Cubical block on a cylinder

  • Thread starter Saitama
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  • #51
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$$v_y=\frac{dh}{dt}$$

$$v_x=\frac{d}{dt}\left(\frac{L}{2}\sin\theta\right)$$

where ##v_x## is the horizontal velocity of block and ##v_y## is the vertical velocity of block.

I can replace ##\omega## with ##d\theta/dt##.

Am I correct?

By the way, if you think that the x-position of the CoM is ## \frac{L}{2}\sin\theta ##, you are mistaken (this is obvious from the vector formula).
 
  • #52
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By the way, if you think that the x-position of the CoM is ## \frac{L}{2}\sin\theta ##, you are mistaken

Yes, you are right, very sorry. I got the following x-position:
$$\left(R+\frac{L}{2}\right)\sin\theta-R\theta\cos\theta$$
This gives the same answer as yours.

(this is obvious from the vector formula).
Can you please tell me how do you find the horizontal and vertical components?

Are the following correct?

$$\vec u=\sin\theta \hat{i}+\cos\theta \hat{j}$$
$$\vec t=\cos\theta \hat{i}-\sin\theta \hat{j}$$
 
  • #53
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Are the following correct?

$$\vec u=\sin\theta \hat{i}+\cos\theta \hat{j}$$
$$\vec t=\cos\theta \hat{i}-\sin\theta \hat{j}$$

Yes. Note that ## \vec t = \dot {\vec u} = - \dot {\vec t}##, which I used above.

Have you found the period of small oscillations?
 
  • #54
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Have you found the period of small oscillations?

Here is my attempt at it:

$$E=\frac{1}{2}mv^2+\frac{1}{2}I_{CM}\omega^2+U(\theta)$$
$$\Rightarrow E=\frac{1}{2}m\dot{\theta}^2\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM}\dot{\theta}^2+U(\theta)$$

Since ##I_{CM}=mL^2/6##

$$\Rightarrow E=\frac{1}{8}mL^2\dot{\theta}^2+\frac{1}{2}mR^2\theta^2\dot{\theta}^2+ \frac {1}{12}mL^2\dot{\theta}^2+U(\theta)=\frac{5}{24}mL^2\dot{\theta}^2+ \frac {1}{2}mR^2\theta^2\dot{\theta}^2+U(\theta)$$

Differentiating wrt time,
$$\frac{dE}{dt}=\frac{5}{24}mL^2(2 \theta \dot{\theta})+\frac{1}{2}mR^2(2\theta\dot{\theta}^3+2\theta^2 \dot{ \theta }\ddot{\theta})+U'(\theta)$$
Using the small angle approximation, I neglect the terms involving ##\theta^2## and
$$U'(\theta)=mg\left(R\theta\cos\theta-\frac{L}{2}\sin\theta\right)\dot{\theta} \approx mg\left(R \theta \left(1-\frac{\theta^2}{2}\right)-\frac{L}{2}\theta\right)\dot{\theta}$$
$$\Rightarrow U'(\theta)\approx mg\left(R-\frac{L}{2}\right)\dot{\theta}$$
I feel I am doing it wrong. I have a ##\dot{\theta}^3## when I differentiate the energy equation which I think is wrong. :(
 
  • #55
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The usual approach is this: form the exact equation for conservation of energy: KE + PE = KE(0) + PE(0). Use the small angle approx at this stage. KE has the form ## K(\theta) \dot \theta^2/2##. It is approximated with the lowest non vanishing order, which is simply ## K(0) \dot \theta^2/2 ##. PE is expanded as ## U(0) + U''(0) \theta^2 / 2 + ... ## We need more than just the lowest term of PE, because this will be differentiated later. So we end up with ## K(0) \dot \theta^2 + U''(0) \theta^2 /2 = K(0) \dot \theta (0) ##. Note ##U''## is differentiated with respect to its argument, not time (time diff is denoted by dots).

Now differentiate w.r.t. time: ## K(0)\ddot \theta + U''(0) \theta = 0 ##.
 
  • #56
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KE has the form ## K(\theta) \dot \theta^2/2##. It is approximated with the lowest non vanishing order, which is simply ## K(0) \dot \theta^2/2 ##.

Sorry voko, I can't follow this. I understand what you did to PE, can you please elaborate this one?

Thank you for being so patient!
 
  • #57
BruceW
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Is ##K(0)## like an 'effective moment of inertia', evaluated at ##\theta=0## ?
 
