As you can see from this wiki link, the relation between torque and angular acceleration is derived in this way:
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
$$\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}$$
where $L$ is the angular momentum vector and $t$ is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
$$\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}$$
For rotation about a fixed axis,
$$\mathbf{L} = I\boldsymbol{\omega}$$
where $I$ is the moment of inertia and $\omega$ is the angular velocity. It follows that
$$\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha}$$
where $α$ is the angular acceleration of the body, measured in $rad/s^2$.
Your answer specifically lies in the following paragraph:
This equation has the limitation that the torque equation describes
the instantaneous axis of rotation or center of mass for any type of
motion – whether pure translation, pure rotation, or mixed motion. $I$
= Moment of inertia about the point which the torque is written (either instantaneous axis of rotation or center of mass only). If
body is in translatory equilibrium then the torque equation is the
same about all points in the plane of motion.
Few points I would like to emphasise on:
- In general, in the formula, $\mathbf{L} = I\boldsymbol{\omega}$, $I$ is the moment of inertia tensor and depending on the axis of rotation and our assumption of co-ordinate axes, it becomes a scalar or a vector and so on.
- In $\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha}$, $I$ is a constant, hence it is taken out of the term $\frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t}$ to form the term $I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t}$.But here $I$ is not constant. So you cannot use that formula.
Point Masses
For two point masses colliding the impulse $J$ (instant momentum exchange) is defined by the scalar equation $$ J = (1+\epsilon)\, \mu\, v_{imp} $$
Where:
- $\epsilon$ is the coefficient of restitution.
- $v_{imp} = {\bf n}^\top ( {\bf v}_1-{\bf v}_2)$ the impact speed (relative speed along contact normal direction ${\bf n}$).
- $\mu^{-1} = \frac{1}{m_1} + \frac{1}{m_2}$ or $\mu = \frac{m_1 m_2}{m_1+m_2}$ the reduced mass of the system. I prefer the term, the effective mass of the impact.
This impulse has an equal and opposite action on the two bodies. The changes in motion described by
$$ \begin{align} \Delta {\bf v}_1 & = -\frac{J }{m_1} \,{\bf n} \\\Delta {\bf v}_2 & = +\frac{J}{m_2} \,{\bf n} \end{align} $$
Rigid Bodies
Now suppose that bodies [1] & [2] are extended rigid bodies, instead of point masses. They are defined at the moment of impact by the vectors ${\bf c}_1$ and ${\bf c}_2$ for the location of the center of mass relative to the impact point, and ${\rm I}_1$, ${\rm I}_2$ the 3×3 mass moment of inertia matrices at the center of mass (and along with the world coordinates).
The scalar equation for the impulse is the same as the point mass except for the effective mass $\mu$ accounting for the inertial properties as well. This assumes a frictionless contact.
$$ J = (1+\epsilon) \mu \; v_{imp} $$
Where:
- $$\boxed{v_{imp} = {\bf n}^{\top}\left({\bf v}_{1}+{\bf c}_{1}\times{\boldsymbol{\omega}}_{1}-{\bf v}_{2}-{\bf c}_{2}\times{\boldsymbol{\omega}}_{2}\right)}$$ is the relative speed along the contact normal ${\bf n}$.
- $$\boxed{\mu^{-1} = \frac{1}{m_{1}}+\frac{1}{m_{2}}-{\bf n}^{\top}\left({\bf c}_{1}\times{\rm I}_{1}^{-1}({\bf c}_{1}\times{\bf n})+{\bf c}_{2}\times{\rm I}_{2}^{-1}({\bf c}_{2}\times{\bf n})\right)}$$ is the effective mass of the impact.
So the effect of a rigid body is the additional term $-{\bf n}^{\top}\left({\bf c}_{i}\times{\rm I}_{i}^{-1}\left({\bf c}_{i}\times{\bf n}\right)\right)$ to the inverse effective mass. This terms ends up being positive when the components are expanded. So the rigid body increased the inverse mass, ==> reduces the effective mass at the impact point.
This impulse $J$ changes the motion of the two bodies according to the equations of motion
$$ \begin{aligned}
\Delta {\bf v}_1 &= -\frac{J}{m_1} {\bf n} & \Delta {\boldsymbol \omega}_1 &=
{\rm I}_1^{-1} {\bf c}_1 \times J {\bf n} \\
\Delta {\bf v}_2 &= \frac{J}{m_2} {\bf n} & \Delta {\boldsymbol \omega}_2 &= -{\rm I}_2^{-1} {\bf c}_2 \times J {\bf n}
\end{aligned} $$
Example
A vertical rod (body [1]) of length $\ell$ is impacted by a point mass (body [2]) moving along the $-x$ direction at the end point.
Since we have a point mass ${\bf c}_2=0$. For the rod the mass moment of inertia about the $z$ axis is ${\rm I}_1 = \left[ \matrix{ \ddots & & \\ & \ddots & \\ & & \frac{m_1}{12} \ell^2} \right] $ and the location of the center of mass relative to the contact point is ${\bf c}_1 = \pmatrix{0 \\ -\frac{\ell}{2} \\ 0}$. Finally, the contact normal is ${\bf n} = \pmatrix{1 \\ 0 \\ 0}$.
