A board of length $\mathbf{2}\mathit{d}\mathbf{=}\mathbf{6}\mathit{m}$rests on a cylinder (the “pivot”). A ball of mass $\mathbf{5}\mathit{k}\mathit{g}$is placed on the end of the board. Figure 11.104 shows the objects at a particular instant. (a) On a free-body diagram, show the forces acting on the ball + board system, in their correct locations. (b) Take the point at which the board touches the cylinder as location $\mathit{A}\mathbf{.}$ What is the magnitude of the torque on the system of (ball + board) about location$\mathit{A}\mathbf{?}$(c) Which of the following statements are correct? (1) Because there is a torque, the angular momentum of the system will change in the next tenth of a second. (2) The forces balances, so the angular momentum of the system about location $\mathit{A}$will not change. (3) The forces by the cylinder on the board contributes nothing to the torque about the location $\mathit{A}$.

Torque is the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.

(a)The point at which the board touches the cylinder will be at rest. Due to the gravitational force on the ball there is a torque that rotates the ball-board system in anti-clock wise direction.

The following figure represents the free body diagram of the ball-board system.

In the figure, $m$is mass of the ball and$g$ is acceleration due to gravity and localid="1668602571675" $mg$represents the gravitational force on the ball which is downward direction. The point at which the board touches the cylinder will be stationary because the normal reaction force and gravitational force$Mg$on the ball-board system are balanced. Thus, there will be net force acting on the system which causes rotation of the system.

(b)The expression for torque acting on the ball board system about the point $A$is given as follows:

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Here, represents the perpendicular distance of the ball from point $A$and $\overrightarrow{F}$is the applied force.

The force is applied perpendicular to the board thus the torque is maximum and it is directed along the negative localid="1668602702078" $y-$axis.

Replace$\overrightarrow{F}$by then the magnitude of the torque is given as follows:

$\tau =r\left(mg\right)$

Substitute $3m$for localid="1668602640934" $r,5kg$for $m$and $9.8m/s^2$for $g.$

$$\begin{gathered} \tau =\left(3m\right)\left(5kg\right)(9.8m/s^2) \\ =147N.m \end{gathered}$$

Therefore, the magnitude of the torque acting on the system about the point$A$is$147N.m.$

(c) The net torque acting on the system is non zero and hence there will be a change in the angular momentum of the system and as discussed in part b, there is no effect of force of cylinder on the board in contribution of torque of the system about point$A.$

Even the forces are balanced at point$A$ there is net downward force acting on the system, then the torque rotates the system and hence there will be a change in the angular momentum of the system.

Hence, the statements (1) and (3) are correct.