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ParthKohli
 one year ago
Physics...
ParthKohli
 one year ago
Physics...

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2OK, 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 workenergy theorem and some don't do that. When is it OK to not include the rotational kinetic energy?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I'll do a followup question as an example to explain why I'm confused here.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3If I remember correctly, workenergy 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Could you provide examples to contrast where rotational kinetic energy is included/excluded in work energy equation

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2This 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443076205726:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Wow, sorry about the formatting in that post. I think that 1 is wrong and 2 is right.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Work done "by a force" = change in kinetic energy "due to that force"

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Oh, so what exactly is happening here? Is it not the gravitational force that is also assisting it to gain the rotational kinetic energy?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443076331186:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Is that thing spinning and falling freely under gravitation ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Arrey... I meant conservation of energy, not the workenergy 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443076456909:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[E_i = E_f\]\[mgL = \frac{1}{2}mv_f^2\]or\[mgL = mv_f^2\]Which one do you think is correct? The second one also includes rotational kinetic energy.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3This is like pendulum/SHO right ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3SHO = simple harmonic oscillation

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes, I drew that arrow in the diagram too...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Kind of like how a gransfather clockworks.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I think the problem is these are both two optional ways of representing the same thing:\[\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2 \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3good question, the situation as same as that of earthsun system : earth spins and orbits around the sun

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Exactly, Empty. I see that some solutions add both rotational and translational kinetic energies, and others say that they're the same thing.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I personally feel that you should be adding those two. They're not the same, and are two components of rotation.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0do u want to apply this kinda concept  \[\omega _{f}^2 =\omega_{i}^2 +2\alpha \theta\]?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ok, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443076723125:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0helps if you needed to understand the relationships I guess? Haha.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think so, if there is a rotation, total kinetic energy equals the sum of translational KE and rotational KE

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2OK, so (2) is right, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3but with pendulums, we assume the balls are not spinning

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3depends, if you assume the ball is spinning, add \(\frac{1}{2}I\omega^2\) otherwise forget it

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2The rotational component isn't about the balls spinning, it's about the ball rotating about the hinge point.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Does circular motion qualify as rotational motion? Great, I've reduced this conversation to a single question.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that is accounted as translational kinetic energy along circumferencce of circle, right ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2That's the point  it's not...

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0it depends upon the motion of system... but if the particle is jst doing circular motion then we cant say its undergoing rotational motion too..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw: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..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Try 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3A 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1For your question, you're just writing the same energy twice, so (1) is correct.\[\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2 \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3If the force is perpendicular to the motion, it does no work. so you can forget about it

Empty
 one year ago
Best ResponseYou've already chosen the best response.1This is pointless to talk about unless you show us a real problem where you see both arising, then we can explain why both arise.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2This is what you guys suggest. Oh, and also, \(L = 25 ~cm\).\[mg\left(\frac{L}{2} + \frac{L}{2}\cos \theta \right)=\frac{1}{2}mv^2 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[gL (1 + \cos \theta) = v^2\]\[\Rightarrow 9 = 10\cdot 0.25 \cdot (1 + \cos \theta) \]Lo and behold...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2But there has to be a theta, right? According to this, there is none.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443077457196:dw lmao... it's an easy one I guess.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[mg\left(L + \frac{L}{2}\cos \theta \right) = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\]\[3mgL(1 + \cos \theta) = mv^2 + \frac{mL^2}{3}\cdot \frac{v^2}{L^2} = \frac{4mv^2}{3}\]\[\Rightarrow 3\cdot 10 \cdot 0.25 (1 + \cos \theta) = 4\cdot 9/3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Forgot something in it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443077907086:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3are you taking 0 for potential energy at L/2 ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Rotational 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, and I'm concentrating the entire system at its center of mass which is legal in physics.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443078285888:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It's been so many years, let me try and catch up on these..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Oh, great catch here... the 3m/s is for the bottommost point, not the middle one. For the middle one, it's should be 3/2. Sorry 'bout that.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3potential energy = \(mg\left(\dfrac{L}{2}  \dfrac{L}{2}\cos\phi\right) \) dw:1443078517457:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Fixed:\[mgL/2 (1 + \cos \theta) = 1/2 m(3/2)^2 + 1/2 \cdot mL^2/3 \cdot 3^2/L^2 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2That looks ugly.\[10\cdot 0.25 \cdot (1 + \cos \theta) = 9/8 + 3\]\[1 + \cos \theta = 33/(8\cdot2.5) = 33/20 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Something doesn't feel right about this. Ugh.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\phi =\theta+90 \) so shouldn't the potential energy be \[mg\left(\dfrac{L}{2}  \dfrac{L}{2}\cos\phi\right) = mg\left(\dfrac{L}{2}  \dfrac{L}{2}\cos(\theta+90)\right)\\~\\= mg\left(\dfrac{L}{2} +\dfrac{L}{2}\color{red}{\sin}\theta\right)\] ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Absolutely. Sorry, I was too busy worrying about the other aspect of the problem that I actually forgot to solve it correctly.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2That gives \(\sin \theta = 0.65\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Ok so we're good with potential energy let me try and understand whats going with rotational ke

