Chapter 11, Problem 11.49AP

A rigid, massless rod has three particles with equal masses attached to it as shown in Figure P11.37. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through the point P and is released from rest in the horizontal position at t = 0. Assuming m and d are known, find (a) the moment of inertia of the system of three particles about the pivot, (b) the torque acting on the system at t = 0, (c) the angular acceleration of the system at t = 0, (d) the linear acceleration of the particle labeled 3 at t = 0, (e) the maximum kinetic energy of the system, (f) the maximum angular speed reached by the rod, (g) the maximum angular momentum of the system, and (h) the maximum speed reached by the particle labeled 2.

Figure P11.37

To determine

The moment of inertia of the system of three particles about the pivot.

Answer to Problem 11.49AP

The moment of inertia of the system of three particles about the pivot is $7m\left(\frac{d^2}{3}\right)$.

Explanation of Solution

The mass of three particles is $m$, the distance between mass 2 to mass 1 and 3 is $d$ and the distance between point P and mass 3 is $\frac{2d}{3}$.

The formula to calculate moment of inertia is,

    $\begin{array}{c}I={\displaystyle \sum m_{\text{i}}{r}_{\text{i}}^2}\\ =m_{\text{1}}{r}_{\text{1}}^2+m_{\text{2}}{r}_{\text{2}}^2+m_{\text{3}}{r}_{\text{3}}^2\end{array}$

The distance of the particle 1 from point P is,

    $r_1=\frac{4d}{3}$

The distance of the particle 2 from point P is,

    $r_2=\frac{d}{3}$

Substitute $m$ for $m_{\text{1}}$, $m_{\text{2}}$ and $m_{\text{3}}$, $\frac{4d}{3}$ for $r_1$, $\frac{d}{3}$ for $r_2$ and $\frac{2d}{3}$ for $r_3$ in above equation to find $I$.

    $\begin{array}{c}I=m{\left(\frac{4d}{3}\right)}^2+m{\left(\frac{d}{3}\right)}^2+m{\left(\frac{2d}{3}\right)}^2\\ =m\left(\frac{16d^2+d^2+4d^2}{9}\right)\\ =m\left(\frac{21d^2}{9}\right)\\ =7m\left(\frac{d^2}{3}\right)\end{array}$

Conclusion:

Therefore, the moment of inertia of the system of three particles about the pivot is $7m\left(\frac{d^2}{3}\right)$.

To determine

The torque acting on the system at $t=0$.

Answer to Problem 11.49AP

The torque acting on the system at $t=0$ is $\left(mgd\right)\widehat{\text{k}}$.

Explanation of Solution

Consider that the whole weight, $3mg$ acting at the centre of gravity.

The formula to calculate torque is,

    $\overrightarrow{\text{τ}}=\overrightarrow{\text{r}}\times \overrightarrow{\text{F}}$

Substitute $\frac{d}{3}\left(-\widehat{\text{i}}\right)$ for $\overrightarrow{\text{r}}$ and $3mg\left(\widehat{\text{j}}\right)$ for $\overrightarrow{\text{F}}$ in above equation to find $\overrightarrow{\text{τ}}$.

    $\begin{array}{c}\overrightarrow{\text{τ}}=\frac{d}{3}\left(-\widehat{\text{i}}\right)\times 3mg\left(\widehat{\text{j}}\right)\\ =\left(mgd\right)\widehat{\text{k}}\end{array}$

Conclusion:

Therefore, the torque acting on the system at $t=0$ is $\left(mgd\right)\widehat{\text{k}}$.

To determine

The angular acceleration of the system at $t=0$.

Answer to Problem 11.49AP

The angular acceleration of the system at $t=0$ is $\frac{3g}{7d}$ counterclockwise.

Explanation of Solution

The formula to calculate angular acceleration is,

    $\alpha =\frac{\text{τ}}{I}$

Substitute $7m\left(\frac{d^2}{3}\right)$ for $I$ and $mgd$ for $\tau$ in above equation to find $\alpha$.

    $\begin{array}{c}\alpha =\frac{mgd}{7m\left(\frac{d^2}{3}\right)}\\ =\frac{3g}{7d}\end{array}$

Conclusion:

Therefore, the angular acceleration of the system at $t=0$ is $\frac{3g}{7d}$ counterclockwise.

To determine

The linear acceleration of the particle 3 at $t=0$.

Answer to Problem 11.49AP

The linear acceleration of the particle 3 at $t=0$ is $\frac{2g}{7}$ upward.

