Acceleration in uniform circular motion is uniform or non-uniform?

[kuruman]

You have made responding a bit awkward by embedding your responses in your quotes of my posts. It means I cannot use the quote button on them.

"My teacher told me so, sir. He said they both counterbalances and cancel themselves."
Oh dear. Your teacher is incompetent. You have my sympathy.
"Never both, why sir? we do have centripetal and centrifugal force, right?"
As I wrote, in an inertial frame there is no centrifugal force; in the noninertial frame locked to the body, the body is not moving, so there is no centripetal acceleration nor force. In some other noninertial frame there can be both, but it would be a very odd choice of frame.

This got me thinking. In our part of the world we teach that the force exerted on the stone by the string is "centripetal", i.e. directed towards the center of rotation. Centripetal defines a direction, not a special kind of force and has an orthogonal counterpart that is given the name "tangential." We also teach, as shown in the quote above, that "centrifugal" is a force that appears in a non-inertial frame when the right hand side of Newton's second law is moved to the left hand side with a change in sign. In either frame, when the system is the stone, the tension acts towards the center of the circle. We can write unambiguous equations using radial unit vectors for the two cases
$$\begin{align} & T(-\mathbf{\hat r})=m\omega^2 r(-\mathbf{\hat r}) \\
& T(-\mathbf{\hat r})+m\omega^2 r(\mathbf{\hat r})=0. \end{align}$$Equation (1) is in the inertial frame and says the the centripetal force, which is the net force, is the same as mass times the centripetal acceleration. Equation (2) is in the non-inertial frame and says that there are two forces (let's not give them names) that add to give zero. That's what we teach in our part of the world. It would be much simpler if everybody agreed to drop "centripetal" and "centrifugal" altogether and use "radially in" and "radially out" instead.

So here is my question. What if in OP's part of the world the terms "centrifugal" and "centripetal" are used simply to designate direction, respectively $(\mathbf{+\hat r})$ and $(-\mathbf{\hat r})$, with no reference to inertial or non-inertial frames? Then, in view of equation (2) the teacher's claim that "they both counterbalances and cancel themselves" makes sense. The answer to (iii) that the force on the hand is "centrifugal acting radially outward" also makes sense in that context. Maybe the points that are argued here arise from semantic cultural differences.

I should add that it is here, at PF, that I learned that when someone from India says "I have a doubt about your solution", he does not mean that he questions the validity of my solution. He is simply informing me that he has a question about my solution that needs further clarification. That kept my feathers from being ruffled.

 

[haruspex]

Equation (2) is in the non-inertial frame and says that there are two forces (let's not give them names) that add to give zero. That's what we teach in our part of the world. It would be much simpler if everybody agreed to drop "centripetal" and "centrifugal" altogether and use "radially in" and "radially out" instead.

Nice try, but it doesn’t work. Equation 2 is not just for any non-inertial frame; it is for the frame in which the body is at rest. So how can T be a centripetal or "radially inward" force?

Edit: I shift my ground:
In equation 1, the RHS is the centripetal force, i.e. the required resultant to provide the observed acceleration. The LHS is the applied force normal to the velocity. (More generally it would be something like $(\vec F_{net}\times \vec v)\times \vec v/v^2$.)
So the LHS in equation 2 only becomes describable as the centripetal force as a result of applying equation 1.

To make this clearer, suppose we choose some other non-inertial frame. We can then have
tension + centrifugal force = centripetal force

 

[Darshit Sharma]

Hi @Darshit Sharma. I’d like to add this…

Let's assume that the path of the mass is a circle in the horizontal plane (otherwise a uniform speed is hard to achieve).

.Note that the centre of rotation is a point vertically below the hand (because the string can never be perfectly horizontal). We have to be very careful when we refer to directions as 'radially' inwards or outwards - do we mean with respect to the centre of rotation or with respect to the hand?
_______________

The answer to question (i) is: Uniform speed

Comment: The problem-statement explicitly states that the mass is moving ‘with a uniform speed’. So question (i) seems pointless.
_______________

An acceptable (IMO) answer to question (ii) is:

The magnitude of the acceleration is uniform (constant). But since the direction of the acceleration continually changes, the acceleration (a vector) is non-uniform.

The direction of the mass's acceleration is radially inwards towards the centre of rotation.

Comment: The answer in the answer-key is unsatisfactory at-best, and wrong at-worst.
_______________

An acceptable (IMO) answer to question (iii) is:

The force acting on the hand is the tension at the (inner) end of string. It’s direction is towards the mass.

An alternative acceptable (IMO) answer is that the force acting on the hand is the frictional force of the string on the hand, acting towards the mass.

Comment: The answer in the answer-key is wrong about the force. The hand is effectively stationary in an inertial frame of reference and therefore never experiences a ‘centrifugal force’.

EDIT: Some changes to improve answer.

I agree with all your answers......

 

[kuruman]

Nice try, but it doesn’t work. Equation 2 is not just for any non-inertial frame; it is for the frame in which the body is at rest. So how can T be a centripetal or "radially inward" force?

Is the name the problem? I think one can define "radial" in a non-inertial frame where the body is at rest. There is a unique line, whatever name you choose to give it, along which the string is at rest and under tension which means that two equal and opposite forces are acting on it. An observer (of zero height) can crawl around a rotating platform holding a plumb bob and verify that, when he is at rest w.r.t. the platform and the bob is at rest with respect to him, the bob will point to the same point on the platform. He calls that direction "centripetal". In short, he detects and maps a force field. (It is assumed that gravity is not in the picture.)

 

[haruspex]

Is the name the problem? I think one can define "radial" in a non-inertial frame where the body is at rest. There is a unique line, whatever name you choose to give it, along which the string is at rest and under tension which means that two equal and opposite forces are acting on it. An observer (of zero height) can crawl around a rotating platform holding a plumb bob and verify that, when he is at rest w.r.t. the platform and the bob is at rest with respect to him, the bob will point to the same point on the platform. He calls that direction "centripetal". In short, he detects and maps a force field. (It is assumed that gravity is not in the picture.)

See my edit to post #52.

 

[Darshit Sharma]

I should add that it is here, at PF, that I learned that when someone from India says "I have a doubt about your solution", he does not mean that he questions the validity of my solution. He is simply informing me that he has a question about my solution that needs further clarification. That kept my feathers from being ruffled.

Haha, sir, you are cent per cent correct.

Anyway, I feel the conversation is a bit out of my syllabus. I am finishing the academic year in March or so. Subsequently, I shall study all this. Sorry for the English errors so far lol.

Thank you, guys, @haruspex @kuruman @berkeman @Steve4Physics @Hill @orodium and if anyone is left out, he/she too.