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.
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.
1 answer
1 answer