Chapter 10, Problem 10.47P

A war-wolf or trebuchet is a device used during the Middle Ages to throw rocks at castles and now sometimes used to fling large vegetables and pianos as a sport. A simple trebuchet is shown in Figure P10.26. Model it as a stiff rod of negligible mass, 3.00 m long, 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 large-mass particle. The operator releases the trebuchet from rest in a horizontal orientation. (a) Find the maximum speed that the small-mass object attains. (b) While the small-mass object is gaining speed, does it move with constant acceleration? (c) Does it move with constant tangential acceleration? (d) Does the trebuchet move with constant angular acceleration? (e) Does it have constant momentum? (f) Does the trebuchet–Earth system have constant mechanical energy?

Figure P10.26

To determine

The maximum speed that the small mass object attains.

Answer to Problem 10.47P

The maximum speed that the small mass object attains is $\text{24}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

The mass of small object is $0.120\text{kg}$, the mass of the heavier object is $60.0\text{kg}$ and the length of the stiff rod is $3.00\text{m}$. The distance of the large mass from the axle is $14.0\text{cm}$.

Formula to calculate the distance of the small mass object from the axle is,

    $r_1=l-r_2$

Here, $r_1$ is the distance of the distance of the small mass object from the axle, $r_2$ is the distance of the distance of the heavier mass object from the axle and $l$ is the length of the stiff rod.

Substitute $3.00\text{m}$ for $l$ and $14.0\text{cm}$ for $r_2$ to find $r_1$.

    $\begin{array}{c}{r}_1=3.00\text{m}-14.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\\ =2.86\text{m}\end{array}$

The small mass attains maximum speed when the stiff rod lies in vertical position by lower the heavier mass down ward. Initially the energy of the system is zero and the system start gaining speed and attains maximum speed at higher gravitational potential energy.

Assume the zero potential line is passes through the axle of the system horizontally.

Formula to calculate the final gravitational potential energy of the small mass is,

    $U_{\text{f1}}=m_{1}g{r}_1$

Here, $U_{\text{f1}}$ is the final gravitational potential energy of the small object, $m_1$ is the mass of the small object, $g$ is the acceleration due to gravity and $r_1$ is the distance of the small object from the axle.

Formula to calculate the final gravitational potential energy of the heavier mass is,

    $U_{\text{f2}}=-m_{2}g{r}_2$

Here, $U_{\text{f2}}$ is the final gravitational potential energy of the small object, $m_2$ is the mass of the small object an d $r_2$ is the distance of the small object from the axle.

Formula to calculate the moment of inertia of the system is,

    $I=m_1{r}_1^2+m_2{r}_2^2$

Formula to calculate the rotational kinetic energy of the system is,

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

Here, $K_{\text{r}}$ is the rotational kinetic energy of the system, $I$ is the moment of inertia of the system and $\omega$ is the angular speed of the system.

Substitute $m_1{r}_1^2+m_2{r}_2^2$ for $I$ in above equation.

    $K_{\text{r}}=\frac{1}{2}\left(m_1{r}_1^2+m_2{r}_2^2\right){\omega }^2$

The law of energy conservation for the system is given as,

    $U_{\text{i}}=U_{\text{f1}}+U_{\text{f2}}+K_{\text{r}}$

Substitute $0$ for $U_{\text{i}}$, $m_{1}g{r}_1$ for $U_{\text{f1}}$, $\left(-m_{2}g{r}_2\right)$ for $U_{\text{f2}}$ and $\frac{1}{2}\left(m_1{r}_1^2+m_2{r}_2^2\right){\omega }^2$ for $K_{\text{r}}$ in above equation.

    $m_{1}g{r}_1+\left(-m_{2}g{r}_2\right)+\frac{1}{2}\left(m_1{r}_1^2+m_2{r}_2^2\right){\omega }^2=0$

Rearrange the above equation for $\omega$.

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

Substitute $0.120\text{kg}$ for $m_1$, $60.0\text{kg}$ for $m_2$, $2.86\text{m}$ for $r_1$, $14.0\text{cm}$ for $r_2$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in above equation.

