Why Is My Calculation of Angular Momentum Incorrect?

AI Thread Summary
The calculation of angular momentum presented is incorrect due to the use of the wrong formula for moment of inertia. The correct moment of inertia for a thin disk is ½Mr², not mr². This affects the final calculation of angular momentum, leading to an inaccurate result. The user should recalculate using the correct moment of inertia to find the accurate angular momentum value. Proper understanding of the formulas is essential for accurate physics calculations.
vorcil
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i seem to have got it wrong,


http://img10.imageshack.us/img10/2373/masteringphysics.jpg

my solving:

L=IW
I=mr^2

L=mr^2w

L=2.7kg* (0.03^2) * w

w= 2*pi * 600rpm /60 =62.83 rad/s

L=2.7*0.0009*62.83 =0.15

and i got it wrong? i was pretty sure it was right,

help please
 
Last edited by a moderator:
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Hi vorcil,

The moment of inertia of the disk is ½Mr².
 
What's the moment of inertia of a thin disk/cylinder about what is in your case the x-axis?
 
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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