Andrew Mason said:
I don't see how the body can "induce a torque" on itself. If there is no external torque there can be no change in the angular momentum vector. What is the source of the (external) torque on the bullet?
Andrew Mason said:
Are you talking about a rigid body? If so, please explain how the moment of inertia can change with time.
Filip is correct, Andrew.
You have probably seen the rotational analog of
F=m
a written as
τ=
I d
ω/dt. As you know,
F=m
a is correct only if the mass is constant. Just as
F=d
p/dt is a more generic way of writing Newton's second law, it is better to write the rotational analog as
τ=d
L/dt, where
L=
Iω. The general expression is thus
τ=d
I/dt
ω +
I d
ω/dt, and this reduces to the simple freshman physics
τ=
I d
ω/dt only if
I is constant from the perspective of an inertial observer. While the inertia tensor of a rigid body is constant in a frame rotating with the rigid body, it is not necessarily constant from the perspective of an inertial observer.
Imagine a body that is not subject to any external torques and is rotating such that the angular momentum vector is not aligned with one of the body's principal axes. The body will undergo a torque-free precession. From the perspective of an inertial observer, the object's angular momentum vector will be constant, but the angular velocity and inertia tensor will vary with time. From the perspective of a body-fixed observer, the object's inertia tensor will be constant, but the angular velocity and angular momentum will vary with time.
Going back to the rotational analog of Newton's second law, I'll write it as
\boldsymbol{\tau} = \left(\frac {d\mathbf L}{dt}\right)_I
Here I have used the subscript I on the derivative to explicitly denote that this is the time derivative of the angular momentum as observed by an inertial observer. Before going on, I need to take a side journey into the relation between the time derivatives of some vector quantities b]q[/b] as observed by inertial and rotating observers. I'll assume a common origin; there is no reason to generalize to affine+rotating transformations. The time derivatives of some vector quantity
q as observed by these inertial and rotating observers are related via
1[/color]
\left(\frac {d\mathbf q}{dt}\right)_I =<br />
\left(\frac {d\mathbf q}{dt}\right)_R + \boldsymbol{\omega} \times \mathbf q
Applying this to the angular momentum vector yields
\left(\frac {d\mathbf L}{dt}\right)_I =<br />
\left(\frac {d\mathbf L}{dt}\right)_R + \boldsymbol{\omega} \times \mathbf L
The left hand side is the external torque. Assuming a rigid body, the time derivative of the inertia tensor in the rotating frame vanishes. With this assumption,
\boldsymbol I \frac{d{\boldsymbol{\omega}}_R}{dt} =<br />
\boldsymbol{\tau} -<br />
\boldsymbol{\omega}_R \times (\boldsymbol I \boldsymbol{\omega}_R)
The second term on the right, -
ω×(
Iω) is the inertial torque. It is a fictitious torque, somwhat analogous to the Coriolis force (but without the factor of 2).-----------------------------------------
1[/color] You can find hand-waving proofs of this in most undergraduate classical physics texts. A non-handwaving proof requires showing that the time derivative of the transformation matrix from the rotating to the inertial frame is
\frac{d}{dt}\boldsymbol T_{R\to I} =<br />
\boldsymbol T_{R\to I}\boldsymbol S(\boldsymbol{\omega}_R)
where
S(
ωR) is the skew symmetric matrix generated from the angular velocity vector
ωR as expressed in rotating frame coordinates.
