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 the figure. Model it as a stiff rod of negligible mass 3.00 m long and joining particles of mass 74.0 kg and 0.280 kg at its ends. It can turn on a frictionless horizontal axle perpendicular to the rod and 25.0 cm from the particle of larger mass. The rod is released from rest in a horizontal orientation. Find the maximum (a) angular speed.
To solve this problem, we can apply the conservation of mechanical energy. The total mechanical energy (E) of the system is given by the sum of the kinetic energy (K) and potential energy (U) of both particles:
E = K + U
At the initial position (horizontal orientation), the particles have no kinetic energy (K = 0) and their potential energy is also 0 (since the height is 0). So, E_initial = 0.
When the trebuchet is at its maximum angular speed, the rod will be vertical, with the 74 kg mass being at its highest and the 0.280 kg mass at its lowest point. Thus, their potential energy is:
U = m1 * g * h1 - m2 * g * h2
Where m1 and m2 are the masses of the larger and smaller particle, g is the acceleration due to gravity (approximately 9.81 m/s²), and h1 and h2 are the heights of the larger and smaller particle when the trebuchet is in the final position.
The height for the larger particle is equal to the distance from the axle (25.0 cm or 0.25 m) plus the distance between the axle and the smaller particle (3 m - 0.25 m = 2.75 m). So, h1 = 0.25 m + 2.75 m = 3 m.
The height for the smaller particle is equal to the distance between the axle and the larger particle (0.25 m). So, h2 = 0.25 m.
The potential energy of both particles is:
U = (74 kg)(9.81 m/s²)(3 m) - (0.280 kg)(9.81 m/s²)(0.25 m)
U = 2176.46 J - 0.6837 J
U = 2175.7763 J
This potential energy is fully converted into kinetic energy (K = U) in the final position. The kinetic energy of a rotating system is given by:
K = 0.5 * I * ω^2
Where I is the moment of inertia of the system and ω is the angular speed.
The moment of inertia for the system can be found using the formula:
I = m1 * r1^2 + m2 * r2^2
Where r1 and r2 are the distances from the axle to the larger and smaller particle. r1 = 0.25 m and r2 = 2.75 m.
I = (74 kg)(0.25 m)^2 + (0.280 kg)(2.75 m)^2
I = 4.625 kg * m²
Now, we can find the maximum angular speed using the kinetic energy expression:
0.5 * I * ω^2 = 2175.7763 J
ω^2 = (2 * 2175.7763 J) / (4.625 kg * m²)
ω^2 = 1886.39 1/s²
ω = √(1886.39 1/s²)
ω = 43.41 rad/s
So, the maximum angular speed of the trebuchet is 43.41 rad/s.
To find the maximum angular speed of the trebuchet, we need to analyze the conservation of angular momentum.
Angular momentum is the product of the moment of inertia and the angular velocity. In this case, the moment of inertia of the system is the sum of the moments of inertia of the two particles rotating about the axis.
The moment of inertia for a point particle rotating about an axis is given by I = mr^2, where m is the mass and r is the distance from the axis.
For the particle with mass 74.0 kg:
I1 = m1 * r1^2 = 74.0 kg * (0.25 m)^2
For the particle with mass 0.280 kg:
I2 = m2 * r2^2 = 0.280 kg * (2.75 m)^2
Now, let's calculate the total moment of inertia of the system:
I_total = I1 + I2
Next, we need to analyze the conservation of angular momentum. The initial angular momentum is zero since the trebuchet is released from rest:
L_initial = 0
The final angular momentum can be calculated as the product of the total moment of inertia and the maximum angular speed of the system:
L_final = I_total * w_max
Since angular momentum is conserved, we can set the initial and final angular momentum equal to each other and solve for the maximum angular speed:
0 = I_total * w_max
Solving for w_max:
w_max = 0
Therefore, the maximum angular speed of the trebuchet is 0 rad/s.