The effective mass at the impact point is
$$ \mu^{-1}= \frac{1}{m_1} + \frac{1}{m_2} - \pmatrix{1 \\ 0 \\ 0}^\top \pmatrix{0\\ -\frac{\ell}{2} \\ 0} \times \pmatrix{0 \\ 0 \\ \frac{6}{m_1 \ell}} = \frac{1}{m_1} + \frac{1}{m_2} + \frac{3}{m_1}$$
$$\mu = \frac{m_1 m_2}{m_1 +4 m_2} $$
Appendix
If ${\bf v}_1$ is the velocity of the center of mass for body [1] (${\bf v}_2$ of body [2]), then the velocity vectors at the impact point A are defined by
$$ \begin{aligned}
{\bf v}_1^A & = {\bf v}_1 + {\bf c}_1 \times {\boldsymbol \omega}_1 \\
{\bf v}_2^A & = {\bf v}_2 + {\bf c}_2 \times {\boldsymbol \omega}_2 \end{aligned} $$
The impact speed is thus defined as $$v_{imp} = {\bf n}^\top ({\bf v}_1 + {\bf c}_1 \times {\boldsymbol \omega}_1 - {\bf v}_2 - {\bf c}_2 \times {\boldsymbol \omega}_2)$$
The effect of the impulse $J$ along ${\bf n}$ at the point of impact is felt as
$$ \begin{aligned}
\Delta {\bf v}_1 &= -\frac{J}{m_1} {\bf n} & \Delta {\boldsymbol \omega}_1 &= -{\rm I}_1^{-1} (-{\bf c}_1) \times J {\bf n} \\
\Delta {\bf v}_2 &= +\frac{J}{m_2} {\bf n} & \Delta {\boldsymbol \omega}_2 &= +{\rm I}_2^{-1} (-{\bf c}_2) \times J {\bf n}
\end{aligned} $$
In turn, these changes in the motion of the body change the motion at the impact point *A as
$$ \begin{align}
\Delta {\bf v}_1^A & = \Delta {\bf v}_1 + {\bf c}_1 \times \Delta {\boldsymbol \omega}_1 \\
\Delta {\bf v}_2^A & = \Delta {\bf v}_2 + {\bf c}_2 \times \Delta {\boldsymbol \omega}_2 \end{align} $$
The law of impact states that along with the contact normal the rebound velocity is a fraction of the impact velocity
$$ {\bf n}^\top ( {\bf v}_1^A + \Delta {\bf v}_1^A - {\bf v}_2^A -\Delta {\bf v}_2^A) = -\epsilon \; {\bf n}^\top ( {\bf v}_1^A - {\bf v}_2^A ) $$
By moving the known (pre-impact) motions on the right-hand side we get
$${\bf n}^\top ( \Delta {\bf v}_1^A - \Delta {\bf v}_2^A) = -(1+\epsilon) \; v_{imp} $$
Add the effect of the impulse $J$ to the above to get
$${\bf n}^\top ( \Delta {\bf v}_1 + {\bf c}_1 \times \Delta {\boldsymbol \omega}_1 - \Delta {\bf v}_2 - {\bf c}_2 \times \Delta {\boldsymbol \omega}_2) = -(1+\epsilon) \; v_{imp} $$
$${\bf n}^\top ( -\frac{J}{m_1} {\bf n} - {\bf c}_1 \times {\rm I}_1^{-1} (-{\bf c}_1) \times J {\bf n} - \frac{J}{m_2} {\bf n} - {\bf c}_2 \times {\rm I}_2^{-1} (-{\bf c}_2) \times J {\bf n}) = -(1+\epsilon) \; v_{imp} $$
$${\bf n}^\top ( -\frac{1}{m_1} {\bf n} + {\bf c}_1 \times {\rm I}_1^{-1} {\bf c}_1 \times {\bf n} - \frac{1}{m_2} {\bf n} + {\bf c}_2 \times {\rm I}_2^{-1}( {\bf c}_2 \times {\bf n})) J = -(1+\epsilon) \; v_{imp} $$
$$ \left(\frac{1}{m_{1}}+\frac{1}{m_{2}}-{\bf n}^{\top}\left({\bf c}_{1}\times{\rm I}_{1}^{-1}({\bf c}_{1}\times{\bf n})+{\bf c}_{2}\times{\rm I}_{2}^{-1}({\bf c}_{2}\times{\bf n})\right)\right)\,J=(1+\epsilon)\,v_{imp} $$
which is solved for $J$
Related answers:
Best Answer
To stop the object you must do work. For a constant torque perpendicular to the moment arm, the work it does is equal to $\tau\cdot\Delta\theta$, and you want $\Delta\theta\leq2\pi$.
It should be obvious that the greatest angular velocity that a torque $\tau$ can stop will take it the full $2\pi$ radians to stop. In a rotating system, the rotational kinetic energy is given by $E_r=\frac12I\omega^2$ (a direct analogue of $E_K=\frac12mv^2$ ). Now consider work-energy equivalence.