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
Angular momentum will conserved, and the total moment of inertia (I) is tripled. Since angular momentum is I*w, the angular spped w must be reduced to 1/3 of the original value.
With w reduced to 1/3 of the original and I tripled, the kinetic energy (1/2) I w^2 will be reduced to 1/3 the former value.
To solve this problem, we need to apply the principles of conservation of angular momentum and conservation of energy.
(a) Let's start with finding the angular speed of the resultant combination of the shaft and two wheels.
The angular momentum of an object is given by the product of its moment of inertia (I) and angular velocity (ω). Mathematically, it can be written as L = Iω.
We are given that the first wheel is rotating freely with an angular speed of 800 rev/min. To find the angular velocity (ω1) in radians per second, we need to convert rev/min to radians/s. We know that 1 revolution = 2π radians. Therefore:
ω1 = (800 rev/min) * (2π radians/1 rev) * (1 min/60 s)
= (800 * 2π / 60) radians/s
= (80/3)π radians/s (approx. 83.78 radians/s)
Since the shaft has a negligible moment of inertia, the angular momentum of the first wheel is I1 * ω1.
Now, we have a second wheel with twice the moment of inertia (2I1) initially at rest. When it is coupled to the same shaft, the total angular momentum (L_total) would be the sum of individual angular momenta.
L_total = (I1 * ω1) + (2I1 * 0)
= I1 * ω1
Since the mechanics of the system would remain the same, the angular speed of the resultant combination of the shaft and two wheels (ω_total) will remain the same as ω1.
Hence, the angular speed of the resultant combination of the shaft and two wheels is (80/3)π radians/s.
(b) Now, let's find the fraction of the original rotational kinetic energy that is lost.
The rotational kinetic energy (KE) of an object can be calculated using the formula KE = (1/2) I ω^2.
The initial rotational kinetic energy (KE_initial) is the sum of the kinetic energies of the two wheels, which can be written as:
KE_initial = (1/2) I1 ω1^2 + (1/2) (2I1) * 0^2
= (1/2) I1 ω1^2
The final rotational kinetic energy (KE_final) is the rotational energy of the resultant combination of the shaft and two wheels, which can be written as:
KE_final = (1/2) (I1 + 2I1) ω_total^2
= (1/2) 3I1 ω_total^2
The fraction of the original rotational kinetic energy lost is given by:
Fraction lost = (KE_initial - KE_final) / KE_initial
Substituting the values, we get:
Fraction lost = [(1/2) I1 ω1^2 - (1/2) 3I1 ω_total^2] / [(1/2) I1 ω1^2]
Simplifying further, we have:
Fraction lost = (ω1^2 - 3ω_total^2) / ω1^2
Now, substituting the values of ω1 and ω_total, we can calculate the fraction of the original rotational kinetic energy lost.
Please note that the values I1 and ω1 were given in the problem statement and ω_total was calculated in part (a).