Determining angular momentum of a point mass.

AI Thread Summary
To calculate the angular momentum of a point mass, the correct formula involves the vector cross product, L = r x p, where p is the momentum (p = mv). The position vector r is given as (2.00i - 1.00j) m and the velocity vector v as (7.00i - 2.00j) m/s. The initial attempt incorrectly used scalar multiplication instead of the cross product, leading to an incorrect result. The proper calculation using the cross product yields an angular momentum of 6.00k kgm²/s. Understanding vector operations is crucial for solving this problem accurately.
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Homework Statement



The velocity of a 2.00 kg body moving in the xy plane is given by v=(7.00i-2.00j) m/s. Its position vector is r=(2.00I-1.00j) m. Calculate its angular momentum (magnitude and direction) about the origin (kgm2/s). Express your answer in unit vector notation.


Homework Equations



L=rp
p=mv
..
L=Iω?


The Attempt at a Solution



I attempted to use L=rp=rmv and then substituted in my conditions. I got (28.0i + 4.00j) kgm2/s but the answer is apparently 6.00k. Where am I going wrong?
 
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You are dealing with vectors, not scalars.
Therefore the product rv should really be r x v. (x being the vector cross product).

A 2 dimensional cross product can be written as
r x v = r1 v2 - r2 v1 , where 1 and 2 simply denote the components of the vector: 1 for the i component, 2 for the j.

R.
 
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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