## ParthKohli one year ago Physics...

1. ParthKohli

@imqwerty

2. ParthKohli

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?

3. ParthKohli

I'll do a followup question as an example to explain why I'm confused here.

4. ganeshie8

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.

5. ganeshie8

Could you provide examples to contrast where rotational kinetic energy is included/excluded in work energy equation

6. ParthKohli

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.

7. ParthKohli

|dw:1443076205726:dw|

8. ParthKohli

Wow, sorry about the formatting in that post. I think that 1 is wrong and 2 is right.

9. ganeshie8

Work done "by a force" = change in kinetic energy "due to that force"

10. ParthKohli

Oh, so what exactly is happening here? Is it not the gravitational force that is also assisting it to gain the rotational kinetic energy?

11. ganeshie8

|dw:1443076331186:dw|

12. ganeshie8

Is that thing spinning and falling freely under gravitation ?

13. ParthKohli

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.

14. ParthKohli

Yes.

15. ganeshie8

|dw:1443076456909:dw|

16. ParthKohli

$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.

17. ganeshie8

This is like pendulum/SHO right ?

18. ganeshie8

SHO = simple harmonic oscillation

19. ParthKohli

Yes, I drew that arrow in the diagram too...

20. ParthKohli

Yup.

21. anonymous

Kind of like how a gransfather clockworks.

22. Empty

I think the problem is these are both two optional ways of representing the same thing:$\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2$

23. ganeshie8

good question, the situation as same as that of earth-sun system : earth spins and orbits around the sun

24. ParthKohli

Exactly, Empty. I see that some solutions add both rotational and translational kinetic energies, and others say that they're the same thing.

25. ParthKohli

I personally feel that you should be adding those two. They're not the same, and are two components of rotation.

26. imqwerty

do u want to apply this kinda concept - $\omega _{f}^2 =\omega_{i}^2 +2\alpha \theta$?

27. Empty

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.

28. anonymous

|dw:1443076723125:dw|

29. anonymous

helps if you needed to understand the relationships I guess? Haha.

30. ganeshie8

I think so, if there is a rotation, total kinetic energy equals the sum of translational KE and rotational KE

31. ParthKohli

OK, so (2) is right, right?

32. ganeshie8

but with pendulums, we assume the balls are not spinning

33. ganeshie8

depends, if you assume the ball is spinning, add $$\frac{1}{2}I\omega^2$$ otherwise forget it

34. ParthKohli

The rotational component isn't about the balls spinning, it's about the ball rotating about the hinge point.

35. ParthKohli

Does circular motion qualify as rotational motion? Great, I've reduced this conversation to a single question.

36. ganeshie8

that is accounted as translational kinetic energy along circumferencce of circle, right ?

37. ParthKohli

That's the point - it's not...

38. ParthKohli

Or is it?

39. imqwerty

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..

40. anonymous

|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..

41. ParthKohli

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|

42. ganeshie8

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.

43. Empty

For your question, you're just writing the same energy twice, so (1) is correct.$\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2$

44. ganeshie8

If the force is perpendicular to the motion, it does no work. so you can forget about it

45. Empty

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.

46. ParthKohli

This is what you guys suggest. Oh, and also, $$L = 25 ~cm$$.$mg\left(\frac{L}{2} + \frac{L}{2}\cos \theta \right)=\frac{1}{2}mv^2$

47. ParthKohli

$gL (1 + \cos \theta) = v^2$$\Rightarrow 9 = 10\cdot 0.25 \cdot (1 + \cos \theta)$Lo and behold...

48. ParthKohli

But there has to be a theta, right? According to this, there is none.

49. anonymous

|dw:1443077457196:dw| lmao... it's an easy one I guess.

50. ParthKohli

$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$

51. anonymous

Forgot something in it

52. anonymous

|dw:1443077907086:dw|

53. ganeshie8

are you taking 0 for potential energy at L/2 ?

54. ParthKohli

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.

55. ParthKohli

Yeah, and I'm concentrating the entire system at its center of mass which is legal in physics.

56. ganeshie8

|dw:1443078285888:dw|

57. ganeshie8

It's been so many years, let me try and catch up on these..

58. ParthKohli

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.

59. ganeshie8

potential energy = $$mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos\phi\right)$$ |dw:1443078517457:dw|

60. ParthKohli

Fixed:$mgL/2 (1 + \cos \theta) = 1/2 m(3/2)^2 + 1/2 \cdot mL^2/3 \cdot 3^2/L^2$

61. ParthKohli

That looks ugly.$10\cdot 0.25 \cdot (1 + \cos \theta) = 9/8 + 3$$1 + \cos \theta = 33/(8\cdot2.5) = 33/20$

62. ParthKohli

Still dunno.

63. ParthKohli

64. ganeshie8

$$\phi =\theta+90$$ so shouldn't the potential energy be $mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos\phi\right) = mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos(\theta+90)\right)\\~\\= mg\left(\dfrac{L}{2} +\dfrac{L}{2}\color{red}{\sin}\theta\right)$ ?

