ChiuRotationEx

```補充例題 4-1
Figure A shows a thin, uniform rod of mass M and length L.
(a) What is its rotational inertia about an axis perpendicular to the rod,
thought its center of mass?
Solution: We place the rod on an axis, with its center of mass at the origin,
and we choose as a mass element a slice dx of the rod. The center of the
slice is a distance x from the rotation axis. The mass per unit length of the rod
is M L , so the mass dm of the element dx is
M 
dm   dx.
L
we have
I   r 2 dm  
x L


 
M 3
x
3L
L
L
2
M 
x 2  dx
2
 L
x L

2
2
3
3
M  L   L  


 
  
3L  2   2  
1
ML2 .
12
(b) What is the rotational inertia of the rod about an axis perpendicular to the
rod through one end?
Solution: We combine the result in (a) with the parallel-axis theorem,
obtaining
I  I cm  Mh 2
2
1
1
1 
 ML2  M  L   ML2 ,
12
3
2 

Find the moment of inertia of a thin rod of mass M and length L about an axis
at one end and perpendicular to the rod, as shown in Fig. B.
Solution: The mass of an element of length dx is dm  dx, where   M L
is the linear mass density. The moment of inertia of the mass element that is at
a distance x from the axis is dI  r 2 dm  x 2 dx . For the entire rod the
moment of inertia is
L
L3
0
3
I   x 2 dx 
Since M  L,
(rod)
I END
ML2

3

Find the moment of inertia of a circular disk or solid cylinder of radius R about
the following axes: (a) through the center and perpendicular the flat surface; (b)
at the rim and perpendicular to the flat surface.
Solution: (a) Figure C shows that the appropriate mass element is a circular
ring of radius r and width dr. Its area is dA  2rdr and its mass is dm  dA,
where   M / A is the area mass density. The moment of inertia of this
element is
dI  r 2 dm  2r 3 dr
For the whole body,.
R
I  2  r 3 dr 
0
1
R 4
2
The mass of the whole disk or cylinder is M  A  R 2 , and so
(Disk or solid cylinder)
ICM 
1
MR 2
2
(b) The evaluation of the moment of inertia about an axis at the rim., as in Fig.
C (b), by integration is difficult. The parallel axis theorem, with h=R, provides
I  ICM  Mh 2 
1
3
MR 2  MR 2  MR 2
2
2

Find the moment of inertia of a uniform solid sphere of mass M and radius R
Solution: The sphere may be divided into disks perpendicular to the given
axis, as in Fig. 11-8. The disk at a distance x from the center of the sphere has
a radius r  (R 2  x 2 )1 2 and a thickness dx. If   M / V is the volume
mass density (mass per unit volume), the mass of this elemental disk is
dm  dV  r 2 dx , or
dm   R 2  x 2 dx


From Eq. 11-19 the moment of inertia of this elemental disk is
dI 


2
1
1
dmr 2   R 2  x 2 dx
2
2
The total moment of inertia is
I


R
1
  R 4  2R 2 x 2  x 4 dx
0
2
R
1 
2
1 
  R 4 x  R 2 x 3  x 5 
2
3
5 0


8
R 5
15
4

The total mass of the sphere is M    R 3 , so the moment of inertia may
3

be written as
(Solid sphere)
I
3
MR 2
5
(11-20a)
This is the moment of inertia of a solid sphere about an axis through its center.
In Problem 6 you are asked to show that the moment of inertia of a thin
spherical shell. Of mass M and radius R. about an axis through its center is
(Spherical Shell)
I
2
MR 2
3
(11-20b)

(a) Show that the moment of inertia of a thin spherical shell of mass M and
2
MR 2 . (Hint: Break the shell into rings
3
and use the angle to the axis as the variable.)
(b) Use part (a) to find the moment of inertia of a solid sphere of mass M and
Solution:
(a) dI  r 2dm  (R sin 2 )2 sinRd 

c 3  8R 4
I  2R 4  1  c 2 dc  2R 4 c   
3
3



2MR 2
M  4R , so I 
3
2


2dmr 2 2 4r 2 dr r 2
4R 3

(b) dI 
and M 
.
