Physics Asked by Andrew Tomazos on March 9, 2021
Suppose I have a torque $tau in mathbb{R}^3$ on a rigid body, about its center of gravity, where the direction of $tau$ is the axis of the torque and the magnitude of $tau$ is the strength of the torque.
Let’s suppose the rigid body is on some infinitely strong fixed axel, through its center of gravity, with unit vector axis $alpha in mathbb{R}^3, |alpha| = 1$. Notice that $tau$ and $alpha$ may be in different directions.
The torque $tau$ is therefore equivalent to some torque $kalpha$ where $k in mathbb{R}$.
If $tau$ and $alpha$ are perpendicular ($tau cdot alpha = 0$) then $k$ is zero because the torque would be entirely absorbed by the axel.
If $tau$ and $alpha$ are collinear, then there exists some $c in mathbb{R}$ such that $calpha = tau$. In such a case none of the torque is absorbed by the axel, and $k = c$
For the remaining cases what is $k$ in terms of $tau$ and $alpha$ ?
Saying that the torque $vectau$ and the rotation axis $vecalpha$ forms an angle $theta$ in between. The projection of $vectau$ on $vecalpha$ is
$$ tau_parallel=tau costheta $$
Hence, the angular acceleration is
$$ c alpha = tau_parallel = tau cos theta k alpha = frac{c}{costheta}alpha=tau k = frac{c}{costheta} $$
Answered by ytlu on March 9, 2021
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