What is the moment of inertia of the particle at the end of the meter stick?

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a particle attached to a meter stick. The particle has a mass of 0.400 kg and is located at the 100-cm mark of a meter stick weighing 0.100 kg. The moment of inertia for the particle is determined using the formula I = mr², where 'm' is the mass and 'r' is the distance from the pivot point. The calculations involve two pivot points: one at the 50.0-cm mark and another at the 0-cm mark. The thread emphasizes the importance of understanding both the particle's and the meter stick's contributions to the system's overall moment of inertia.
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Moment of Inertia

A particle of mass 0.400 kg is attached to the 100-cm mark of a meter stick of mass 0.100 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 4.00 rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the 50.0-cm mark and (b) perpendicular to the table through the 0-cm mark.

What is the moment of intertia of the particle at the end of the meter stick? I know how to find the moment of intertia of the meter stick.
 
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The moment of inertia of a particle is mr^2
 
thank you
 
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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