65. ParthKohli

Absolutely. Sorry, I was too busy worrying about the other aspect of the problem that I actually forgot to solve it correctly.

66. ParthKohli

That gives $$\sin \theta = 0.65$$

67. ganeshie8

Ok so we're good with potential energy let me try and understand whats going with rotational ke

68. ganeshie8

are you sure the moment of inertia of that rod is $$\dfrac{mL^2}{3}$$ ?

69. ParthKohli

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.

70. ParthKohli

Yes.

71. ganeshie8

also the speed of center of mass will not be 3 m/s it should be way less than that

72. ParthKohli

Yeah, I wrote that correction above. It's 3/2 and not 3.

73. ParthKohli

And I wrote $$\cos \theta$$ again <_<$10\cdot 0.25 \cdot (1 + \sin \theta) = 9/4$$\Rightarrow 1 + \sin \theta = 9/10$

74. ParthKohli

And so $$\sin \theta = -1/10$$ Which is invalid.

75. ParthKohli

So we definitely have to add rotational kinetic energy to that!

76. ganeshie8

yes replace $$m$$ by $$I$$ and $$v$$ by $$\omega$$

77. ParthKohli

so is $$mgL = mv^2$$ correct in my first question?

78. ganeshie8

Total mechanical energy, $$mgL/2(1+\sin\theta) + \frac{1}{2} I\omega^2$$, is conserved

79. ParthKohli

$mgL/2(1 + \sin \theta) = \frac{1}{2}I \omega^2 + \frac{1}{2}mv^2$

80. ParthKohli

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?

81. ganeshie8

|dw:1443079851015:dw|

82. ganeshie8

the object is rotating about that point A, so the moment of inertia is about that point A. it wont change right ?

83. ganeshie8

see, im rusty on these, im just going by intuition, i can be wrong...

84. ParthKohli

Yup. And $$I = mL^2/3$$.

85. ganeshie8

so it is the property of that rotating rod about that point A it should not change

86. ParthKohli

Yes, indeed.

87. ParthKohli

$mgL/2 ( 1 + \sin \theta ) = \frac{1}{2}m\left(3/2\right)^2 + \frac{1}{2}\frac{mL^2}{3} \left(\frac{3^2}{L^2}\right)$

88. ParthKohli

$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!

89. ParthKohli

wait, what if this problem is wrong?

90. ParthKohli

let's reverse the thing and find the maximum velocity at the bottom

91. ParthKohli

$mgL = \frac{1}{2}mv_{CM}^2 + \frac{1}{2}I\omega^2$ (for max. velocity)$2mgL =mv_{CM}^2 + mL^2/3 \cdot (v_{CM}/2)^2/ (L/2)^2$

LMAO!! We just ignored @Jhannybean

93. ParthKohli

$2mgL = 4mv_{CM}^2/3$$3gL/2 = v_{CM}^{2}$

94. ParthKohli

$v_{bottom}^2 = 4 v^2_{CM} = 6gL = 6\cdot 10 \cdot 0.25 = \cdots$So I guess we're fine.

95. ganeshie8

$$3g(1+\sin\theta) =L\omega^2$$

96. ParthKohli

What's that? Just the rotational kinetic energy?

97. ParthKohli

So $$v_{bottom} = \sqrt{15} > 3$$

98. ParthKohli

Yeah, that's a good idea... what if we just include the rotational kinetic energy?

99. ParthKohli

$3gL/2( 1 + \sin \theta) = v_{CM}^2 = 9/4$$gL(1 + \sin \theta ) = 3/2$$1 + \sin \theta = 3/5$Oh lol, again...

100. ParthKohli

So absolutely nothing works at this point. =_=_=_=_=_=

101. ganeshie8

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 $$\theta$$

102. ParthKohli

Could you solve the rest? I just am not able to.

103. ganeshie8

plugin $$\omega = \dfrac{v}{L} =\dfrac{3}{25}$$

104. ganeshie8

I'm getting $1+\sin\theta = \dfrac{9}{125g}$

105. ParthKohli

so that's wrong, isn't it?

106. ganeshie8
107. ParthKohli

But 0<x<pi/2

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 $$E= \frac 1 2 I \omega^2$$ :(

110. ParthKohli

Amazing. I wonder why I forgot that.

111. ParthKohli

Thank you so very much!

It a pleasure ;)

113. ParthKohli

I mean you didn't really need to type all of that, but thanks!

114. ParthKohli

But we do need to add the linear component in plane motion.

115. ParthKohli

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.

116. ParthKohli

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 ?

118. ParthKohli

Nothing, just rotational+translational motion.

Ok. Btw to which group do you take this original question?

120. ParthKohli

?

I mean to which group among the three catagories does the scenario - "particle rotating around a point " falls

122. ParthKohli

Oh, it's pure rotation. I got it. :)

123. ParthKohli

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?

125. ParthKohli

Exactly. But here, the axis is stationary and the original question is a kind of pure rotation.