We can use conservation of energy to find the maximum speed of the small-mass object. Initially, the trebuchet is at rest and all of its energy is potential energy due to the gravitational forces on the two masses. At the point of maximum launch, all of the potential energy has been converted to kinetic energy of the small-mass object.
The potential energy of the system is given by:
U = m1gh1 + m2gh2
where g is the acceleration due to gravity (9.81 m/s^2), h1 is the height of the small-mass object above its initial position, and h2 is the height of the center of mass of the system above its initial position.
At the point of maximum launch, h1 = 0 and h2 = d/2 = 1.30 m. The potential energy of the system is:
U = m2gd/2
The kinetic energy of the small-mass object is:
K = (1/2)m1v^2
where v is the speed of the small-mass object at the point of maximum launch.
By conservation of energy:
U = K
m2gd/2 = (1/2)m1v^2
Solving for v:
v = sqrt((2m2gd)/(m1))
Substituting the given values:
v = sqrt((2*68.5*9.81*1.30)/(0.115)) = 68.9 m/s
Therefore, the maximum speed that the small-mass object attains when it leaves the trebuchet horizontally is 68.9 m/s.
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 = 2.60 m long, joining particles of mass m1 = 0.115 kg and m2 = 68.5 kg at its ends. It can turn on a frictionless, horizontal axle perpendicular to the rod and 13.0 cm from the large-mass particle. The operator releases the trebuchet from rest in a horizontal orientation.
Find the maximum speed that the small-mass object attains when it leaves the trebuchet horizontally. (in m/s)
1 answer
0 answers