Finding out the rotational speed of a mass

[kuruman]

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Is it okay sir? @kuruman

On the left diagram, you show two vertical forces. These forces arise from physical entities, one entity for one force. Specifically, you have the Earth that exerts the "down" force and the string that exerts the "up" force.

Now that you understand this, please explain to me what two physical entities are responsible for the two horizontal forces that you drew in the diagram on the right. They must be items that you can see and touch.

 

[nazmulhasanshipon]

On the left diagram, you show two vertical forces. These forces arise from physical entities, one entity for one force. Specifically, you have the Earth that exerts the "down" force and the string that exerts the "up" force.

Now that you understand this, please explain to me what two physical entities are responsible for the two horizontal forces that you drew in the diagram on the right. They must be items that you can see and touch.

@kuruman sir Maybe the physical entities are: the string that provides the centripetal force and the rotation which provides centrifugal but should we consider centrifugal force too although it's a fictions force?

 

[kuruman]

@kuruman sir Maybe the physical entities are: the string that provides the centripetal force and the rotation which provides centrifugal but should we consider centrifugal force too although it's a fictions force?

The string is something that you and I can see and touch. I cannot see or touch "the rotation which provides centrifugal", can you? If not, then you should not consider it in the free body diagram drawn in an inertial frame. Fictitious forces get their name because they don't exist in inertial free body diagrams but appear in non-inertial frames. They are not forces generated by things that you can see and touch; they are mass times acceleration which is moved over to the "other side" of Newton's second law and just looks like a force.

The time has come for you to write Newton's second law for the hanging mass and the rotating mass. Two equations, please.

 

[nazmulhasanshipon]

The string is something that you and I can see and touch. I cannot see or touch "the rotation which provides centrifugal", can you? If not, then you should not consider it in the free body diagram drawn in an inertial frame. Fictitious forces get their name because they don't exist in inertial free body diagrams but appear in non-inertial frames. They are not forces generated by things that you can see and touch; they are mass times acceleration which is moved over to the "other side" of Newton's second law and just looks like a force.

The time has come for you to write Newton's second law for the hanging mass and the rotating mass. Two equations, please.

I think for the hanging mass there is gravitational force which is $F_g=m_1g$ and the tension the string provides is $T$ but I don't know how to write $T$ in equation just yet. For the rotating mass, I think the centripetal force is provided by the tension of the string which means $$T=\frac{m_2v^{2}}{r}$$ but I don't know what balances this force. My brain tells me centrifugal force balances it.

 

[kuruman]

Please tell your brain to start thinking. Where is it written that the force on the rotating mass needs to be balanced? What does Newton's second law have to say on the matter? Remember, you are doing this in the inertial frame.

 

[nazmulhasanshipon]

Please tell your brain to start thinking. Where is it written that the force on the rotating mass needs to be balanced? What does Newton's second law have to say on the matter? Remember, you are doing this in the inertial frame.

@kuruman sir, I don't know what Newton's second law says about this. Would you please tell me?

 

[kuruman]

@kuruman sir, I don't know what Newton's second law says about this. Would you please tell me?

Do you know what Newton's second law says in general, not necessarily about this?

 

[nazmulhasanshipon]

@kuruman sir, The second law states that the rate of change of momentum of an object is directly proportional to the force applied.

 

[kuruman]

@kuruman sir, The second law states that the rate of change of momentum of an object is directly proportional to the force applied.

Have you ever solved any problems requiring the use of the second law? If so, is that what you used?

 

[nazmulhasanshipon]

Do you know what Newton's second law says in general, not necessarily about this?

And @kuruman sir one of my seniors was saying that if the table isn't frictionless then at first the rotating mass would try to get away and feel frictional force in the inward direction then $$T+f_k-\frac{m_2v^2}{r}=0$$ would be true for the rotating body. He also added after sometime it will try to go the inward direction towards center so $$T-f_k-\frac{m_2v^2}{r}=0$$ would be true. So, if you were given $m_2, T, f_k$ and $r$ then you will find the speed of the rotating body $v$ ranges between some value. But I disagree, if something like this happen then the radius vector $r$ will not be a constant. I mean how can there be friction in the radial direction if $r$ is constant. What will you say about it sir?

 

[nazmulhasanshipon]

Have you ever solved any problems requiring the use of the second law? If so, is that what you used?

I used $$\sum \vec{F}=m \vec{a}$$

 

[kuruman]

I used $$\sum \vec{F}=m \vec{a}$$

Very good. That's what I was looking for. Now answer this. What is $\Sigma \vec F$ for the rotating mass?

