Vertical energy of a pendulum - is there one?

AI Thread Summary
The discussion centers on the concept of vertical kinetic energy in the context of a simple pendulum. The original poster questions whether vertical kinetic energy exists as a separate entity or if it is merely a component of total kinetic energy. Responses clarify that while the pendulum's velocity can be broken down into vertical and horizontal components, kinetic energy itself is a scalar quantity that encompasses the entire velocity vector. Ultimately, the distinction between vertical kinetic energy and total kinetic energy is acknowledged, but it is emphasized that kinetic energy does not have direction. The conversation concludes with a clear understanding that kinetic energy is defined by the total velocity, not its components.
jojotank
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Homework Statement


Hello. I was just wondering. You have a simple pendulum . string and a ball with mass at the end. We usually solve examples and we think about conversation of kinetic energy and potentional energy. I was just wandering if there is also something like vertical kinetic energy. Or it is just a part of kinetic energy itself (just the sine part of velocity vector). If it exist, i know it is very small. I was just wondering... Thank you.

Homework Equations


no need for them

The Attempt at a Solution


thinking
 
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Not quite sure what you mean. You can certainly express the velocity of the pendulum in terms of vertical and horizontal components, but the kinetic energy uses the full velocity. Velocity has a direction, but kinetic energy does not.
 
It is totaly clear now. 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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