3
3
3
2MR 2
Integrate to find I 
.
5

A bowing bull, whose radius R is 11cm and whose mass M is 7.2 kg, rolls from
rest down a ramp whose length L is 2.1m. The ramp is inclined at an angle 
of 34 to the horizontal; see the sphere in Fig. 12-7. How fast is the ball
moving when it reaches the bottom of the ramp? Assume the ball is uniform in
density.
Solution: The center of the ball falls a vertical distance h  L sin ; so the
decrease in gravitational potential energy is MgL sin . This loss of potential
energy equals the gain in kinetic energy. Thus we can write (see Eq.12-5)
MgL s i n 
1
1
2
I cm 2  Mvcm
2
2
From Table 11-2(g) we see that, for a solid sphere, I cm 
replace  with its equal,
v cm
R
(12-6)
2
MR 2 . We can also
5
Substituting both these quantities into Eq.
12-6 yields


2
1
v 
 1  2 
2
MgL s i n     MR 2  cm   Mv cm
2
 2  3 
 R 
(12-6)
Solving for
 10 
 10 
v cm   gL sin    9.8 m 2 2.1m sin 34  4.1m
s
s 
 7
 7 
Note that the answer does not depend on the mass or radius of the ball.

Figure 12-7 Sample Problem 12-2 and 12-3. A hoop, a disk, and a sphere roll
from rest down a ramp of angle  . Although released from rest at the same
position and time, they arrive at the bottom in the order shown.

Halliday 5th 12-2

Here we generalize the result of Sample Problem 12-2. A uniform hoop, disk,
and sphere, having the same mass M and the same radius R, are released
simultaneously from rest at the top of a ramp whose length L is 2.5m and
whose ramp angle  is 12 (Fig. 12-7)
(a) Which object reaches the bottom first?
Solution: Table 12-1 gives us the answer. The sphere outs the largest share of
its kinetic energy (71%) into translational motion, so it wins the race. Next
comes the disk and then the hoop.
(b) How fast are the objects moving at the ramp’s bottom?
Solution: The center of mass of each object rolling down the ramp falls the
same vertical distance h. Like a body in free fall, the object loses potential
energy in amount Mgh and thus gains this amount of kinetic energy. At the
bottom of the ramp then, the total kinetic energies of all three objects are the
same. How these kinetic energies are divided between the translational and
rotational forms depends on each object’s distribution of mass.
From Eq.12-5 we can write (putting  
v cm
)
R
1
1
2
I cm 2  Mv cm
2
2
2
 1
1  v cm
2

 I cm  2   Mv cm
2 R  2
Mgh 
1  I cm  2
1
2
 2 v cm  Mv cm
2R 
2
Putting h  L sin and solving for v CM , we obtain

(12-7)
v cm 
2gL s i n
I
1  cm
MR 2
which is the symbolic answer to the question.
Note that the speed depends not on the mass or the radius of the rolling
object, but only on the distribution of its mass about its central axis, which
enters through the term
I cm
MR 2
. A marble and a bowling ball will have the
same speed at the bottom of the ramp and will thus roll down the ramp in the
same time. A bowling ball will beat a disk of any mass or radius, and almost
anything that rolls will beat a hoop.
For the rolling hoop (see the hoop listing in Table 12-1), we have
I cm
 1 , so Eq.12-8 yields
MR 2
v cm 

2gL sin
I
1  cm
MR 2
2 9.8 m s 2 2.5m sin12

 2 .3 m

1 1
s
From a similar calculation, we obtain v cm  2.6 m
s
for the disk (
I cm
1
 )
2
2
MR
I cm
2
 ). This supports our prediction of (a)
2
5
MR
that the win, place, and show sequence in this race will be sphere, disk, and
hoop.
and 2.7 m
s
for the sphere (
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