Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d.
A)Find an expression for the magnitude L of the angular momentum of the two-particle system around a point midway between the two lines. Use m for the mass, v for the speed and d for the distance of separation.
B)Does the expression change if we change the point about which L is calculated?
C)Now reverse the direction of travel for one of the particles and repeat (A)
D)Does the expression change if we change the point about which L is calculated?
I will be happy to critique your thinking on this.
The answer is 5
A) To find the expression for the magnitude L of the angular momentum of the two-particle system around a point midway between the two lines, we can use the formula for angular momentum:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since the particles are moving in opposite directions, their angular velocities will have opposite signs. Let's assume that the particle moving in the positive direction has an angular velocity ω1, and the particle moving in the negative direction has an angular velocity ω2.
The moment of inertia I for each particle can be calculated using the parallel-axis theorem, which states that the moment of inertia of a particle about an axis parallel to and a distance d away from an axis through its center of mass is equal to the sum of the moment of inertia about its center of mass and the product of its mass and the square of the distance d.
For each particle:
I = (1/12)m(2r)^2 (moment of inertia about its center of mass)
+ m(d/2)^2 (product of its mass and the square of the distance d)
where r is the radius of each particle.
Since the particles are identical, r is the same for both of them.
The angular velocities ω1 and ω2 can be related to the speed v by the formula:
v = ωr
where r is the radius of each particle.
Substituting the expressions for I and ω in the formula for angular momentum, we have:
L = 2I(ω1) - 2I(ω2)
= 2(I(ω1) + I(ω2))
= 2(m(1/12)(2r)^2(ω1) + m(d/2)^2(ω1) + m(1/12)(2r)^2(-ω2) + m(d/2)^2(-ω2))
Simplifying this expression will give us the final answer for part A.
B) The expression for the magnitude L of the angular momentum does change if we change the point about which it is calculated. The expression we derived in part A is specifically for the point midway between the two lines. If we calculate the angular momentum around a different point, the moment of inertia will change, resulting in a different expression for L.
C) Now if we reverse the direction of travel for one of the particles, the angular velocity of that particle will change sign. Let's call its new angular velocity ω2'. The other particle still has the same angular velocity ω1.
Using the same approach as in part A, but with the new angular velocities, the expression for the magnitude L of the angular momentum is:
L' = 2(m(1/12)(2r)^2(ω1) + m(d/2)^2(ω1) + m(1/12)(2r)^2(-ω2') + m(d/2)^2(-ω2'))
D) Similar to part B, the expression for the magnitude L' of the angular momentum will change if we change the point about which it is calculated. The moment of inertia will be different, resulting in a different expression for L'.