Physics...

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

OK, so I have a really specific question. Whenever we study circular motion and aspects of circular motion, I see some solutions including the extra rotational kinetic energy when applying the work-energy theorem and some don't do that. When is it OK to not include the rotational kinetic energy?
I'll do a followup question as an example to explain why I'm confused here.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

If I remember correctly, work-energy theorem is like universal... it works for all kinds of forces; It doesn't matther if the forces are conservative or not. Conservation of energy requires the forces to be conservative, though.
Could you provide examples to contrast where rotational kinetic energy is included/excluded in work energy equation
This isn't the real question yet, but here's where the confusion arises.|dw:1443076050348:dw|\[mgL = \frac{1}{2}mv^2\tag{1} \]\[mgL = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mL^2 \frac{v^2}{L^2} = mv^2\tag{2}\]I think \(\tag 1\) is wrong and \(\tag 2\) is right.
|dw:1443076205726:dw|
Wow, sorry about the formatting in that post. I think that 1 is wrong and 2 is right.
Work done "by a force" = change in kinetic energy "due to that force"
Oh, so what exactly is happening here? Is it not the gravitational force that is also assisting it to gain the rotational kinetic energy?
|dw:1443076331186:dw|
Is that thing spinning and falling freely under gravitation ?
Arrey... I meant conservation of energy, not the work-energy theorem. It's nothing, really. A particle is just fixed at the end of a massless rod/string and it starts to go down. Its path is circular.
Yes.
|dw:1443076456909:dw|
\[E_i = E_f\]\[mgL = \frac{1}{2}mv_f^2\]or\[mgL = mv_f^2\]Which one do you think is correct? The second one also includes rotational kinetic energy.
This is like pendulum/SHO right ?
SHO = simple harmonic oscillation
Yes, I drew that arrow in the diagram too...
Yup.
Kind of like how a gransfather clockworks.
I think the problem is these are both two optional ways of representing the same thing:\[\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2 \]
good question, the situation as same as that of earth-sun system : earth spins and orbits around the sun
Exactly, Empty. I see that some solutions add both rotational and translational kinetic energies, and others say that they're the same thing.
I personally feel that you should be adding those two. They're not the same, and are two components of rotation.
do u want to apply this kinda concept - \[\omega _{f}^2 =\omega_{i}^2 +2\alpha \theta\]?
Ok, show me an example where they have used both like this, so we can determine what's going on. For this problem you've given, it's clear that only one is necessary because there's only one degree of freedom here.
|dw:1443076723125:dw|
helps if you needed to understand the relationships I guess? Haha.
I think so, if there is a rotation, total kinetic energy equals the sum of translational KE and rotational KE
OK, so (2) is right, right?
but with pendulums, we assume the balls are not spinning
depends, if you assume the ball is spinning, add \(\frac{1}{2}I\omega^2\) otherwise forget it
The rotational component isn't about the balls spinning, it's about the ball rotating about the hinge point.
Does circular motion qualify as rotational motion? Great, I've reduced this conversation to a single question.
that is accounted as translational kinetic energy along circumferencce of circle, right ?
That's the point - it's not...
Or is it?
it depends upon the motion of system... but if the particle is jst doing circular motion then we cant say its undergoing rotational motion too..
|dw:1443076982465:dw| the circular movement would just be applied to the pulley/lever thing holding it, but not to the ball itself, that's what I think..
Try doing this question, for example. Find the value of \(\theta\). Here, we have a rod instead of a point mass. If we do not account for the rotational kinetic energy, we'll be in trouble (you'll get \(\cos \theta > 1\) if I recall corrrectly). If we account for the rod's rotation, why not a point mass'? |dw:1443077039882:dw|
