Chapter 10, Problem 73AP

(a)

To determine

The angular speed of the rod.

Answer to Problem 73AP

The angular speed of the rod is $\sqrt{\frac{3g}{L}}$.

Explanation of Solution

For calculation of gravitational energy, rigid body can be modeled as particle at center of mass.

Only conservative forces are acting on the bar, we have conservation of energy of the bar-Earth system. This can be solving by Conservation of energy.

Write the expression for conservation of energy as.

    $\left(K_f-K_i\right)+\left(U_f-U_i\right)=0$                                                                             (I)

Here, $K_f$ is final kinetic energy, $K_i$ is initial kinetic energy, $U_f$ is final potential energy and $U_i$ is initial potential energy.

Substitute $0$ for $U_f$ and $0$  for $K_i$ in equation (I).

    $\left(K_f-0\right)+\left(0-U_i\right)=0$

Rearrange the above equation as.

    $K_f-U_i=0$                                                                                                  (II)

Write the expression for final kinetic energy of the rod as.

    $K_f=\frac{1}{2}I{\omega }_f{}^2$

Here, $I$ is the moment of inertia of rod and ${\omega }_f$ is final angular speed.

The moment of inertia of rod is $\frac{1}{3}M{L}^2$.

Substitute $\frac{1}{3}M{L}^2$ for $I$ in above equation as.

    $K_f=\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){\omega }_f{}^2$

Here, $M$ is the mass of rod and $L$ is the length of rod.

Rearrange the above equation as.

    $K_f=\frac{1}{6}M{L}^2{\omega }_f{}^2$

Write the expression for initial potential energy as.

    $U_i=\frac{1}{2}MgL$

Here, $g$ is the acceleration under gravity.

Conclusion:

Substitute $\frac{1}{2}MgL$ for $U_i$ and $\frac{1}{6}M{L}^2{\omega }_f{}^2$ for $K_f$ in equation (II)

    $\frac{1}{6}M{L}^2{\omega }_f{}^2-\frac{1}{2}MgL=0$

Simplify the above equation as.

    ${\omega }_f=\sqrt{\frac{3g}{L}}$

Thus, the angular speed of the rod is $\sqrt{\frac{3g}{L}}$.

(b)

To determine

The magnitude of angular acceleration.

Answer to Problem 73AP

The magnitude of the angular acceleration is $\frac{3g}{2L}$. $\begin{array}{l}Mg\\ L/2\end{array}$

Explanation of Solution

Redraw the figure P10.73 as.

Angular acceleration can be found using torque analysis.

Write the expression for torque of the rod as.

    ${\displaystyle \sum \tau =I\alpha }$                                                                                                    (III)

Here, $\alpha$ is the angular acceleration and $\tau$ is the torque.

Write the expression for torque in terms of force and distance as.

    $\tau =Fd$                                                                                                        (IV)

Here, $F$ is the force acting on rod and $d$ is the length at which the pivot point is present.

The length is between a center of rotation and the pivot point where a force is applied that means distance for the pivot is $\frac{L}{2}$. The force acting on rod is $Mg$.

Substitute $\frac{L}{2}$ for $d$ and $Mg$ for $F$ in equation (IV)

  $\tau =Mg\left(\frac{L}{2}\right)$                                                                                                  (V)

Substitute $\frac{1}{3}M{L}^2$ for $I$ and $Mg\left(\frac{L}{2}\right)$ for $\tau$ in equation (III).

    $Mg\left(\frac{L}{2}\right)=\frac{1}{3}M{L}^2\alpha$

Rearrange the above expression for angular acceleration as.

    $\alpha =\frac{3g}{2L}$

Thus, the magnitude of the angular acceleration is $\frac{3g}{2L}$.

Conclusion:

(c)

To determine

The $x$ and $y$ components of the acceleration of its center of mass.

Answer to Problem 73AP

The net acceleration for the system is $-\frac{3g}{2}\widehat{i}-\frac{3g}{4}\widehat{j}$.

Explanation of Solution

Redraw the figure P10.73 for calculate the components of the acceleration as

Refer to figure, the horizontal acceleration is directed towards the centre. If an object is free to move due to fixed point. The object will follow the circular path. Since, the horizontal acceleration will act as centripetal acceleration.

Write the expression for horizontal acceleration as.

    ${\overrightarrow{a}}_x=-a_c$                                                                                                      (VI)

Here, ${\overrightarrow{a}}_x$ is horizontal acceleration and $a_c$ is centripetal acceleration.

Negative sign shows the direction of acceleration towards the negative $x$ axis on $xy$ plane.

Write the expression for centripetal acceleration for the rod as.

    $a_c=r{\omega }_f{}^2$

Here, $r$ is the radius of circular path.

The radius of circular path for this system is defined as the half length of the rod that is $\left(\frac{L}{2}\right)$ for $r$.

