Finding out the rotational speed of a 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, $T=mr^2\omega$.