A rotating ball has a speed, so it clearly has a momentum and kinetic energy. Since it is in circular motion, it is under the presense of centripetal acceleration which always points "perpendicular" to the direction of motion. So you can forget about the rotational part of the kinetic energy.
For your question, you're just writing the same energy twice, so (1) is correct.\[\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2 \]
If the force is perpendicular to the motion, it does no work. so you can forget about it
This is pointless to talk about unless you show us a real problem where you see both arising, then we can explain why both arise.
This is what you guys suggest. Oh, and also, \(L = 25 ~cm\).\[mg\left(\frac{L}{2} + \frac{L}{2}\cos \theta \right)=\frac{1}{2}mv^2 \]
\[gL (1 + \cos \theta) = v^2\]\[\Rightarrow 9 = 10\cdot 0.25 \cdot (1 + \cos \theta) \]Lo and behold...
But there has to be a theta, right? According to this, there is none.
|dw:1443077457196:dw| lmao... it's an easy one I guess.
\[mg\left(L + \frac{L}{2}\cos \theta \right) = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\]\[3mgL(1 + \cos \theta) = mv^2 + \frac{mL^2}{3}\cdot \frac{v^2}{L^2} = \frac{4mv^2}{3}\]\[\Rightarrow 3\cdot 10 \cdot 0.25 (1 + \cos \theta) = 4\cdot 9/3\]
Forgot something in it
|dw:1443077907086:dw|
are you taking 0 for potential energy at L/2 ?
Rotational and translational kinetic energies are not the same. They just turn out to be the same here because the moment of inertia is mL^2, which could have been anything.
Yeah, and I'm concentrating the entire system at its center of mass which is legal in physics.
|dw:1443078285888:dw|
It's been so many years, let me try and catch up on these..
Oh, great catch here... the 3m/s is for the bottom-most point, not the middle one. For the middle one, it's should be 3/2. Sorry 'bout that.
potential energy = \(mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos\phi\right) \) |dw:1443078517457:dw|
Fixed:\[mgL/2 (1 + \cos \theta) = 1/2 m(3/2)^2 + 1/2 \cdot mL^2/3 \cdot 3^2/L^2 \]
That looks ugly.\[10\cdot 0.25 \cdot (1 + \cos \theta) = 9/8 + 3\]\[1 + \cos \theta = 33/(8\cdot2.5) = 33/20 \]
Still dunno.
Something doesn't feel right about this. Ugh.
\(\phi =\theta+90 \) so shouldn't the potential energy be \[mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos\phi\right) = mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos(\theta+90)\right)\\~\\= mg\left(\dfrac{L}{2} +\dfrac{L}{2}\color{red}{\sin}\theta\right)\] ?
Absolutely. Sorry, I was too busy worrying about the other aspect of the problem that I actually forgot to solve it correctly.
That gives \(\sin \theta = 0.65\)
Ok so we're good with potential energy let me try and understand whats going with rotational ke
are you sure the moment of inertia of that rod is \(\dfrac{mL^2}{3}\) ?
If you try to solve it like this...\[\rm mgL/2 (1 + \cos \theta) = \frac{1}{2}mv^2\]then \(v = 3/2 \), \(L = 0.25 m\) and so on.
Yes.
also the speed of center of mass will not be 3 m/s it should be way less than that
Yeah, I wrote that correction above. It's 3/2 and not 3.
And I wrote \(\cos \theta\) again <_<\[10\cdot 0.25 \cdot (1 + \sin \theta) = 9/4\]\[\Rightarrow 1 + \sin \theta = 9/10\]
And so \(\sin \theta = -1/10\) Which is invalid.
So we definitely have to add rotational kinetic energy to that!
yes replace \(m\) by \(I\) and \(v\) by \(\omega\)
so is \(mgL = mv^2\) correct in my first question?
Total mechanical energy, \(mgL/2(1+\sin\theta) + \frac{1}{2} I\omega^2\), is conserved
\[mgL/2(1 + \sin \theta) = \frac{1}{2}I \omega^2 + \frac{1}{2}mv^2\]
Another question: if I take the whole rod to be concentrated at the center of mass, I still take \(I \) to be the moment of inertia of the whole rod, right? And not just the point-mass?
|dw:1443079851015:dw|
the object is rotating about that point A, so the moment of inertia is about that point A. it wont change right ?
see, im rusty on these, im just going by intuition, i can be wrong...
Yup. And \(I = mL^2/3\).
so it is the property of that rotating rod about that point A it should not change
Yes, indeed.
\[mgL/2 ( 1 + \sin \theta ) = \frac{1}{2}m\left(3/2\right)^2 + \frac{1}{2}\frac{mL^2}{3} \left(\frac{3^2}{L^2}\right) \]
