Acceleration in a circular path

[Mr Genius]

What is the acceleration of a body (you can consider it as a particle) moving in a circular path placed in the vertical plane as it passes through the bottom of the path? (Provided that the body is under its weight and the path normal only).

 

[Andrew Mason]

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)

AM

 

[Mr Genius]

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)

AM

I mean that the forces acting on the body are only its weight and the normal reaction of the support

 

[CWatters]

In order to move in a circle the net radial force on the object must be equal to the centripetal force (mv²/r) regardless of what actual individual forces act on the object.

Then use Newtons F=ma to find the acceleration.

 

[Dale]

In order to move in a circle the net radial force on the object must be equal to the centripetal force (mv2/r)

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.

 

[Mr Genius]

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 I'm not familiar with that method

 

[Mr Genius]

At the bottom of the circular path the summation of weight and normal reaction is equal to zero??

 

[Andrew Mason]

At the bottom of the circular path the summation of weight and normal reaction is equal to zero??

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?

AM

 

[CWatters]

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.

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.

 

[Dale]

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.

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.

 

[jbriggs444]

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]

Sorry for my lazy use of velocity when i meant speed.

 

[nasu]

The circular path is vertical. Both velocity and speed change along the path .

 

[Andrew Mason]

The circular path is vertical. Both velocity and speed change along the path .

I believe jbriggs was referring to the speed at the very bottom where the rate of change of velocity is 0.

AM

 

[jbriggs444]

I believe jbriggs was referring to the speed at the very bottom where the rate of change of velocity is 0.

Yes. At the bottom of the arc the first derivative of speed is zero but the first derivative of velocity is not.

 

[jaumzaum]

Have a look at this op
https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf
You can use the formula of the centripetal and tangential acceleration (polar coordinates), in you example r is constant so it’s first and second derivative is zero. You can find (for an initial theta zero) the centripetal and tangential acceleration with respect to theta. You will see in the bottom there is no tangential acceleration, so a = v^2/r as usual

 

[Andrew Mason]

Quite right. I should have said the rate of change of speed or $|\vec{v}|$ is zero.

AM