Calculating moments of intertia

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The discussion focuses on calculating the moments of inertia for two particles with respect to different axes. For particle a, located at (3r, r, 0), the moments of inertia are derived as I_x = 1/3 * mr^3, I_y = 3mr^3, and I_z = 10/3 * mr^3, but there is confusion regarding the correct formula and units. The participants clarify that for point masses, the moment of inertia is calculated using I = m * r^2, simplifying the integration process. The second particle, b, is located at (r, -4r), and the total moment of inertia for both particles about the y-axis is debated, with calculations leading to a total of 10mr^2. The discussion highlights the importance of correctly applying the formulas and understanding the geometry involved in moment of inertia calculations.
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1. Find the moment of inertia I_x of particle a with respect to the x-axis (that is, if the x-axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes).

Particle a is located at x=3r, y=r, z=o



2. I = Integral(x^2 + y^2)



3. I_x = 1/3 * mr^3
I_y = 3mr^3
I_z = 10/3 * mr^3
 
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Surely the formula is I = m*r^2 ?
so I_x = mr^2.
No integration needed when you have a point charge. The particle is a distance r from the x-axis. Your r^3 answers don't even have the right units.
 
Ah, ok thanks. Now I am supposed to consider a second particle, b, located at x=r and y=-4r, and calculate the total moment of inertia for the two particles of this system, with the y-axis as the axis of rotation.

Why 10mr^2 not the answer? I obtain this my summing I_ya + I_yb = 9mr^2 + mr^2
 
This is is driving me crazy :cry:
 
10mr^2 looks good to me, assuming the two particles have the same mass.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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