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 0.110 kg at its ends. It can turn on a frictionless horizontal axle perpendicular to the rod and 13.0 cm from the particle of larger mass. The rod is released from rest in a horizontal orientation. Find the maximum (a) angular speed. (b) linear speed that the object of smaller mass attains.
http://www.physics.umd.edu/~richman/phys141/hw8.pdf
Step 1:
Start by finding the moment of inertia of the system. The moment of inertia is given by the formula:
I = Σmr^2
where m is the mass of each particle and r is the distance of each particle from the axis of rotation.
For the particle of mass 74.0 kg:
I1 = m1 * r1^2 = 74.0 kg * (0.13 m)^2 = 1.1376 kg⋅m^2
For the particle of mass 0.110 kg:
I2 = m2 * r2^2 = 0.110 kg * (2.87 m)^2 = 0.0911 kg⋅m^2
Step 2:
Next, calculate the total moment of inertia of the system by summing the individual moments of inertia:
I_total = I1 + I2 = 1.1376 kg⋅m^2 + 0.0911 kg⋅m^2 = 1.2287 kg⋅m^2
Step 3:
The law of conservation of angular momentum states that the initial angular momentum of the system is equal to the final angular momentum. Since the rod is released from rest, its initial angular momentum is zero.
L_initial = 0
The final angular momentum is given by the formula:
L_final = I_total * ω
where ω is the angular speed.
Setting L_initial = L_final, we have:
0 = I_total * ω
0 = 1.2287 kg⋅m^2 * ω
Therefore, ω = 0 rad/s.
Step 4:
To find the maximum angular speed, we need to consider the conservation of energy. The potential energy at the initial position is equal to the kinetic energy at the maximum angular speed.
The potential energy at the initial position is given by:
PE_initial = m1 * g * h
where m1 is the mass of the heavier particle, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the center of mass of the system.
h = r1 / 2 = 0.13 m / 2 = 0.065 m
PE_initial = 74.0 kg * 9.8 m/s^2 * 0.065 m = 47.8784 J
The kinetic energy at the maximum angular speed is given by:
KE_final = (1/2) * I_total * ω_max^2
where ω_max is the maximum angular speed.
Since the rod is released from rest, the initial kinetic energy is zero:
KE_initial = 0
Therefore, the potential energy at the initial position is equal to the kinetic energy at the maximum angular speed:
PE_initial = KE_final
47.8784 J = (1/2) * 1.2287 kg⋅m^2 * ω_max^2
Solving for ω_max, we have:
ω_max^2 = (2 * 47.8784 J) / 1.2287 kg⋅m^2
ω_max = √(97.2464 J / kg⋅m^2) ≈ 4.962 rad/s
Therefore, the maximum angular speed is approximately 4.962 rad/s.
Step 5:
To find the linear speed of the smaller mass, we can use the relation between linear speed and angular speed:
v = r * ω
where v is the linear speed and r is the distance of the smaller mass from the axis of rotation.
r = 3.00 m - 0.13 m = 2.87 m
v = 2.87 m * 4.962 rad/s ≈ 14.26 m/s
Therefore, the linear speed that the object of smaller mass attains is approximately 14.26 m/s.
To find the maximum angular speed and linear speed attained by the object of smaller mass, we can use the principle of conservation of angular momentum and conservation of linear momentum.
(a) First, let's find the moment of inertia of the system about the axis of rotation. Since the rod is a straight line, the moment of inertia of a rod rotating about one end is given by: I = (1/3) * m * L^2, where m is the total mass of the rod and L is its length.
Using the given values, the moment of inertia I = (1/3) * (74.0 kg + 0.110 kg) * (3.00 m)^2 = 836 kg·m^2.
According to the conservation of angular momentum, the initial angular momentum (L_initial) of the system is zero since the rod is released from rest. At the maximum angular speed, the angular momentum (L_max) is given by the product of moment of inertia (I) and angular speed (ω).
L_initial = 0
L_max = I * ω_max
Since L_initial = L_max, we can set up the equation:
0 = I * ω_max
Solving for ω_max, we get:
ω_max = 0 rad/s
Therefore, the maximum angular speed of the system is 0 rad/s.
(b) To find the linear speed of the object of smaller mass, we can use the conservation of linear momentum. Since the system is initially at rest, the initial linear momentum is zero.
The linear momentum of the system is given by the product of the linear speed (v) of the object of smaller mass and its mass (m).
Initial linear momentum = 0
Final linear momentum = m * v
According to the conservation of linear momentum, the initial and final linear momenta should be equal, so we can write:
0 = m * v
Solving for v, we get:
v = 0 m/s
Therefore, the linear speed of the object of smaller mass is 0 m/s.
In summary:
(a) The maximum angular speed is 0 rad/s.
(b) The linear speed of the object of smaller mass is 0 m/s.