How is the following relation true
$$\tau = \large\frac{I}{g} \times \alpha$$
where $\tau$ is torque,
$I$ is moment of inertia,
$g= 9.8ms^{-2}$,
and $\alpha=$ angular acceleration.
[Physics] How is torque equal to moment of inertia times angular acceleration divided by g
angular momentumhomework-and-exercisesrotational-dynamicsrotational-kinematicstorque
Related Solutions
This issue is a bit confusing because there are two types of angular momentum. There's spin, where a rigid body rotates about an axis through its center of mass, and there's orbital, where the center of mass of a rigid body rotates about an axis. For example, the Earth spins about its axis and rotates around the Sun. The total angular momentum can always be decomposed into a sum of these two terms.
You probably have never heard of this, because the definition of the torque is just $\mathbf{r} \times \mathbf{F}$, which doesn't have any reference to "spin or orbital". But when you write the equation $\tau = I \alpha$, you're implicitly choosing to talk about one or the other, or else $\tau$, $I$, and $\alpha$ are meaningless. (For example, the Earth takes 1 day to spin but 1 year to orbit. So you can't just say "the" angular velocity of the Earth.)
If you're talking about spin, $I$ means the moment of inertia about the center of mass, $\alpha$ means the angular acceleration about an axis through the center of mass, and $\tau$ means the torque about the center of mass. There's no meaningful way to change axes or origin because it's always the center of mass.
Now let's talk about the orbital part. The instantaneous angular velocity can always be defined, even if the object isn't moving in a circle about some point, using the equation $$\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$$ However, you can see this quantity is frame dependent: if we just move the origin, $\mathbf{r}$ will change but $\mathbf{v}$ won't, so $\boldsymbol{\omega}$ will change. As an example, if you see an airplane from the ground, its angular velocity appears very low, but if you're hovering next to it, it zips around really fast. So in this case, everything does change.
Okay, maybe neither of these examples were really what you wanted: the first was trivial, and the second didn't say much. We can get more insight by not doing the spin-orbital decomposition at all, which requires tossing out $\boldsymbol{\omega}$ and $\alpha$. The only rotational equation we have left is $$\tau = \frac{dL}{dt}$$ which expanded out is $$ \sum \mathbf{r} \times \mathbf{F} = \frac{d}{dt}\left( \sum \mathbf{r} \times \mathbf{p}\right)$$ Let's transform to another frame. To make it easy, let's just shift the origin by $\mathbf{r}_0$. Now if physics works, the resulting equation should be equivalent.
Let's confirm it. In the other frame, we have $$ \sum (\mathbf{r} + \mathbf{r}_0) \times \mathbf{F} = \frac{d}{dt}\left(\sum (\mathbf{r} + \mathbf{r}_0) \times \mathbf{p}\right)$$ If we subtract out our previous equation, we're left with $$ \sum \mathbf{r}_0 \times \mathbf{F} = \frac{d}{dt} \left( \sum \mathbf{r}_0 \times \mathbf{p} \right)$$ Since $\mathbf{r}_0$ is constant, the derivative only acts on $\mathbf{p}$, giving $\sum \mathbf{r}_0 \times \mathbf{F}$ on the right hand side. So the equation is true. Physics works!
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.
Best Answer
This is only true for engineering units which have $I$ in ${\rm lbf\,in^2}$. In the metric system the units of $I$ are ${\rm kg\, m^2}$. So to convert force ${\rm lbf}$ to mass you divide by $g$.