Chapter 8, Problem 89AP

A war-wolf, or trebuchet, is a device used during the Middle Ages to throw rocks at castles and now sometimes used to fling pumpkins and pianos. A simple trebuchet is shown in Figure P8.89. Model it as a stiff rod of negligible mass 5.00 m long and joining particles of mass m₁ = 0.120 kg and m₂ = 60.0 kg at its ends. It can turn on a frictionless horizontal axle perpendicular to the rod and 14.0 cm from the particle of larger mass. The rod is released from rest in a horizontal orientation. Find the maximum speed dial the object of smaller mass attains.

FigureP8.89

To determine

The magnitude of maximum speed of the smaller mass.

Answer to Problem 89AP

Answer The maximum speed of the smaller particle is $24.5\text{m}\cdot {\text{s}}^{-1}$

Explanation of Solution

Given Info:

Length of the stiff rod is $3.00\text{m}$ , mass of the small particle $0.120\text{kg}$ , mass of the big particle $60.0\text{kg}$, the distance of the smaller particle from the axle is $2.86\text{m}$ , the distance of the larger particle from the axle is $0.14\text{m}$.

Formula to calculate initial total mechanical energy is,

$E_{\text{T,i}}=P{E}_{\text{i}}+K{E}_{\text{i}}$

• $E_{\text{T,i}}$ is the initial total mechanical energy of the system
• $P{E}_{\text{i}}$ is the initial potential energy of the system
• $K{E}_{\text{i}}$ is the initial kinetic energy of the system

Consider the reference plane as the horizontal rotational axis. The potential energy of the system is zero at the reference plane. Initially both kinetic and potential energy of the system will be zero

$E_{\text{T,i}}=0$

Formula to calculate final total mechanical energy is,

$E_{\text{T,f}}=P{E}_{\text{f}}+K{E}_{\text{f}}$

• $E_{\text{T,f}}$ is the final total mechanical energy of the system
• $P{E}_{\text{f}}$ is the final potential energy of the system
• $K{E}_{\text{f}}$ is the final kinetic energy of the system

Formula to calculate the final kinetic energy of the system is given by

$K{E}_{\text{f}}=\frac{1}{2}{I}_1{\omega }^2{}_{\max }+\frac{1}{2}{I}_2{\omega }^2{}_{\max }$

• $I_1$ is the moment of inertia of the small mass
• $I_2$ is the moment of inertia of the big mass
• ${\omega }_{\max }$ is the maximum  angular speed

The formula to calculate $I_1$ is,

$I_1=m_1{r}^2{}_1$

• $m_1$ is the mass of small particle
• $r_1$ is the distance of the small mass from the axle

The formula to calculate $I_2$ is,

$I_2=m_2{r}^2{}_2$

• $m_2$ is the mass of big particle
• $r_2$ is the distance of the big particle from the axle

The formula to calculate the total potential energy of the system is,

$P{E}_{\text{f}}=m_{1}g{r}_1-m_{2}g{r}_2$

• g is the acceleration due to gravity

Use $\frac{1}{2}{I}_1{\omega }^2{}_{\max }+\frac{1}{2}{I}_2{\omega }^2{}_{\max }$ for $K{E}_{\text{f}}$ and $m_{1}g{r}_1-m_{2}g{r}_2$ for $P{E}_{\text{f}}$ in $E_{\text{T,f}}=P{E}_{\text{f}}+K{E}_{\text{f}}$ to rewrite $E_{\text{T,f}}$.

$E_{\text{T,f}}=\frac{1}{2}\left(m_1{r}^2{}_1\right){\omega }^2{}_{\max }+\frac{1}{2}\left(m_2{r}^2{}_2\right){\omega }^2{}_{\max }+m_{1}g{r}_1-m_{2}g{r}_2$

According to conservation of mechanical energy of the system sum of both kinetic and potential energy of the system before and after releasing of the small particle will be same. The smaller particle will attain maximum velocity when the rod becomes vertical with respect to the ground.

$E_{\text{T,f}}=0$

Substitute 0 for $E_{\text{T,f}}$ in the above equation and rewrite in terms of ${\omega }_{\max }$.

$\begin{array}{c}0=\frac{1}{2}\left(m_1{r}^2{}_1\right){\omega }^2{}_{\max }+\frac{1}{2}\left(m_2{r}^2{}_2\right){\omega }^2{}_{\max }+m_{1}g{r}_1-m_{2}g{r}_2\\ {\omega }^2{}_{\max }=2\left(\frac{m_{2}g{r}_2-m_{1}g{r}_1}{\left(m_1{r}^2{}_1\right)+\left(m_2{r}^2{}_2\right)}\right)\\ {\omega }_{\max }=\sqrt{2\left(\frac{m_{2}g{r}_2-m_{1}g{r}_1}{\left(m_1{r}^2{}_1\right)+\left(m_2{r}^2{}_2\right)}\right)}\end{array}$

Substitute $60.0\text{kg}$ for $m_2$ , $0.120\text{kg}$ for $m_1$ , $0.14\text{m}$ for $r_2$ , $2.86\text{m}$ for $r_1$ ,and $9.8\text{m}\cdot {\text{s}}^{-2}$ for g to calculate ${\omega }_{\max }$

$\begin{array}{c}{\omega }_{\max }=\sqrt{2\left(\frac{\left(60.0\text{kg}\right)\left(9.8\text{m}\cdot {\text{s}}^{-2}\right)\left(0.14\text{m}\right)-\left(0.120\text{kg}\right)\left(9.8\text{m}\cdot {\text{s}}^{-2}\right)\left(2.86\text{m}\right)}{\left(\left(0.120\text{kg}\right){\left(2.86\text{m}\right)}^2\right)+\left(\left(60.0\text{kg}\right){\left(0.14\text{m}\right)}^2\right)}\right)}\\ =8.56\text{rad}\cdot {\text{s}}^{-1}\end{array}$

The formula to calculate maximum speed of the smaller particle is given by

$v=r_1{\omega }_{\max }$

Substitute $8.56\text{rad}\cdot {\text{s}}^{-1}$ and $2.86\text{m}$ for $r_1$ in the above equation to calculate v

$\begin{array}{c}v=\left(2.86\text{m}\right)\left(8.56\text{rad}\cdot {\text{s}}^{-1}\right)\\ =24.5\text{m}\cdot {\text{s}}^{-1}\end{array}$

Thus the maximum speed of the smallest particle is $24.5\text{m}\cdot {\text{s}}^{-1}$

Conclusion:

Therefore the maximum speed of the smallest particle is $24.5\text{m}\cdot {\text{s}}^{-1}$