Two masses, mA = 35.0 kg and mB = 42.0 kg, are connected by a rope that hangs over a pulley (as in the figure). The pulley is a uniform cylinder of radius 0.50 m and mass 4.8 kg. Initially mA is on the ground and mB rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of mB just before it strikes the ground. Assume the pulley bearing is frictionless.
To determine the speed of mB just before it strikes the ground, we can use the principle of conservation of energy. The total mechanical energy of the system remains constant throughout the motion.
The initial position of mA on the ground and mB above it can be considered as the zero reference level for potential energy. The potential energy of mB is given by:
PE(B) = mB * g * h
Where mB is the mass of object B, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of object B above the ground.
The initial potential energy of mB is:
PE(B initial) = mB * g * h
The initial kinetic energy of the system is zero, as both masses are initially at rest.
As the system falls, the potential energy of mB is converted into its kinetic energy, given by:
KE(B final) = (1/2) * mB * v²
Where v is the final velocity of mB just before it strikes the ground.
Since the pulley is a uniform cylinder, we also need to consider its rotational kinetic energy. The rotational kinetic energy of the pulley is given by:
KE(pulley final) = (1/2) * I * ω²
Where I is the moment of inertia of the pulley and ω is its angular velocity. Since the pulley is a cylindrical shape, the moment of inertia can be given by:
I = (1/2) * m * r²
Where m is the mass of the pulley and r is its radius.
The total energy of the system is the sum of the individual energies:
Total energy = PE(B initial) + KE(B final) + KE(pulley final)
Since the total mechanical energy is conserved, the initial energy and the final energy of the system must be equal:
PE(B initial) = KE(B final) + KE(pulley final)
Substituting the equations for potential energy and kinetic energy and simplifying, we get:
mB * g * h = (1/2) * mB * v² + (1/2) * (1/2) * m * r² * ω²
Since the rope connecting mA and mB is assumed to be inextensible, the magnitudes of their accelerations must be the same. Therefore, we can relate the linear acceleration of mB (given by a) to the angular acceleration of the pulley (given by α) as:
a = r * α
We can also relate the linear speed of mB (v) and the angular speed of the pulley (ω) as:
v = r * ω
Substituting these relationships into the previous equation, we get:
mB * g * h = (1/2) * mB * v² + (1/2) * (1/2) * m * r² * (v/r)²
Simplifying further, we have:
mB * g * h = (1/2) * mB * v² + (1/4) * m * v²
Solving for v, we can find the final speed of mB just before it strikes the ground.
v² = (2 * mB * g * h) / (mB + 0.25 * m)
Finally, taking the square root of both sides, we get:
v = √[(2 * mB * g * h) / (mB + 0.25 * m)]