  • #58
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In #54, kinetic energy was shown to be ## \frac{1}{2}m\dot{\theta}^2\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM}\dot{\theta}^2 ##. This can be written as ## \left[\frac{1}{2}m\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM}\right]\dot{\theta}^2 = K(\theta)\dot \theta^2##, so ## K(\theta) = \frac{1}{2}m\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM} ##.

Now we approximate ## K(\theta) ## with just ## K(0) ##.
 
  • #59
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Sorry for not being clear.

In #54, kinetic energy was shown to be ## \frac{1}{2}m\dot{\theta}^2\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM}\dot{\theta}^2 ##. This can be written as ## \left[\frac{1}{2}m\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM}\right]\dot{\theta}^2 = K(\theta)\dot \theta^2##, so ## K(\theta) = \frac{1}{2}m\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM} ##.
I already understood this but what I don't get is the following:
Now we approximate ## K(\theta) ## with just ## K(0) ##.
You approximate using Taylor expansion at ##\theta=0##, right?

$$K(0+\theta)=K(0)+K'(0)\theta+K''(0)\frac{\theta^2}{2}+....$$
Why don't you take K'(0) while approximating? :confused:
 
  • #60
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Yes, the Taylor expansion is used for the approximation.

For both KE and PE we take the lowest-order terms, except that in PE we "ignore" the constant term.

We "ignore" for two reasons.

First, potential energy is defined up to an additive constant, so the constant term is meaningless physically.

Second, unlike the full expression for kinetic energy, there is no further multiplication with something varying, so if we retained only the constant term of PE, it would occur on both sides of the equation. Then it can be cancelled out, leaving no trace of PE in the equation at all (this is the mathematical consequence of the physical meaninglessness of PE's constant term).

As to why we take only the lowest terms, this is again for two reasons.

First, we want to obtain a linear equation in the end. This is probably the main reason for all the shortcuts.

Second, higher-order terms for small angles are much smaller. For example, 10 degree corresponds to 0.17 radian. That thing squared is 0.03 radian, almost an order of magnitude difference. This justifies our desire to neglect anything but the convenient terms :)
 
  • #61
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I have checked what Kleppner has to say on small oscillations, and found that lacking. The form of the kinetic energy they have has no dependency on the coordinates, only on velocities, so there is no demonstration how to proceed in a case like you have here, even though it is taken from that same book.

Symon and Goldstein do a far better job at explaining that; Landau is also noteworthy.
 
  • #62
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Thank you very much voko for such nice explanation. :smile:

I had
$$U(\theta)=mg((R+L/2)\cos\theta+R\theta\sin\theta)$$
$$\Rightarrow U''(\theta)=mg((R+L/2)(-\cos\theta)+R\cos\theta+R\cos\theta-R\theta\sin\theta)$$
For ##\theta=0##,
$$U''(0)=mg\left(R-\frac{L}{2}\right)$$
Since
$$K(\theta) = \frac{1}{2}m\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM} \Rightarrow K(0)=\frac{1}{8}mL^2+\frac{1}{2}I_{CM}$$
Substituting ##I_{CM}=mL^2/6##,
$$K(0)=\frac{5}{24}mL^2$$
As you stated before, from energy conservation we have,
$$K(0)\dot{\theta}^2+U''(0)\theta^2/2=K(0)\dot{\theta}(0)$$
Differentiating wrt time,
$$K(0)(2\dot{\theta}\ddot{\theta})+U''(0)\theta\dot{\theta}=0$$
$$\Rightarrow \ddot{\theta}=-\frac{U''(0)}{2K(0)}\theta$$
Substituting the values,
$$\ddot{\theta}=-\frac{12g}{5L^2}\left(R-\frac{L}{2}\right)\theta$$
The time period can be easily calculated from the above equation.

Is the above correct?

I have checked what Kleppner has to say on small oscillations, and found that lacking. The form of the kinetic energy they have has no dependency on the coordinates, only on velocities, so there is no demonstration how to proceed in a case like you have here, even though it is taken from that same book.

Symon and Goldstein do a far better job at explaining that; Landau is also noteworthy.
I will have a look at them soon if they are available in my country. :)
 
  • #63
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Looking good!
 
  • #64
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Looking good!

Thanks a lot voko! :smile:

That was quite long for a single line problem. :tongue:
 
  • #65
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You did a lot more than the problem required.
 
  • #66
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You did a lot more than the problem required.

Yes, but it was a great learning experience. :)
 

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