A wheel is roatating freely at an angular speed of 800 rev/min on a shaft of negligible moment of inertia. A second wheel initially at rest and with twice the moment of inertia of the first, is suddenly coupled to the same shaft. (a) what is the angular speed of the resultant combination of the shaft and two wheels? (b) What fraction of the original rotational kinetic energy is lost? Treat the wheels as hoops in your analysis
Apply the law of conservation of angular momentum. Show work for further assistance.
To solve this problem, we need to apply the principle of conservation of angular momentum. Angular momentum is given by the formula:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
(a) Angular speed of the resultant combination:
Since the initial wheel is rotating freely with an angular speed of 800 rev/min, we can convert it to radians per second by using the conversion factor: 1 revolution = 2π radians.
So, ω1 = 800 rev/min * 2π rad/rev * 1 min/60 s = 800 * 2π / 60 rad/s
The moment of inertia of the first wheel, I1, is given.
The second wheel initially at rest has twice the moment of inertia of the first wheel, which means I2 = 2 * I1.
When the second wheel is coupled to the same shaft, the total angular momentum of the system should be conserved.
Ltotal = L1 + L2
Since the moment of inertia of the shaft is negligible, the angular momentum of the system is essentially the sum of the angular momenta of the two wheels.
Ltotal = I1 * ω1 + I2 * ω2
At the initial moment, the second wheel is at rest with an angular speed of ω2 = 0.
So, Ltotal = I1 * ω1 + I2 * 0
Ltotal = I1 * ω1
Since angular momentum is conserved, the total angular momentum after coupling should be equal to the initial angular momentum.
Ltotal = Linitial
I1 * ω1 = I1 * ωf
where ωf is the angular speed of the resultant combination.
Simplifying the equation:
I1 * ω1 = I1 * ωf
ωf = ω1
So, the angular speed of the resultant combination of the shaft and two wheels is 800 * 2π / 60 rad/s, which is the same as the initial angular speed of the first wheel.
(b) Fraction of the original rotational kinetic energy lost:
The rotational kinetic energy of a rotating object is given by the formula: KE = (1/2) * I * ω^2, where KE is the rotational kinetic energy.
The initial rotational kinetic energy of the first wheel is:
KE_initial = (1/2) * I1 * ω1^2
The final rotational kinetic energy of the combined system is:
KE_final = (1/2) * (I1 + I2) * ωf^2
Substituting the value of ωf as ω1 (from part a):
KE_final = (1/2) * (I1 + I2) * ω1^2
The fraction of the original rotational kinetic energy lost is given by:
Fraction_lost = (KE_initial - KE_final) / KE_initial
Fraction_lost = (KE_initial - (1/2) * (I1 + I2) * ω1^2) / KE_initial
Simplifying this expression will give the required fraction.