318
CHAPTE R 11 ROLLI NG, TORQU E, AN D ANG U L AR M OM E NTU M
M, mass cancels from Eq. 11-46; thus ⍀ is independent of the mass.
Equation 11-46 also applies if the shaft of a spinning gyroscope is at an angle
to the horizontal. It holds as well for a spinning top, which is essentially a spinning
gyroscope at an angle to the horizontal.
According to Eq. 11-41, torque causes an incremental change
in the
angular momentum of the gyroscope in an incremental time interval dt; that is,
(11-44)
However, for a rapidly spinning gyroscope, the magnitude of
is fixed by
Eq. 11-43. Thus the torque can change only the direction of
not its magnitude.
From Eq. 11-44 we see that the direction of
is in the direction of perpendicular to . The only way that
can be changed in the direction of
without the magnitude L being changed is for to rotate around the z axis as
shown in Fig. 11-22c. maintains its magnitude,the head of the vector follows
a circular path, and
is always tangent to that path. Since
must always
point along the shaft, the shaft must rotate about the z axis in the direction of .
Thus we have precession. Because the spinning gyroscope must obey Newton’s
law in angular form in response to any change in its initial angular momentum, it
must precess instead of merely toppling over.
Precession. We can find the precession rate ⍀ by first using Eqs. 11-44 and
11-42 to get the magnitude of
:
dL ϭ t dt ϭ Mgr dt.
( 1 1 - 4 5 )
As changes by an incremental amount in an incremental time interval dt, the shaft
and precess around the z axis through incremental angle df. (In Fig. 11-22c, angle
df is exaggerated for clarity.) With the aid of Eqs. 11-43 and 11-45, we find that df is
given by
Dividing this expression by dt and setting the rate ⍀ ϭ df/dt, we obtain
(precession rate).
(11-46)
This result is valid under the assumption that the spin rate v is rapid. Note that ⍀
decreases as v is increased. Note also that there would be no precession if the
gravitational force
did not act on the gyroscope, but because I is a function of
Mg
:
⍀ ϭ
Mgr
Iv
df ϭ
dL
L
ϭ
Mgr dt
Iv
.
L
:
L
:
dL
:
t
:
L
:
t
:
L
:
L
:
L
:
t
:
L
:
L
:
t
:
,
dL
:
L
:
,
L
:
dL
: ϭ t
: dt.
dL
:
t
:
Rolling Bodies For a wheel of radius R rolling smoothly,
v com ϭ vR,
( 1 1 - 2 )
where v com is the linear speed of the wheel’s center of mass and v is
the angular speed of the wheel about its center. The wheel may
also be viewed as rotating instantaneously about the point P of the
“road” that is in contact with the wheel. The angular speed of the
wheel about this point is the same as the angular speed of
the wheel about its center. The rolling wheel has kinetic energy
(11-5)
where I com is the rotational inertia of the wheel about its center of
mass and M is the mass of the wheel. If the wheel is being accelerated
but is still rolling smoothly, the acceleration of the center of mass
is related to the angular acceleration a about the center with
a com ϭ aR.
( 1 1 - 6 )
a
:
com
K ϭ
1
2 I com v
2 ϩ
1
2 ⌴v
2
com ,
Review & Summary
If the wheel rolls smoothly down a ramp of angle u, its acceleration
along an x axis extending up the ramp is
(11-10)
Torque as a Vector In three dimensions, torque is a vector
quantity defined relative to a fixed point (usually an origin); it is
(11-14)
where is a force applied to a particle and is a position vector locating the particle relative to the fixed point.The magnitude of is
(11-15, 11-16, 11-17)
where f is the angle between and
is the component of
perpendicular to and is the moment arm of . The direction
of is given by the right-hand rule.
t
:
F
:
r Ќ
r
:
,
F
:
F Ќ
r
:
,
F
:
t ϭ rF sin f ϭ rF Ќ ϭ r Ќ F,
t
:
r
:
F
:
t
: ϭ r
: ϫ F
:
,
t
:
a com, x ϭ Ϫ
g sin u
1 ϩ I com /MR
2
.
