Chapter 8, Problem 59P

A rigid, massless rod has three particles with equal masses attached to it as shown in Figure P8.59. 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 (rod plus 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, (0 the maximum angular speed reached by the rod, (g) the maximum angular momentum of the system, and (h) the maximum translational speed reached by the particle labeled 2.

  

To determine

The moment of inertia of the system about the pivot

Answer to Problem 59P

Solution: The moment of inertia of the three particle system about the pivot is $\frac{7}{3}m{d}^2$.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of the discrete particle system is given below.

The moment of inertia of a system of discrete particles is given by,

$I={\displaystyle \sum m{r}^2}$

• $I$ is the moment of inertial of a system
• $m$ is the mass of the particles
• $r$ is the distance of the particles from the axis of rotation

From the above diagram the distance of the first particle from the pivot is $\frac{4d}{3}$. Distance of the second particle from the pivot is $\frac{d}{3}$. The distance of the third particle from the pivot is $\frac{2d}{3}$.

The moment of inertia about the pivot of the three particle system will be,

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

Conclusion:

The moment of inertia of the three particle system about the pivot is $\frac{7}{3}m{d}^2$.

To determine

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

Answer to Problem 59P

Solution: The torque of the system about the pivot is $mgd\text{ counterclockwise direction}$.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of discrete particle system is given below.

The torque of a system of discrete particles is defined as,

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

• $\overrightarrow{\tau }$ is the torque.
• $\overrightarrow{r}$ is the position vector of the particle with respect to the axis of rotation
• $\overrightarrow{F}$ is the force

Let the counter clockwise torque is considered as positive and clockwise torque is considered as negative.

From the above diagram the distance of the first particle from the pivot is $\frac{4d}{3}$. Distance of the second particle from the pivot is $\frac{d}{3}$. The distance of the third particle from the pivot is $\frac{2d}{3}$.

The force on the particles is the weight of the particle.

$F=mg$

• $g$ is the free fall acceleration

The torque of the system about the pivot of the three particle system will be,

$\begin{array}{c}\tau =mg\left(\frac{4d}{3}\right)+mg\left(\frac{d}{3}\right)-mg\left(\frac{2d}{3}\right)\\ =\frac{3}{3}mgd\\ =mgd\end{array}$

Conclusion:

The torque of the system about the pivot is $mgd\text{ counterclockwise direction}$.

To determine

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

Answer to Problem 59P

Solution: The angular acceleration of the system is $\frac{3g}{7d}$.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of discrete particle system is given below.

The torque of a system of discrete particles is defined as,

$\tau =I\alpha$

• $\alpha$ is the angular acceleration

Re-write the above equation in order to get an expression for the angular acceleration.

$\alpha =\frac{\tau }{I}$

From section (a) the moment of inertia of the system is $\frac{7}{3}m{d}^2$.

From section (b) the torque acting on the system is $mgd$.

Substitute $\frac{7}{3}m{d}^2$ for $I$ and $mgd$ for $\tau$ to determine $\alpha$.

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

Conclusion:

The angular acceleration of the system is $\frac{3g}{7d}$.

To determine

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

Answer to Problem 59P

Solution: The linear acceleration of the third particle of the system is $\frac{2g}{7}$ and directed upward.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of discrete particle system is given below.

The relation between the angular acceleration and linear acceleration of a particle is given by,

$a=r\alpha$

From section (c) the angular acceleration is given by $\frac{3g}{7d}$.

The distance of the third particle from the pivot is $\frac{2d}{3}$.

Substitute $\frac{3g}{7d}$ for $\alpha$ and $\frac{2d}{3}$ for $r$ to determine $a$.

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

The direction of the linear acceleration of the third particle is directed upward.

Conclusion:

The linear acceleration of the third particle of the system is $\frac{2g}{7}$ and directed upward.

To determine

The maximum kinetic energy of the system.

Answer to Problem 59P

Solution: The linear acceleration of the third particle of the system is $\frac{2g}{7}$ and directed upward.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of discrete particle system is given below.

From the law of conservation of energy the total energy of an isolated system with only conservative forces is conserved.