    $\begin{array}{c}\omega =\sqrt{\frac{2\left(\left(60.0\text{kg}\right)\left(14\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)-\left(0.120\text{kg}\right)\left(2.86\text{m}\right)\right)\left(9.8\text{m}/\text{s}\right)}{\left(\left(0.120\text{kg}\right){\left(2.86\text{m}\right)}^2+\left(60.0\text{kg}\right){\left(14\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}}\\ =\sqrt{\frac{157.91\text{kg}\cdot {\text{m}}^2/{\text{s}}^{\text{2}}}{2.157\text{kg}\cdot {\text{m}}^{\text{2}}}}\\ =8.55\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Formula to calculate the maximum speed that the small mass object attains is,

    $v_1=\omega r_1$

Here, $v_1$ is the maximum speed that the small mass object attains.

Substitute $8.55\text{rad}/{\text{s}}^{\text{2}}$ for $\omega$ and $2.86\text{m}$ for $r_1$ to find $v_1$.

    $\begin{array}{c}{v}_1=8.55\text{rad}/{\text{s}}^{\text{2}}\times 2.86\text{m}\\ =\text{24}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, maximum speed that the small mass object attains is $\text{24}\text{m}/{\text{s}}^{\text{2}}$.

To determine

Whether the small object move with constant acceleration while gaining speed.

Answer to Problem 10.47P

The acceleration of the small object changes while gaining speed.

Explanation of Solution

The rotational body has two components of acceleration one is radial acceleration that act towards the radius of the rotation circle while the tangential acceleration is tangent to the rotational circle at any point.

If the smaller mass moves under constant acceleration, it either moves in a straight line or in a parabolic path. However, its circular path signifies that the acceleration is not constant.

Conclusion:

Therefore, the acceleration of the small object changes while gaining speed.

To determine

Whether the small object move with constant tangential acceleration.

Answer to Problem 10.47P

The tangential acceleration of the object is not constant.

Explanation of Solution

Write the expression for tangential acceleration

  $a_{\text{t}}=r\alpha$

Here, $a_{\text{t}}$ is the tangential acceleration, $r$ is the radius of the circular path and $\alpha$ is the angular acceleration.

Since angular acceleration is not constant, tangential acceleration is also not constant. Refer part (d) for the reason for variable angular acceleration.

Conclusion:

Therefore, tangential acceleration of the object is not constant.

To determine

Whether the trebuchet move with constant angular acceleration.

Answer to Problem 10.47P

The trebuchet does not move with constant angular acceleration.

Explanation of Solution

Formula to calculate the angular acceleration of the trebuchet and the torque is,

    ${\tau }_{\text{net}}=I\alpha$

Here, $I$ is the moment of inertia and ${\tau }_{\text{net}}$  is the net torque.

The net torque is not constant as the lever arm of the gravitational force changes. Since torque changes, the angular acceleration must also change as given by the above equation.

Conclusion:

Therefore, the trebuchet doesn’t move with constant angular acceleration.

To determine

Whether the trebuchet have constant momentum.

Answer to Problem 10.47P

The trebuchet doesn’t have constant momentum.

Explanation of Solution

The rotational angular momentum is the product of the linear momentum to the distance from the rotational axis.

Formula to calculate the angular momentum of the system is,

    $L=M\omega$

Here, $M$ is the total mass of the system and $L$ is the angular momentum of the system.

Since the angular velocity changes, the angular momentum must also change as they are directly proportional.

Conclusion:

Therefore, the trebuchet doesn’t have constant momentum.

To determine

Whether the trebuchet-Earth system have constant mechanical energy.

Answer to Problem 10.47P

The total mechanical energy of the trebuchet earth system is constant

Explanation of Solution

The total mechanical energy of the system is the sum of the potential energy and the kinetic energy of the body. Potential energy cause due to its position while the kinetic energy cause due to virtue of its motion.

The trebuchet-Earth system is rotated in clockwise direction that produces rotational kinetic energy as well as the potential energies of the two mass. Initially, both the masses are at rest and the system has no mechanical energy. But when the heavier mass lower down, the mechanical energy applied on the heavier mass is converted into the rotational energy of the system hence the total energy of the system is constant.

Conclusion:

Therefore, the total mechanical energy of the trebuchet earth system is constant.