Rotational Force

on . Posted in Classical Mechanics

Rotational force, abbreviated as \(\tau\), also known as torque, is the twisting or turning force that acts on an object due to the application of a force at a distance from the object's axis of rotation.  The torque produced by a force depends on the magnitude of the force, the distance from the axis of rotation at which the force is applied, and the direction of the force relative to the axis of rotation.  The cross product operator indicates that the torque is perpendicular to both the force vector and the radius vector.

This is important in the study of mechanics and the design of machines, such as engines, motors, and turbines.  It is also used in the analysis of rotational motion and the development of new technologies, such as robotics and automation systems.

 

Rotational Force formula

\( \tau = I \; \alpha  \)     (Rotational Force)

\( I =  \tau \;/\; \alpha \)

\( \alpha =  \tau \;/\; I \)

Symbol English Metric
\( \tau \)  (Greek symbol tau) = rotational force \(lbf - ft\) \(N-m\)
\( I  \) = moment of inertia \(in^4\) \(mm^4\)
\( \alpha \)  (Greek symbol alpha) = angular acceleration \(deg\;/\;sec^2\) \(rad\;/\;s^2\)

 

Rotational Force formula

\( \tau = r \; x \; F_a \)     (Rotational Force)

\( r =  \tau \;/\; x \; F_a \)

\( x =  \tau \;/\; r \; F_a \)

\( F_a =  \tau \;/\; r \; x \)

Symbol English Metric
\( \tau \)  (Greek symbol tau) = rotational force \(lbf - ft\) \(N-m\)
\( r  \) = distance between the axis of rotation and the point where the force is applied \(ft\) \(m\)
\( x \) = cross product operator \(dimensionless\)
\( F_a  \) = applied force \(lbf\) \(N\)

 

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Tags: Torque Rotational