Using law of conservation of energy, the angular speed of the spool is 15.96 rad/s
What is the law of conservation of energy?
The law of conservation of energy states that for a system, the total mechanical energy is conserved.
To use conservation of energy to determine the angular speed of the spool shown in the figure below after the 3.00-kg bucket has fallen 3.90 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds. The spool may be treated as a uniform solid cylinder, we proceed as follows.
Now using the law of conservation of energy,
Initial energy of system = final energy of system
or
Potential energy lost by bucket = rotational kinetic energy gained by spool
mgh = 1/2Iω²
where
- m = mass of bucket = 3.00 kg
- g = acceleration due to gravity = 9.8 m/s²
- h = drop in height of bucket = 3.90 m
- I = rotational inertial of spool (solid cylinder) = 1/2MR² where
- M = mass of spool = 5.00 kg
- R = radius of spool = 0.600 m
- ω = angular speed of spool
Making the ω subject of the fprmula, we have that
ω = √(2mgh/I)
= √(2mgh/1/2MR²)
= √(4mgh/MR²)
So, substituting the values of the variables into the equation, we have that
ω = √(4mgh/MR²)
= √[4 × 3.00 kg × 9.8 m/s² × 3.90 m/(5.00 kg × (0.600 m)²)]
= √[458.64 kgm²/s²/(5.00 kg × 0.3600 m²)]
= √[458.64 kgm²/s²/1.8 kgm²]
= √[254.8/s²]
= 15.96 rad/s