Use conservation of energy to determine the angular speed of the spool shown in Figure P8.36 after the 3.00 kg bucket has fallen 3.20 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.
The Figure shows the spool to have a mass of 5.00kg and a radius of 0.600m.
Conservation of energy tells us that the initial potential energy of the bucket is turned into the kinetic energy of the falling bucket and the rotational kinetic energy of the spool. We can write this as:
m_b * g * h = 0.5 * m_b * v^2 + 0.5 * I * ω^2
where m_b is the mass of the bucket (3.00 kg), g is the gravitational acceleration (9.81 m/s^2), h is the height the bucket falls (3.20 m), v is the final linear speed of the falling bucket, I is the moment of inertia of the spool, and ω is the final angular speed of the spool.
The spool has the shape of a disk, so its moment of inertia is given by:
I = 0.5 * m_s * r^2
where m_s is the mass of the spool (5.00 kg) and r is its radius (0.600 m). Calculate I:
I = 0.5 * 5.00 kg * (0.600 m)^2 I = 0.9 kg m^2
As the string unwinds without slipping, the linear speed of the bucket is related to the angular speed of the spool by
v = r*ω
Squaring both sides of the equation, we have:
v^2 = r^2 * ω^2
Now we can plug this into the conservation of energy equation:
3.00 kg * 9.81 m/s^2 * 3.20 m = 0.5 * 3.00 kg * r^2 * ω^2 + 0.5 * 0.9 kg m^2 * ω^2
Solve for ω^2:
ω^2 = (3.00 kg * 9.81 m/s^2 * 3.20 m) / (0.5 * 3.00 kg * r^2 + 0.5 * 0.9 kg m^2)
ω^2 = (3.00 kg * 9.81 m/s^2 * 3.20 m) / (0.5 * 3.00 kg * (0.600 m)^2 + 0.5 * 0.9 kg m^2)
ω^2 ≈ 74.526
Take the square root of both sides:
ω ≈ 8.63 rad/s
The angular speed of the spool after the bucket has fallen 3.20 m is approximately 8.63 rad/s.
To determine the angular speed of the spool, we will use the principle of conservation of energy.
The conservation of energy states that the total mechanical energy of a system remains constant if there are no external forces doing work on the system.
In this case, the only external force acting on the system is gravity. As the bucket falls, the gravitational potential energy is converted into kinetic energy and rotational kinetic energy of the spool.
Let's break down the conservation of energy equation into two parts:
1. calculate the change in gravitational potential energy of the bucket
2. calculate the change in rotational kinetic energy of the spool
1. Change in gravitational potential energy of the bucket:
The gravitational potential energy of an object is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
Given:
Mass of the bucket (m1) = 3.00 kg
Height (h) = 3.20 m
Change in gravitational potential energy (ΔPE) of the bucket = m1 * g * h
2. Change in rotational kinetic energy of the spool:
The rotational kinetic energy of an object is given by the formula: KE = (1/2) * I * ω^2, where I is the moment of inertia and ω is the angular speed.
Given:
Mass of the spool (m2) = 5.00 kg
Radius (r) = 0.600 m
The moment of inertia (I) of a solid cylinder can be calculated as: I = (1/2) * m * r^2
Change in rotational kinetic energy (ΔKE) = (1/2) * I * ω^2
Now, using the conservation of energy, we can equate the change in gravitational potential energy to the change in rotational kinetic energy:
ΔPE = ΔKE
m1 * g * h = (1/2) * I * ω^2
Substituting the values:
3.00 kg * 9.81 m/s^2 * 3.20 m = (1/2) * (1/2) * 5.00 kg * (0.600 m)^2 * ω^2
Solving for ω:
ω^2 = [3.00 kg * 9.81 m/s^2 * 3.20 m] / [(1/2) * (1/2) * 5.00 kg * (0.600 m)^2]
ω^2 = 124.025 rad^2/s^2
Taking the square root of both sides:
ω = √(124.025) rad/s
Therefore, the angular speed of the spool is approximately 11.13 rad/s.