\[gL ( 1 + \sin \theta ) = 9/4 + 3 = 21/4\]\[\Rightarrow 1 + \sin \theta = 21/4\cdot 10\cdot 0.25 = 21/10\]\[\sin \theta = 11/10\]JEEEEEZUZ!
wait, what if this problem is wrong?
let's reverse the thing and find the maximum velocity at the bottom
\[mgL = \frac{1}{2}mv_{CM}^2 + \frac{1}{2}I\omega^2 \] (for max. velocity)\[2mgL =mv_{CM}^2 + mL^2/3 \cdot (v_{CM}/2)^2/ (L/2)^2 \]
LMAO!! We just ignored @Jhannybean
\[2mgL = 4mv_{CM}^2/3\]\[3gL/2 = v_{CM}^{2}\]
\[v_{bottom}^2 = 4 v^2_{CM} = 6gL = 6\cdot 10 \cdot 0.25 = \cdots\]So I guess we're fine.
\(3g(1+\sin\theta) =L\omega^2\)
What's that? Just the rotational kinetic energy?
So \(v_{bottom} = \sqrt{15} > 3\)
Yeah, that's a good idea... what if we just include the rotational kinetic energy?
\[3gL/2( 1 + \sin \theta) = v_{CM}^2 = 9/4\]\[ gL(1 + \sin \theta ) = 3/2 \]\[ 1 + \sin \theta = 3/5\]Oh lol, again...
So absolutely nothing works at this point. =_=_=_=_=_=
Here is my best attempt : Total energy when the rod is at angle \(\theta\) is given by \(\dfrac{1}{2}I\omega^2 + mg\frac{L}{2}(1+\sin\theta) \) since the rod hasn't started rotating, \(\omega = 0\), so the total energy becomes \(mg\frac{L}{2}(1+\sin\theta)\tag{1} \) Total energy when the rod is at its final position is given by \(\dfrac{1}{2}I\omega^2 +mg*0 \tag{2}\) since the forces are conservative, the total energy is conserved : \[mg\frac{L}{2}(1+\sin\theta)=\dfrac{1}{2}I\omega^2\] we can solve \(\theta\)
Could you solve the rest? I just am not able to.
plugin \(\omega = \dfrac{v}{L} =\dfrac{3}{25} \)
I'm getting \[1+\sin\theta = \dfrac{9}{125g}\]
so that's wrong, isn't it?
not so sure http://www.wolframalpha.com/input/?i=solve+1%2Bsinx+%3D+9%2F%28125*10%29%2C0%3Cx%3C2pi
But 0
To answer the problem of whether the rotional energy should be applied to the linear motion energy, we should go into the point where the basic concepts of the rational parameters are defined. As Ive learned, the moment of inertia was defined as following, |dw:1443082442329:dw| So as the diagram shows the object (placed horizontally so that theres no potential energy changes when rotate) rotates around the O axis and we can take small masses \(\delta m_i\) on it at a distance of \(r\). When the object starts to rotate that small mass will have v velocity . So the kinetic energy the small mass will have is, \(\delta E _i= \frac{1}{2}\delta m_iv_i^2 \) with \(v_i = r_i \omega\) \(\delta E_i = \frac{1}{2}\delta m_ir_i^2\omega^2 \) Since we need the kinetic energy of the whole object which is rotating, \[\begin{align*} E &= \sum \limits_{i=0}^{i=N} E_i \\ &= \sum \limits_{i=0}^{i=N} \delta m_ir_i^2\omega^2\\ &=\omega^2 \underbrace{\sum \limits_{i=0}^{i=N} \delta m_ir_i^2}_{I}\\ &= I \omega^2 \end{align*}\] So I is taken as that large sum. Whats important to see here is that eventually the rotational energy is taken from the linear kinetic energy. So at the end of the day they are both the same.So There is no need to consider a rotational kintic energy if we take linear kinetic energy
sorry missed the 1/2 in the middle for the E, it should be \(E= \frac 1 2 I \omega^2\) :(
Amazing. I wonder why I forgot that.
Thank you so very much!
It a pleasure ;)
I mean you didn't really need to type all of that, but thanks!
But we do need to add the linear component in plane motion.
Yup, so the verdict: - In pure rotation, rotational kinetic energy is the kinetic energy. - In pure translational, translational kinetic energy is the kinetic energy. - In plane motion (a superposition of the two) we have to add both.
That's great, I wonder why I totally forgot the origin. Sometimes you just have to look at the basics. =_=
umm what do you mean by the plane motion ?
Nothing, just rotational+translational motion.
Ok. Btw to which group do you take this original question?
?
I mean to which group among the three catagories does the scenario - "particle rotating around a point " falls
Oh, it's pure rotation. I got it. :)
Basically in plane motion, the axis of rotation is translating.
you mean the third catagory where you should sum the linear motion energy and the ration energy?
Exactly. But here, the axis is stationary and the original question is a kind of pure rotation.

Not the answer you are looking for?

Search for more explanations.

Ask your own question