The acceleration of a body moving in a vertical circular path at the bottom is determined by the centripetal force, which is equal to mv²/r, where m is mass, v is velocity, and r is the radius of the circular path. The forces acting on the body are its weight and the normal reaction from the support. At the bottom of the path, the net radial force is the sum of these forces, and while the speed remains constant, the direction of the velocity changes, indicating a non-zero acceleration. The Lagrangian method is suggested for simplifying the analysis of such constrained motion.
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I mean that the forces acting on the body are only its weight and the normal reaction of the supportAndrew Mason said:I am not sure what you mean by the part in parentheses.
What are the sources of the forces on the body? (Hint: there are two)
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I believe that is only true if it is moving at constant speed. I think the OP is describing a situation where the motion is constrained to a circle, but not moving at constant speed.CWatters said:In order to move in a circle the net radial force on the object must be equal to the centripetal force (mv2/r)
No I'm not familiar with that methodDale said:I believe that is only true if it is moving at constant speed. I think the OP is describing a situation where the motion is constrained to a circle, but not moving at constant speed.
@Mr Genius are you familiar with the Lagrangian method? For constrained problems like this, it makes solving the motion very easy.
No. That is the net force, which is equal to the body's _________ x ____________ (fill in the blanks). Can we assume that there is no tangential acceleration except as provided by gravity? If so, since the question asks what the acceleration is at the bottom of the path, is there non-zero acceleration there and, if so, what is the direction of the acceleration? How is that acceleration related to the tangential speed of the body?Mr Genius said:At the bottom of the circular path the summation of weight and normal reaction is equal to zero??
Dale said:I believe that is only true if it is moving at constant speed. I think the OP is describing a situation where the motion is constrained to a circle, but not moving at constant speed.
No, I think you are fine and it is my mistake. I got in my head that he wanted to solve for the whole motion. You correctly identified that he was asking only about the bottom.CWatters said:That is a mistake I make but if it's acting "under it's weight" (like a pendulum) then right at the bottom I believe the velocity is constant. Perhaps I've missed something.
Well, the speed is constant. If it is traveling in a circular arc with a non-zero speed, its velocity is not.CWatters said:That is a mistake I make but if it's acting "under it's weight" (like a pendulum) then right at the bottom I believe the velocity is constant. Perhaps I've missed something.
I believe jbriggs was referring to the speed at the very bottom where the rate of change of velocity is 0.nasu said:The circular path is vertical. Both velocity and speed change along the path .
Yes. At the bottom of the arc the first derivative of speed is zero but the first derivative of velocity is not.Andrew Mason said:I believe jbriggs was referring to the speed at the very bottom where the rate of change of velocity is 0.