$E=KE+PE$

• $E$ is the total energy of the isolated system
• $KE$ is the kinetic energy of the system
• $PE$ is the potential energy of the system

The kinetic energy of the system will be maximum when the gravitational potential energy of the system is minimum which will occur when the centre of gravity is at minimum height or when the rod is vertical position.

The initial kinetic energy of the system is zero as it is in rest initially. The kinetic energy of a particle will be the difference between the initial potential energy and the final potential energy.

$KE=mg\left(y_i-y_f\right)$

• $KE$ is the kinetic energy of a particle
• $g$ is the acceleration due to gravity
• $y_i$ is the initial height of the centre of gravity
• $y_f$ is the final height of the centre of gravity

The kinetic energy of the entire system will be,

$KE=3mg\left(y_i-y_f\right)$

When the rod is vertical, consider the level of the pivot is taken to be zero, then the height of the centre of gravity at vertical position will be $-\frac{d}{3}$. The initial height of the centre of gravity is $0$.

Substitute $-\frac{d}{3}$ for $y_f$ and $0$ for $y_i$ to determine the kinetic energy of the system.

$\begin{array}{c}KE=3mg\left(0-\left(-\frac{d}{3}\right)\right)\\ =mgd\end{array}$

Conclusion:

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

To determine

The maximum angular speed reached by the rod.

Answer to Problem 59P

Solution: The maximum angular speed of the system is $\sqrt{\frac{6g}{7d}}$.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of discrete particle system is given below.

From the law of conservation of energy the total energy of an isolated system with only conservative forces is conserved.

$E=KE+PE$

Angular speed will be maximum when the entire energy of the system is in the form of rotational kinetic energy.

The total energy of the given system is $mgd$. Thus,

$\frac{1}{2}I{\omega }_{\mathrm{m}}^2=mgd$

• ${\omega }_m$ is the maximum angular speed

Re-write the above equation to get an expression for maximum angular

${\omega }_m=\sqrt{\frac{2mgd}{I}}$

From section (a) the moment of inertia of the system is $\frac{7m{d}^2}{3}$.

Hence the maximum angular speed becomes,

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

Conclusion:

The maximum angular speed of the system is $\sqrt{\frac{6g}{7d}}$.

To determine

The maximum angular momentum of the system.

Answer to Problem 59P

Solution: The maximum angular momentum of the system is $m\sqrt{\frac{14g{d}^3}{3}}$.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of discrete particle system is given below.

The maximum angular momentum of the system is defines as,

$L_m=I{\omega }_m$

• $L_m$ is the angular momentum of the system

From section (a) the moment of inertia of the system is $\frac{7m{d}^2}{3}$.

From section (f) the maximum angular speed is $\sqrt{\frac{6g}{7d}}$.

Substitute $\frac{7m{d}^2}{3}$ for $I$, $\frac{6g}{7d}$ for ${\omega }_m$ to determine the maximum angular momentum.

$\begin{array}{c}{L}_m=\left(\frac{7m{d}^2}{3}\right)\left(\sqrt{\frac{6g}{7d}}\right)\\ =m\sqrt{\frac{14g{d}^3}{3}}\end{array}$

Conclusion:

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

To determine

The maximum translational speed reached by the particle two.

Answer to Problem 59P

Solution: The maximum linear speed reached by the second particle is $\sqrt{\frac{2gd}{21}}$.

Explanation of Solution

Given info: The mass of the particles are $m$. Distance between the consecutive particles is $d$.

The representative diagram of discrete particle system is given below.

The relation between the maximum linear speed and the maximum angular speed is given by,

$v_m=r{\omega }_m$

• $v_m$ is the maximum linear speed

The maximum angular speed of the system is $\sqrt{\frac{6g}{7d}}$.

The distance of the second particle from the pivot is $\frac{d}{3}$.

Substitute $\sqrt{\frac{6g}{7d}}$ for ${\omega }_m$ and $\frac{d}{3}$ for $r$ to determine $v_m$.

$\begin{array}{c}{v}_m=\left(\frac{d}{3}\right)\left(\sqrt{\frac{6g}{7d}}\right)\\ =\sqrt{\frac{2gd}{21}}\end{array}$

Conclusion:

The maximum linear speed reached by the second particle is $\sqrt{\frac{2gd}{21}}$.