CHAPTE R 11 ROLLI NG, TORQU E, AN D ANG U L AR M OM E NTU M
M, mass cancels from Eq. 11-46; thus ⍀ is independent of the mass.
Equation 11-46 also applies if the shaft of a spinning gyroscope is at an angle
to the horizontal. It holds as well for a spinning top, which is essentially a spinning
gyroscope at an angle to the horizontal.
According to Eq. 11-41, torque causes an incremental change
in the
angular momentum of the gyroscope in an incremental time interval dt; that is,
(11-44)
However, for a rapidly spinning gyroscope, the magnitude of
is fixed by
Eq. 11-43. Thus the torque can change only the direction of
not its magnitude.
From Eq. 11-44 we see that the direction of
is in the direction of perpendicular to . The only way that
can be changed in the direction of
without the magnitude L being changed is for to rotate around the z axis as
shown in Fig. 11-22c. maintains its magnitude,the head of the vector follows
a circular path, and
is always tangent to that path. Since
must always
point along the shaft, the shaft must rotate about the z axis in the direction of .
Thus we have precession. Because the spinning gyroscope must obey Newton’s
law in angular form in response to any change in its initial angular momentum, it
must precess instead of merely toppling over.
Precession. We can find the precession rate ⍀ by first using Eqs. 11-44 and
11-42 to get the magnitude of
:
dL ϭ t dt ϭ Mgr dt.
( 1 1 - 4 5 )
As changes by an incremental amount in an incremental time interval dt, the shaft
and precess around the z axis through incremental angle df. (In Fig. 11-22c, angle
df is exaggerated for clarity.) With the aid of Eqs. 11-43 and 11-45, we find that df is
given by
Dividing this expression by dt and setting the rate ⍀ ϭ df/dt, we obtain
(precession rate).
(11-46)
This result is valid under the assumption that the spin rate v is rapid. Note that ⍀
decreases as v is increased. Note also that there would be no precession if the
gravitational force
did not act on the gyroscope, but because I is a function of
Mg
:
⍀ ϭ
Mgr
Iv
df ϭ
dL
L
ϭ
Mgr dt
Iv
.
L
:
L
:
dL
:
t
:
L
:
t
:
L
:
L
:
L
:
t
:
L
:
L
:
t
:
,
dL
:
L
:
,
L
:
dL
: ϭ t
: dt.
dL
:
t
:
Rolling Bodies For a wheel of radius R rolling smoothly,
v com ϭ vR,
( 1 1 - 2 )
where v com is the linear speed of the wheel’s center of mass and v is
the angular speed of the wheel about its center. The wheel may
also be viewed as rotating instantaneously about the point P of the
“road” that is in contact with the wheel. The angular speed of the
wheel about this point is the same as the angular speed of
the wheel about its center. The rolling wheel has kinetic energy
(11-5)
where I com is the rotational inertia of the wheel about its center of
mass and M is the mass of the wheel. If the wheel is being accelerated
but is still rolling smoothly, the acceleration of the center of mass
is related to the angular acceleration a about the center with
a com ϭ aR.
( 1 1 - 6 )
a
:
com
K ϭ
1
2 I com v
2 ϩ
1
2 ⌴v
2
com ,
Review & Summary
If the wheel rolls smoothly down a ramp of angle u, its acceleration
along an x axis extending up the ramp is
(11-10)
Torque as a Vector In three dimensions, torque is a vector
quantity defined relative to a fixed point (usually an origin); it is
(11-14)
where is a force applied to a particle and is a position vector locating the particle relative to the fixed point.The magnitude of is
(11-15, 11-16, 11-17)
where f is the angle between and
is the component of
perpendicular to and is the moment arm of . The direction
of is given by the right-hand rule.
t
:
F
:
r Ќ
r
:
,
F
:
F Ќ
r
:
,
F
:
t ϭ rF sin f ϭ rF Ќ ϭ r Ќ F,
t
:
r
:
F
:
t
: ϭ r
: ϫ F
:
,
t
:
a com, x ϭ Ϫ
g sin u
1 ϩ I com /MR
2
.