Explanation of Solution

The formula to calculate linear acceleration is,

    $a=\alpha r_3$

Substitute $\frac{3g}{7d}$ for $\alpha$ and $\frac{2d}{3}$ for $r_3$ in above equation to find $a$.

    $\begin{array}{c}a=\left(\frac{3g}{7d}\right)\left(\frac{2d}{3}\right)\\ =\frac{2g}{7}\end{array}$

Conclusion:

Therefore, the linear acceleration of the particle 3 at $t=0$ is $\frac{2g}{7}$ upward.

To determine

The maximum kinetic energy of the system.

Answer to Problem 11.49AP

The maximum kinetic energy of the system is $mgd$.

Explanation of Solution

Because the axle is fixed, no external work is performed on the system of the earth and three particles, so the total mechanical energy is conserved.

The rotation kinetic energy is maximum when rod has swung to a vertical orientation with the centre of gravity directly under the axle.

The expression for the energy is,

    $\begin{array}{c}{\left(K+U\right)}_{\text{i, horizontal}}={\left(K+U\right)}_{\text{f, vertical}}\\ 0+3mg\left(\frac{d}{3}\right)=K_{\text{f}}+0\\ K_{\text{f}}=mgd\end{array}$

Conclusion:

Therefore, the maximum kinetic energy of the system is $mgd$.

To determine

The maximum angular speed reached by the rod.

Answer to Problem 11.49AP

The maximum angular speed reached by the rod is $\sqrt{\frac{6g}{7d}}$.

Explanation of Solution

In the vertical orientation, the rod has the greatest rotational kinetic energy.

The expression for the kinetic energy is,

    $K_{\text{f}}=\frac{1}{2}I{\omega }_{\text{f}}^2$

Substitute $mgd$ for $K_{\text{f}}$ and $7m\left(\frac{d^2}{3}\right)$ for $I$ in above equation to find ${\omega }_{\text{f}}$.

    $\begin{array}{c}mgd=\frac{1}{2}7m\left(\frac{d^2}{3}\right){\omega }_{\text{f}}^2\\ {\omega }_{\text{f}}^2=\frac{6g}{7d}\\ {\omega }_{\text{f}}=\sqrt{\frac{6g}{7d}}\end{array}$

Conclusion:

Therefore, the maximum angular speed reached by the rod is $\sqrt{\frac{6g}{7d}}$.

To determine

The maximum angular momentum of the system.

Answer to Problem 11.49AP

The maximum angular momentum of the system is $\left(m{d}^{\frac{3}{2}}\right)\sqrt{\frac{14g}{3}}$.

Explanation of Solution

The expression for the angular momentum is,

    $L_{\text{f}}=I{\omega }_{\text{f}}$

Substitute $\sqrt{\frac{6g}{7d}}$ for ${\omega }_{\text{f}}$ and $7m\left(\frac{d^2}{3}\right)$ for $I$ in above equation to find $L_{\text{f}}$.

    $\begin{array}{c}{L}_{\text{f}}=7m\left(\frac{d^2}{3}\right)\sqrt{\frac{6g}{7d}}\\ =\left(m\right)\sqrt{\frac{\left(49\times 6\right)g{d}^4}{7d\times 9}}\\ =\left(m\right)\sqrt{\frac{14g{d}^3}{3}}\\ =\left(m{d}^{\frac{3}{2}}\right)\sqrt{\frac{14g}{3}}\end{array}$

Conclusion:

Therefore, the maximum angular momentum of the system is $\left(m{d}^{\frac{3}{2}}\right)\sqrt{\frac{14g}{3}}$.

To determine

The maximum speed of particle 2.

Answer to Problem 11.49AP

The maximum speed of particle 2 is $\sqrt{\frac{2gd}{21}}$.

Explanation of Solution

The expression for the speed is,

    $v_{\text{f}}=r_2{\omega }_{\text{f}}$

Substitute $\sqrt{\frac{6g}{7d}}$ for ${\omega }_{\text{f}}$ and $\left(\frac{d}{3}\right)$ for $r_2$ in above equation to find $v_{\text{f}}$.

    $\begin{array}{c}{v}_{\text{f}}=\left(\frac{d}{3}\right)\sqrt{\frac{6g}{7d}}\\ =\sqrt{\frac{6g{d}^2}{7d\times 9}}\\ =\sqrt{\frac{2gd}{21}}\end{array}$

Conclusion:

Therefore, the maximum speed of particle 2 is $\sqrt{\frac{2gd}{21}}$.