18.5 Rotation (check your understanding)  (Page 4/2)

This question is included here with the purpose to highlight conceptual and visualization aspects of the calculation of torque. The torque about the axis of rotation i.e. z-axis is given as :

$$\begin{array}{l}\mathbf{\tau }=\mathbf{r}\mathbf{x}\mathbf{F}\end{array}$$

where “r” and “F” are in the plane perpendicular to the axis of rotation i.e. in “xy”-plane. Explicitly, we can write :

$$\begin{array}{l}{\mathbf{\tau }}_z={\mathbf{r}}_{\mathrm{xy}}\mathbf{x}{\mathbf{F}}_{\mathrm{xy}}\end{array}$$

The figure above shows the force component in “xy”-plane. Have a careful look at the force component in relation to the point of application, "P". The rigid body is deliberately not shown so that we can visualize vector components without any congestion in the figure.

Note here that body is rotating about z-axis. Thus, we need to calculate torque about z-axis only. The z-component of force does not constitute a torque about z-axis. Hence,

$$\begin{array}{l}{\mathbf{F}}_{\mathrm{xy}}=2\mathbf{i}+\mathbf{j}\end{array}$$

The component of radius vector in xy - plane is shown in the figure below. Note that it is measured in the plane of rotation of the point of application of force about the axis. Again, we drop z-component. The “xy” components of force and radius vectors are shown in the red colors in the figure.

$$\begin{array}{l}{\mathbf{r}}_{\mathrm{xy}}=\mathbf{i}+\mathbf{j}\end{array}$$

Therefore, torque about the axis of rotation is :

$$\begin{array}{l}{\mathbf{\tau }}_z={\mathbf{r}}_{\mathrm{xy}}\mathbf{x}{\mathbf{F}}_{\mathrm{xy}}=(\mathbf{i}+\mathbf{j})\mathbf{x}(2\mathbf{i}+\mathbf{j})=\mathbf{k}-2\mathbf{k}=-\mathbf{k}\end{array}$$

The magnitude of torque is :

$$\begin{array}{l}{\tau }_z=1N-m\end{array}$$

Note that the torque is in the direction of “-z”-axis and the body is rotating clockwise.

Alternatively, we can also work this problem without restricting evaluation of vector product in any plane as :

$$\begin{array}{l}\mathbf{\tau }=\mathbf{r}\mathbf{x}\mathbf{F}\end{array}$$

τ = | i j k | | 1 1 1 || 2 1 3 |

$$\begin{array}{l}\mathbf{\tau }=(1x3-1x1)\mathbf{i}+(1x2-1x3)\mathbf{j}+(1x1-2x1)\mathbf{k}\end{array}$$

$$\begin{array}{l}\mathbf{\tau }=2\mathbf{i}-\mathbf{j}-\mathbf{k}\end{array}$$

The component of torque in z-direction is :

$$\begin{array}{l}{\mathbf{\tau }}_z=-\mathbf{k}\end{array}$$

The magnitude of torque is :

$$\begin{array}{l}{\tau }_z=1N-m\end{array}$$

Hence, option (a) is correct.

The rod will not rotate when torques due to the forces are equal in magnitude and opposite in direction :

$$\begin{array}{l}{\tau }_1={\tau }_2\end{array}$$

$$\begin{array}{l}{F}_{1}x{r}_1=F_{2}x{r}_2\end{array}$$

Since $r_1<r_2$ , we can conclude $F_1>F_2$ . The magnitude of torque due to force $F_1$ is given as :

$$\begin{array}{l}{\tau }_1=r_{1}x{F}_1\sin \theta \end{array}$$

Evidently, torque due to $F_1$ is reduced when angle of application of force with respect to rod changes. In order to maintain the balance, the force $F_1$ needs to be increased or the force $F_2$ is decreased or the force $F_2$ is brought closure to the pivot.

Hence, options (a), (b) and (d) are correct.