 

[nazmulhasanshipon]

Very good. That's what I was looking for. Now answer this. What is $\Sigma \vec F$ for the rotating mass?

I can imagine if the table is frictionless then at first someone implied some toque and made it to rotate and then I can see the string provides the centripetal force.

 

[nazmulhasanshipon]

Very good. That's what I was looking for. Now answer this. What is $\Sigma \vec F$ for the rotating mass?

And also sir, can you please recommend me some books where I can learn extensively about motion like Fundamentals of Physics by David Halliday?

 

[kuruman]

And also sir, can you please recommend me some books where I can learn extensively about motion like Fundamentals of Physics by David Halliday?

That's as good as any and you already have the link. I have used it both as a student and as an instructor. So what is $\Sigma \vec F$ for the rotating mass?

 

[kuruman]

And also sir, can you please recommend me some books where I can learn extensively about motion like Fundamentals of Physics by David Halliday?

That's as good as any and you already have the link. I have used it both as a student and as an instructor. So what is $\Sigma \vec F$ for the rotating mass?

 

[nazmulhasanshipon]

That's as good as any and you already have the link. I have used it both as a student and as an instructor. So what is $\Sigma \vec F$ for the rotating mass?

I think only centripetal force provided by the tension.

 

[haruspex]

I think only centripetal force provided by the tension.

Yes, $\Sigma \vec F$ is just the tension.

 

[kuruman]

I think only centripetal force provided by the tension.

Right. Now, what is $m\vec a$ that goes on the other side of the equation for Newton's second law?

 

[nazmulhasanshipon]

@kuruman, I think it's the centripetal force $\frac{mv^2}{r}$.

 

[haruspex]

@kuruman, I think it's the centripetal force $\frac{mv^2}{r}$.

Ok, so what is the equation you get?

 

[nazmulhasanshipon]

Ok, so what is the equation you get?

is it $F=\cfrac{mv^2}{r}$ @haruspex ?

 

[haruspex]

is it $F=\cfrac{mv^2}{r}$ @haruspex ?

Right, but what else is that F equal to? Look at your post #52.

 

[kuruman]

... but I don't know what balances this force. My brain tells me centrifugal force balances it.

Excuse me for interjecting, but I have to complete this train of thought.

To: @nazmulhasanshipon: As you can see from this equation, the tension acting on the rotating mass is unbalanced. That's OK because it means that the mass has to accelerate and indeed it does. On the right side of the equation, you have $\dfrac{mv^2}{r}$ which is mass $m$ times acceleration $\dfrac{v^2}{r}$ according to Newton's second law. You do not have a centrifugal force "balancing" the tension.

Carry on.

 

[haruspex]

I don't know what balances this force. My brain tells me centrifugal force balances it.

You need to pick a reference frame and stick to it. You are confusing yourself by switching between two.

In the ground frame, being an inertial frame, there is no centrifugal force. The tension exerts a force which is not balanced, so it produces the net force and this results in an acceleration. We can choose to resolve the acceleration of a moving object into a component parallel to its velocity and a component normal to its velocity. The component normal to the velocity is known as the centripetal acceleration, and the component of the net force normal to the velocity is known as the centripetal force. If the mass is moving at constant speed then there is no component parallel to the velocity.

In you choose a reference frame rotating about the hole at the same rate as the mass then in this frame there is no acceleration. In order to explain that we need to balance the tension. We do this by inventing a centrifugal force that results from our choice of frame. So, yes, in this choice of frame the centrifugal force balances the tension.

Note that we can have both centripetal and centrifugal forces! This happens if we choose a rotating frame which does not get rid of the component of acceleration normal to the velocity. E.g. we could choose a frame rotating about the hole but at a different rate from the mass.
If the mass is rotating at angular rate $\omega$ in the ground frame and the rotating frame rotates at rate $\omega'$ then the apparent rotation rate of the mass in the rotating frame is $\omega-\omega'$. The centrifugal acceleration for the rotating frame is $r^2\omega'$ and the observed acceleration of the mass in that reference frame is $r^2(\omega-\omega')$.
Taking towards the hole as the positive direction, the F=ma equation in the rotating frame becomes
Net force = real applied force + centrifugal force $= T-mr^2\omega'$
= mass * apparent acceleration $= mr^2(\omega-\omega')$.
Simplifying, $T=mr^2\omega$.