Circular Motion on a Track: Understanding Force and Direction

AI Thread Summary
In circular motion along a track that is a quarter of a circle, the total force acting on an object dropped from 180 degrees (9 o'clock) is directed towards the center of the circle due to centripetal force. As the object moves along the track, the direction of this force differs from the direction of motion at various angles. Initially, at 180 degrees, the force is horizontal, but as the object descends to 270 degrees (6 o'clock), the force shifts downward. The absence of external forces and friction means that only the centripetal force is considered. Understanding this relationship is crucial for analyzing motion in circular paths.
devanlevin
in circular motion, on a track, which is exactly a quarter of a circle, starting at 180 degrees(9 0 clock) and ending at 270(6 o clock) what is the direction of the total force working on an object dropped from the start at 180? no external forces, friction etc... is it always with the direction of motion or does it differ at different angles??
 
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Hi devanlevin,

devanlevin said:
in circular motion, on a track, which is exactly a quarter of a circle, starting at 180 degrees(9 0 clock) and ending at 270(6 o clock) what is the direction of the total force working on an object dropped from the start at 180? no external forces, friction etc... is it always with the direction of motion or does it differ at different angles??

You mentioned some forces that are not acting, but what forces are there that are acting on the object?
 
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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