Using Cartesian vs. Normal/Tangential Coordinates for Centripetal Motion

[Pipsqueakalchemist]

Homework Statement

Question below

Relevant Equations

Newton's 2nd law
Moment equation

So for this problem the solution used Cartesian coordinates but I was wondering wouldn’t it be easier to use Normal and tangential coordinate because the bar is undergoing centripetal motion? Also on the right diagram shouldn’t the acceleration be down and not up. The reason I think that is because of the bar is in centripetal motion then the tangent acceleration should be down and to the right normal to the string. Is that the right reasoning or am I thinking about it wrong?

 

[Pipsqueakalchemist]

Also this was my attempt but it’s wrong. Can anyone tell me where I went wrong please

 

[BvU]

Hi,

undergoing centripetal motion?

What, precisely would that be ? And what coordinates would you choose ?

shouldn’t the acceleration be down and not up

The choice of coordinates seems to be: x to the right and y upwards. Quite reasonable. And the y component of the resulting acceleration is probably going to be negative.

What are your equations explicitly ?
(especially since I don't believe you post #2 at all !)

$\$

 

[BvU]

PS nice exercise to actually do the experiment with a ruler and a cord !

 

[Pipsqueakalchemist]

Hi,

What, precisely would that be ? And what coordinates would you choose ?

The choice of coordinates seems to be: x to the right and y upwards. Quite reasonable. And the y component of the resulting acceleration is probably going to be negative.

What are your equations explicitly ?
(especially since I don't believe you post #2 at all !)

$\$

In my attempt I assumed the bar was undergoing centripetal motion around the fixed end of the rope. I tried using tangential and normal coordinates at that instant shown in the diagram where normal axis in directed toward the centre so pointing to fixed end of string and tangent axis perpendicular to it and directed down and to the right. This question is equation of motion of rigid bodies in general planar motion. So equations needed are Newton’s 2nd law, moments equation about centre of mass, and relative acceleration. I have the required equations attached

 

[Pipsqueakalchemist]

I think I’m struggling to identify whether an rigid body is undergoing translational motion, rotational motion or general plane motion

 

[BvU]

In my attempt I assumed the bar was undergoing centripetal motion around the fixed end of the rope

This is hard to believe. I can understand that point B on the bar is constrained to move on a circle around C (*), but point A isn't going to stay on a line that makes an angle of 30 degrees wrt the cord.

(*) provided the cord stays taut, which I think is a reasonable assumption in this case​

The word centripetal means 'going towards the center'. It is confusing and ambiguous to use it to describe a movement. When there is a centripetal force acting, a point mass can execute a circular motion.

Let me go back to your first attachment, which I assume is the problem statement:

can you find the force that has to be executed at A to keep the bar in position before it is released ? (**)

And your second attachment (looks as if it is part of the solutions manual ?) seems to be missing something on the righthand side

I assume this is at t=0 after releasing. On the left I see two forces, on the right I see two accelerations. Fine, but I don't see how there can be an equals sign in between.

​

[edit] my mistake. The 2 means 2 kg and then the purple arrows are forces too.​

Still I can't explain the equals sign.​

​

​

At t=0 the force you calculated at (**), is instantly removed. You have to ask yourself if it is reasonable to expect that T does not change instantly. If it doesn't, you have two forces that will cause the bar to drop and rotate. Choose an axis of rotation (***) and work out $\alpha$, the angular acceleration.

Re

I can follow the first three lines, but the fourth ?
What is I ? Moment of inertia. Expression ?
$v_{B/A}$ is what ? velocity of B wrt A ? Vector ? Scalar ?
$\omega$ ?

(***) line three makes me think you picked G ? ​

$\$

 

[Pipsqueakalchemist]

The question I got was from an YouTube video. Do you mind watching the part for the solution bc the vid will explain it better bc I’m not 100% sure what I’m doing. The solution for the problem is pretty short. He starts an introduction then goes over the problem. It takes 5 minutes until he finishes the solution. I have the thumbnail. It’s the middle video.

 

[BvU]

Found the link after some googling and Utub commercials. I don't need to digest this for you; the idea is that you come to a point where you understand the whole process. Sitting back and watching may not be enough. Find out where exactly you are unsure:

bc I’m not 100% sure what I’m doing

Write it out step by step and then ask when you run into trouble.

I have to leave now; perhaps @haruspex can take over ?$\$

 

[haruspex]

I think I’m struggling to identify whether an rigid body is undergoing translational motion, rotational motion or general plane motion

So allow for the general case.
It is usually simplest to consider the instantaneous motion as the sum of a linear motion of the mass centre and a rotation about that centre, in respect of both velocity and acceleration.
The initial velocities here are both zero, so that's easy. (And centripetal acceleration is, by definition, orthogonal to the velocity: no linear velocity, no centripetal acceleration.)
You have three scalar acceleration equations available, two linear + one rotational; and four unknowns: the three accelerations and the tension.
You need one more equation. What do you know about the motion of the end of the rod that's tied to the string?

 

[Pipsqueakalchemist]

Found the link after some googling and Utub commercials. I don't need to digest this for you; the idea is that you come to a point where you understand the whole process. Sitting back and watching may not be enough. Find out where exactly you are unsure:
Write it out step by step and then ask when you run into trouble.

I have to leave now; perhaps @haruspex can take over ?$\$

Appreciate the help. Sorry I had to make you search for the vid that was my bad