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 the figure below. Model it as a stiff rod of negligible mass, d 2.75 m long o ning particles of mass m1 0.115 kg and m2 58.5 kg at its ends. It can turn on a frictionless, horizontal axle perpendicular to the rod and 12.0 cm from the large-mass particle. The operator releases the trebuchet from rest in a horizontal orientation. nI (a) Find the maximum speed that the small-mass object attains when it leaves the trebuchet horizontally. 33.8 Your response differs from the correct answer by more than 10%. Double check your calculations.
The Correct Answer and Explanation is :
To solve for the maximum speed of the small mass ( m_1 ) when the trebuchet is released from the horizontal position, we use energy conservation principles.
Given Data:
- ( d = 2.75 ) m (total length of the trebuchet arm)
- ( m_1 = 0.115 ) kg (small mass)
- ( m_2 = 58.5 ) kg (large mass)
- Pivot is located ( d_p = 12.0 ) cm = 0.12 m from ( m_2 )
Step 1: Determine Distances from the Pivot
- Distance of ( m_1 ) from pivot:
[
r_1 = d – d_p = 2.75 – 0.12 = 2.63 \text{ m}
] - Distance of ( m_2 ) from pivot:
[
r_2 = d_p = 0.12 \text{ m}
]
Step 2: Compute Initial Potential Energy
Potential energy is taken relative to the lowest point of rotation. In the initial horizontal position:
[
PE_{\text{initial}} = m_1 g r_1 + m_2 g r_2
]
Step 3: Compute Final Potential Energy
At the lowest point:
- ( m_1 ) moves to ( r_1 ) below the pivot.
- ( m_2 ) moves to ( r_2 ) above the pivot.
[
PE_{\text{final}} = – m_1 g r_1 + m_2 g r_2
]
Step 4: Apply Energy Conservation
Total mechanical energy is conserved:
[
KE_{\text{final}} + PE_{\text{final}} = PE_{\text{initial}}
]
The kinetic energy of the system is rotational:
[
KE = \frac{1}{2} I \omega^2
]
where ( I ) is the moment of inertia:
[
I = m_1 r_1^2 + m_2 r_2^2
]
[
I = (0.115)(2.63)^2 + (58.5)(0.12)^2
]
[
I = 0.791 + 0.843 = 1.634 \text{ kg} \cdot \text{m}^2
]
Solving for ( \omega ):
[
m_1 g r_1 + m_2 g r_2 = \frac{1}{2} I \omega^2 + (- m_1 g r_1 + m_2 g r_2)
]
[
2 m_1 g r_1 = \frac{1}{2} I \omega^2
]
[
\omega = \sqrt{\frac{4 m_1 g r_1}{I}}
]
[
\omega = \sqrt{\frac{4 (0.115)(9.81)(2.63)}{1.634}}
]
[
\omega = \sqrt{7.723} = 2.78 \text{ rad/s}
]
Step 5: Find the Maximum Speed
Velocity of ( m_1 ):
[
v_1 = r_1 \omega
]
[
v_1 = (2.63)(2.78)
]
[
v_1 = 7.31 \text{ m/s}
]
Thus, the correct answer is:
[
\mathbf{7.31 \text{ m/s}}
]
Now, let’s generate the image of the trebuchet.
Here is the generated image of the trebuchet, illustrating the dimensions, pivot point, counterweight, and projectile. Let me know if you need any modifications or further explanations!