
Therefore, its initial mechanical energy is the same for both inclines, and is equal to m g h (relative to point
B). At point B cannonball’s mechanical energy is equal to its kinetic energy,
1
1
7
2
2
2
K = m vcom
+ Icom ω 2 = m vcom
, where vcom = ωr and Icom = m r2 .
2
2
10
5
The only force doing non-zero work during rolling without sliding (smooth rolling) is weight which is conservative force that conserves mechanical energy. Using conservation of mechanical
r energy we can find that
10
g h.
the center of mass velocity at point B is equal for both inclines and is equal to
7
(a) On the less steep incline the cannonball rolls through larger distance with the same average speed of
h
l1
=
, t2 > t1 ).
0.5 vcom and therefore needs longer time to reach point B (t1 =
vcom,avg
vcom,avg sinθ1
(b) The translational kinetic energy at point B is equal for both slopes. The initial mechanical energy
is the same (m g h) and the final kinetic energy of rotation is the same for both because the shape of the
rolling object is the same and the center of mass velocity is the same.
2. Figure 11-29 shows an overhead view of a rectangular slab that can spin like a merry-go-round about its
center at O. Also shown are seven paths along which wads of bubble gum can be thrown (all with the
same speed and mass) to stick onto the stationary slab. (a) Rank the paths according to the angular speed
that the slab (and gum) will have after the gum sticks, greatest first. (b) For which paths will the angular
momentum of the slab (and gum) about O be negative from the view of Fig. 11-29?
After the gum sticks, the slab-gum system will rotate
freely about point O. Therefore, the total angular
momentum of the gum-slab system, about O, is
conserved. The initial angular momentum of the gum
should be calculated relative to the point O:
2, 3, and 5 have zero angular momentum of the gum
relative to O, because their velocities and position
vectors are colinear.
2a
1
-p p p p p p p p p p p p pp
ppp
2
ppp
p
p
pp
ppp
*
©
3 © AK
A
©©
4A
5
A
a
rp O
pp
p
pp
6
¾ 7
6
6
The magnitude of the angular momentum of gum 4 is the largest (position vector and velocity √
are perpendicular), then 6, 7, and then 1, because that is the order of magnitudes of their moment arms (a 2, a, 0.5a,
0.25a).
Before interaction, only gum has angular momentum (~li = m ~ri × ~v ), and after interaction the gum-slab
~ f = Islab,gum ω
system has angular momentum of a rigid body (L
~ f ).
The final rotational inertia of the slab-gum system is the sum of the slab’s rotational inertia about its
5a2 M
1
) and the rotational inertia of the gum stuck on the slab
center of mass (Islab,com = M (4a2 + a2 ) =
12
12
(Igum = m r2 ). Assuming that the mass of each gum m is much smaller than the mass of the slab M we
can neglect the small differences in final rotational inertia of the different gum-slab systems and rank the
paths only relative to the difference in the initial angular momentum of each gum. The larger the moment
arm the larger the final angular speed. Therefore,
(a) Answer: 4 > 6 > 7 > 1 > 2 = 3 = 5 = 0
(b) Paths 1, 4, and 7 have negative angular momenta relative to point O (use right-hand-rule).
3. Figure 11-24 shows two particles A and B at xyz coordinates (1 m, 1 m, 0) and (1 m, 0, 1 m). Acting on
each particle are three numbered forces, all of the same magnitude and each directed parallel to an axis. (a)
Which of the forces produce a torque about the origin that is directed parallel to y? (b) Rank the forces
according to the magnitudes of the torques they produce on the particles about the origin, greatest first.
Fig. 11-24 shows 6 forces (of the same magnitude) acting in
points A = (1, 1, 0) and B = (1, 0, 1), in meters.
(a) Forces 5 and 6 have torque about origin that is parallel to y.
Torque (vector product of ~r and F~ ) is always perpendicular
to the plane defined by vectors ~r and F~ . For torque along
the y-axis vectors ~r and F~ have to be in x-z plane.
(b) Moment arms (perpendicular distance between the origin
and the line of each force) for the forces 2, 3, 5, and 6 are
equal to 1 m. Moment arms for the forces 1 and 4 are
√
equal to 2 m. Therefore, the torques of forces 1 and 4 are
larger than the torques of forces 2, 3, 5, and 6. Therefore,
1 = 4 > 2 = 3 = 5 = 6.
y
A
p p p p p p p p p p p p prp
½
½ pp
pp
½
=
1
pp
?
pp 2
pp
4
p
p
6
p
½
p
p
p 5
½
½
>
pp
½
p p½ ½
p p p p p p p p p p p p pr½
½
½
B
z
3
-
x
6
4. What happens to initially stationary yo-yo in Fig. 11-25 if you pull it via its string with (a) force F~2 (the
line of action passes through the point of contact on the table, as indicated), (b) force F~1 (the line of action
passes above the point of contact), and (c) force F~3 (the line of action passes to the right of the point of
contact)?
~3 ~
F
F2
6¢̧
'$
¢
¶³
pp ¢
p
µ´
~1
pp - F
pp
&%
6
2R1
?
62R
?
Yo-yo, side view
Yo-yo, front view cross section
The yo-yo here is used to produce a rolling motion along the horizontal plane (not in the specific yo-yo
motion along the vertical plane) with a long cord wrapped around the inner cylinder of radius R. The
tension of the cord provides a torque relative to the center of mass of the yo-yo, as well as relative to the
contact point. Depending on the angle of the cord the tension is either assigned as F~1 , or as F~2 , or as F~3 .
Using the description of rolling as pure rotation about the contact point we can answer the questions without
doing any calculation.
(a) The force F~2 does not have any torque relative to the contact point, therefore there will not be any
angular acceleration produced, so α = 0 = acom . If F~2,y is smaller than the weight of the yo-yo and F~2,x is
smaller than the static friction between the yo-yo and the horizontal surface, the yo-yo will be at rest.
(b) The force F~1 has a clockwise torque of magnitude F1 (R1 − R) relative to the contact point, and therefore
will produce a rolling motion with clockwise angular acceleartion, so the yo-yo would roll toward right.
(c) The force F~3 has a counterclockwise tourque of magnitude F3 R relative to the contact point, and therefore
will produce a rolling motion with counterclocwise angular acceleration, so the yo-yo would move toward
left.