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3are you sure the moment of inertia of that rod is \(\dfrac{mL^2}{3}\) ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2If 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3also the speed of center of mass will not be 3 m/s it should be way less than that

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, I wrote that correction above. It's 3/2 and not 3.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2And I wrote \(\cos \theta\) again <_<\[10\cdot 0.25 \cdot (1 + \sin \theta) = 9/4\]\[\Rightarrow 1 + \sin \theta = 9/10\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2And so \(\sin \theta = 1/10\) Which is invalid.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So we definitely have to add rotational kinetic energy to that!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yes replace \(m\) by \(I\) and \(v\) by \(\omega\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2so is \(mgL = mv^2\) correct in my first question?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Total mechanical energy, \(mgL/2(1+\sin\theta) + \frac{1}{2} I\omega^2\), is conserved

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[mgL/2(1 + \sin \theta) = \frac{1}{2}I \omega^2 + \frac{1}{2}mv^2\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Another 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 pointmass?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443079851015:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3the object is rotating about that point A, so the moment of inertia is about that point A. it wont change right ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3see, im rusty on these, im just going by intuition, i can be wrong...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yup. And \(I = mL^2/3\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3so it is the property of that rotating rod about that point A it should not change

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[mgL/2 ( 1 + \sin \theta ) = \frac{1}{2}m\left(3/2\right)^2 + \frac{1}{2}\frac{mL^2}{3} \left(\frac{3^2}{L^2}\right) \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[gL ( 1 + \sin \theta ) = 9/4 + 3 = 21/4\]\[\Rightarrow 1 + \sin \theta = 21/4\cdot 10\cdot 0.25 = 21/10\]\[\sin \theta = 11/10\]JEEEEEZUZ!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2wait, what if this problem is wrong?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2let's reverse the thing and find the maximum velocity at the bottom

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[mgL = \frac{1}{2}mv_{CM}^2 + \frac{1}{2}I\omega^2 \] (for max. velocity)\[2mgL =mv_{CM}^2 + mL^2/3 \cdot (v_{CM}/2)^2/ (L/2)^2 \]

thadyoung
 one year ago
Best ResponseYou've already chosen the best response.0LMAO!! We just ignored @Jhannybean

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[2mgL = 4mv_{CM}^2/3\]\[3gL/2 = v_{CM}^{2}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[v_{bottom}^2 = 4 v^2_{CM} = 6gL = 6\cdot 10 \cdot 0.25 = \cdots\]So I guess we're fine.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(3g(1+\sin\theta) =L\omega^2\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2What's that? Just the rotational kinetic energy?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So \(v_{bottom} = \sqrt{15} > 3\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, that's a good idea... what if we just include the rotational kinetic energy?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[3gL/2( 1 + \sin \theta) = v_{CM}^2 = 9/4\]\[ gL(1 + \sin \theta ) = 3/2 \]\[ 1 + \sin \theta = 3/5\]Oh lol, again...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So absolutely nothing works at this point. =_=_=_=_=_=

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Here 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 \(\theta\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Could you solve the rest? I just am not able to.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3plugin \(\omega = \dfrac{v}{L} =\dfrac{3}{25} \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I'm getting \[1+\sin\theta = \dfrac{9}{125g}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2so that's wrong, isn't it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3not so sure http://www.wolframalpha.com/input/?i=solve+1%2Bsinx+%3D+9%2F%28125*10%29%2C0%3Cx%3C2pi

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2To 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

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2sorry missed the 1/2 in the middle for the E, it should be \(E= \frac 1 2 I \omega^2\) :(

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Amazing. I wonder why I forgot that.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Thank you so very much!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I mean you didn't really need to type all of that, but thanks!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2But we do need to add the linear component in plane motion.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yup, 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2That's great, I wonder why I totally forgot the origin. Sometimes you just have to look at the basics. =_=

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2umm what do you mean by the plane motion ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Nothing, just rotational+translational motion.

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2Ok. Btw to which group do you take this original question?

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2I mean to which group among the three catagories does the scenario  "particle rotating around a point " falls

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Oh, it's pure rotation. I got it. :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Basically in plane motion, the axis of rotation is translating.

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.2you mean the third catagory where you should sum the linear motion energy and the ration energy?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Exactly. But here, the axis is stationary and the original question is a kind of pure rotation.
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