5. Cross Product

e. Applications

1. Balancing Torques

If the object rotates about a fixed rotation axis, \(\hat N\), by a rotation angle, \(\theta\), then its angular velocity is: \[ \vec\omega=\dfrac{d\theta}{dt}\hat N \] and its angular acceleration is: \[ \vec\alpha=\dfrac{d^2\theta}{dt^2}\hat N \]

Newton's Second Law of Motion can be generalized to his Newton's Second Law of Rotational Motion:

If one or more unbalanced forces act on an object, then the object will move with a (linear) acceleration \(\vec a\) satisfying: \[ m\,\vec a=\vec F \] where \(m\) is the mass of the object and \(\vec F\) is the sum of all forces acting on the object.

Analogously:

If one or more unbalanced torques act on an object, then the object will rotate with an angular acceleration \(\vec\alpha\) satisfying: \[ I\,\vec\alpha=\vec\tau \] where \(I\) is the moment of inertia of the object and \(\vec\tau\) is the sum of all torques acting on the object.

If the object is static (not moving) then its angular acceleration is \(\vec\alpha=\vec 0\) and the sum of the torques must be zero. We can use this to find one or more torques when we know the other torques in a static situation.

Since \(\vec\alpha=\vec 0\), we will not need to understand the moment of inertia in this chapter.

The lever arms and torques are: \[\begin{aligned} &\vec r_1=P_1-O=\left\langle 0,3,0\right\rangle &\quad &\vec\tau_1=\vec r_1\times\vec F_1 =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 0 & 3 & 0 \\ 2 & -1 & -1 \end{vmatrix} =\left\langle -3,0,-6\right\rangle \\ &\vec r_2=P_2-O=\left\langle 0,-2,0\right\rangle &\quad &\vec\tau_2=\vec r_2\times\vec F_2 =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 0 & -2 & 0 \\ 3 & 1 & -2 \end{vmatrix} =\left\langle 4,0,6\right\rangle \end{aligned}\] So the sum of the torques is: \[ \vec\tau_1+\vec\tau_2 =\left\langle -3,0,-6\right\rangle+\left\langle 4,0,6\right\rangle =\left\langle 1,0,0\right\rangle \] and the object would start rotating. To counteract this torque, the third torque must be \(\vec\tau_3=\left\langle -1,0,0\right\rangle\). If the force, \(\vec F_3=\left\langle a,b,c\right\rangle\), is applied at the point, \(P_3=(0,1,0)\) then the lever arm and the torque are: \[ \vec r_3=P_3-O=\left\langle 0,1,0\right\rangle \qquad \vec\tau_3=\vec r_3\times\vec F_3 =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 0 & 1 & 0 \\ a & b & c \end{vmatrix} =\left\langle c,0,-a\right\rangle \] Equating the two expressions for \(\vec\tau_3\) we get \(c=-1\), \(a=0\) and \(\vec b\) is abitrary. So the torque is balanced by a force of the form, \(\vec F_3=\left\langle 0,b,-1\right\rangle\), for any \(b\).

Notice that \(b\) is arbitrary because the vector \(\left\langle 0,b,0\right\rangle\) is parallel to the lever arm \(\overrightarrow{OP_3}=\left\langle 0,1,0\right\rangle\) and so has no effect on the torque.