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Figure B.7: Application of torque $ \tau $ about the origin given by a tangential force $ f_t$ on a lever arm of length $ R$.

When twisting things, the rotational force we apply about the center is called a torque (or moment, or moment of force). Informally, we think of the torque as the tangential applied force $ f_t$ times the moment arm (length of the lever arm) $ R$

$\displaystyle \tau \isdefs f_tR \protect$ (B.26)

as depicted in Fig.B.7. The moment arm is the distance from the applied force to the point being twisted. For example, in the case of a wrench turning a bolt, $ f_t$ is the force applied at the end of the wrench by one's hand, orthogonal to the wrench, while the moment arm $ R$ is the length of the wrench. Doubling the length of the wrench doubles the torque. This is an example of leverage. When $ R$ is increased, a given twisting angle $ \theta$ is spread out over a larger arc length $ R\theta$, thereby reducing the tangential force $ f_t$ required to assert a given torque $ \tau $.

For more general applied forces $ \underline{f}\in{\bf R}^3$, we may compute the tangential component $ \underline{f}_t$ by projecting $ \underline{f}$ onto the tangent direction. More precisely, the torque $ \tau $ about the origin $ \underline{0}$ applied at a point $ \underline{x}\in{\bf R}^3$ may be defined by

$\displaystyle \underline{\tau}\isdefs \underline{x}\times \underline{f} \protect$ (B.27)

where $ \underline{f}$ is the applied force (at $ \underline{x}$) and $ \times$ denotes the cross product, introduced above in §B.4.12.

Note that the torque vector $ \underline{\tau}$ is orthogonal to both the lever arm and the tangential-force direction. It thus points in the direction of the angular velocity vector (along the axis of rotation).

The torque magnitude is

$\displaystyle \tau \isdefs \vert\vert\,\tau\,\vert\vert \eqsp \vert\vert\,\unde...
...{x}\,\vert\vert \cdot \vert\vert\,\underline{f}\,\vert\vert \cdot\sin(\theta),

where $ \theta$ denotes the angle from $ \underline{x}$ to $ \underline{f}$. We can interpret $ \vert\vert\,\underline{f}\,\vert\vert \sin(\theta)$ as the length of the projection of $ \underline{f}$ onto the tangent direction (the line orthogonal to $ \underline{x}$ in the direction of the force), so that we can write

$\displaystyle \tau \eqsp \vert\vert\,\underline{x}\,\vert\vert \cdot \vert\vert\,\underline{f}_t\,\vert\vert

where $ \vert\vert\,\underline{f}_t\,\vert\vert = \vert\vert\,\underline{f}\,\vert\vert \sin(\theta)$, thus getting back to Eq.$ \,$(B.26).

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Newton's Second Law for Rotations
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