Friday, December 12, 2014

gravity - When a ball is tossed straight up, does it experience momentary equilibrium at top of its path?


This question has been asked many times all over the Internet and answers can be found on places such as yahoo and ask.com, but I'm not satisfied with those answers and I don't trust the validity of those places. This seems like a more appropriate place to ask. So, here's my train of thought and I would like to ask you if you think I'm right or wrong.


Even though most of the answers point to the fact that the ball does not achieve equilibrium because the force of gravity is constantly acting upon it, thus causing an acceleration, I still think there's a moment of equilibrium and here's why:



When the ball is traveling up, it's accelerating towards the ground and eventually reaches a point at which its speed reaches 0. At this moment, isn't the reason the speed is 0 is due to the fact that net force is 0? The force that made the ball travel in the upward direction was canceled out by the gravitational force. So, the sum of forces at that brief moment is equal to 0, otherwise the ball would be moving. I don't think gravity is the only force acting on the ball at that moment, since the throwing force was acting on it as well, until both canceled each other out for that brief moment.


Am I right or wrong?



Answer



The forces are never balanced, as there is only ever one force - gravity.


The key is to remember Newton's second law: $F = ma$. Force and acceleration are paired, not force and velocity. Knowing just an object's current velocity tells you nothing about what forces are acting on it.


There are two ways to see how the velocity goes to zero. Either the initial impulse (instantaneous transfer of momentum) is depleted by a force applied over time, $$ m \Delta v = \int F \ \mathrm{d}t, $$ or the initial kinetic energy is depleted by a force applied over a distance, $$ \frac{1}{2} m \Delta (v^2) = \int F \ \mathrm{d}x. $$ In both cases, the force of gravity is acting continuously to slowly cancel the initial velocity, and there is nothing that turns off this force at the apex of the trajectory.


No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...