Finding Rotational Inertia with Thin Rod: Explanation and Example

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To find the rotational inertia of a thin rod, the equation I = 1/12 M L^2 is applicable when the rotation axis is at the center. For a rotation axis at one end, the parallel axis theorem should be used, which involves adding the product of mass and the square of the distance from the center of mass to the new axis. The initial attempt yielded a result of 5/3Md^2, but the correct answer is 8/3Md^2. This discrepancy highlights the importance of correctly applying the parallel axis theorem in calculations. Understanding these principles is crucial for accurately determining rotational inertia.
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Homework Statement


How do I find Rotational inertia of this?http://yfrog.com/e4asdfasdfasdfj. Assumed thin rod (ignore r).

Homework Equations


I=1/12 M L2

The Attempt at a Solution


Cause that equation applies if the rotation axis is at the center, so I doubled d (and got like (2d)2 and (4d)2) and used the equation, but I ended up with something like 5/3Md2, but the answer at the back says its 8/3Md2.
 
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If you know the rotational inertia about the center of mass, use the parallel axis theorem to find the rotational inertia about the end.
 
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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