Substitute $\sqrt{\frac{3g}{L}}$ for ${\omega }_f$ and $\left(\frac{L}{2}\right)$ for $r$ in above equation as.

    $a_c=\left(\frac{L}{2}\right){\left(\sqrt{\frac{3g}{L}}\right)}^2$

Simplify the above equation as.

    $a_c=\left(\frac{3g}{2}\right)$

As the vertical acceleration will be same as tangential acceleration the rod follows the circular path.

Write the expression for vertical acceleration as.

    ${\overrightarrow{a}}_y=-a_t$                                                                                                     (VII)

Here, ${\overrightarrow{a}}_y$ is vertical acceleration and $a_t$ is tangential acceleration.

Write the expression for tangential acceleration as.

    $a_t=r\alpha$

Substitute $\left(\frac{L}{2}\right)$ for $r$ and $\frac{3g}{2L}$ for $\alpha$ in above equation as.

    $a_t=\left(\frac{L}{2}\right)\left(\frac{3g}{2L}\right)$

Simplify the above equation as.

    $a_t=\frac{3g}{4}$

Write the expression for net acceleration for the $xy$ plane as.

    $\overrightarrow{a}={\overrightarrow{a}}_x\widehat{i}+{\overrightarrow{a}}_y\widehat{j}$                                                                                           (VIII)

Here, $\overrightarrow{a}$ is the net acceleration for this system.

Conclusion:

Substitute $\left(\frac{3g}{2}\right)$ for $a_c$ in above equation (VI) as.

    ${\overrightarrow{a}}_x=-\frac{3g}{2}$

This is centripetal acceleration; it is directed along the negative horizontal.

Substitute $\frac{3g}{4}$ for $a_t$ in equation (VII).

    ${\overrightarrow{a}}_y=-\frac{3g}{4}$

This is centripetal acceleration; it is directed along the negative vertical.

Substitute $-\frac{3g}{2}$ for ${\overrightarrow{a}}_x$ and $-\frac{3g}{4}$ for ${\overrightarrow{a}}_y$ in equation (VIII)

    $\overrightarrow{a}=\left(-\frac{3g}{2}\right)\widehat{i}+\left(-\frac{3g}{4}\right)\widehat{j}$

Simplify the above equation as.

    $\overrightarrow{a}=-\frac{3g}{2}\widehat{i}-\frac{3g}{4}\widehat{j}$

Thus, the net acceleration for the system is $-\frac{3g}{2}\widehat{i}-\frac{3g}{4}\widehat{j}$.

(d)

To determine

The components of the reaction force at the pivot.

Answer to Problem 73AP

The net force exerts on the rod at the pivot is $-\frac{3Mg}{2}\widehat{i}+\frac{1}{4}Mg\widehat{j}$.

Explanation of Solution

The pivot exerts a force $\overrightarrow{F}$ on the rod. Use Newton’s second law for the calculation of force.

Write the expression for force on the rod as.

    ${\overrightarrow{F}}_{net}={\overrightarrow{F}}_x+{\overrightarrow{F}}_y$                                                                                           (IX)

Here, ${\overrightarrow{F}}_{net}$ is net force, ${\overrightarrow{F}}_x$ is horizontal force and ${\overrightarrow{F}}_y$ is vertical force.

Write the expression for horizontal force as.

    ${\overrightarrow{F}}_x=M{\overrightarrow{a}}_x\widehat{i}$

Substitute $-\frac{3g}{2}$ for ${\overrightarrow{a}}_x$ in above equation as.

    ${\overrightarrow{F}}_x=M\left(-\frac{3g}{2}\right)\widehat{i}$

Rearrange the above equation as.

    ${\overrightarrow{F}}_x=-\frac{3Mg}{2}\widehat{i}$

Write the expression for vertical force as.

    ${\overrightarrow{F}}_y=M\left({\overrightarrow{a}}_y+g\right)\widehat{j}$

Substitute $-\frac{3g}{4}$ for ${\overrightarrow{a}}_y$ in above equation as.

    ${\overrightarrow{F}}_y=M\left(-\frac{3g}{4}+g\right)\widehat{j}$

Simplify the above equation as.

    ${\overrightarrow{F}}_y=\frac{1}{4}Mg\widehat{j}$

Conclusion:

Substitute $-\frac{3Mg}{2}\widehat{i}$ for ${\overrightarrow{F}}_x$ and $\frac{1}{4}Mg\widehat{j}$ for ${\overrightarrow{F}}_y$ in equation (IX).

    ${\overrightarrow{F}}_{net}=-\frac{3Mg}{2}\widehat{i}+\frac{1}{4}Mg\widehat{j}$

Thus, the net force exerts on the rod at the pivot is $-\frac{3Mg}{2}\widehat{i}+\frac{1}{4}Mg\widehat{j}$.