What Is the Maximum Size of a Cube That Can Balance on a Cylinder?

[Saitama]

What are the components of $\vec u$?

$|\vec u| \sin\theta \hat{i}$ in horizontal and $|\vec u| \cos\theta \hat{j}$ in vertical? (where $\hat{i}$ and $\hat{j}$ are unit vectors in horizontal and vertical directions respectively)

 

[gneill]

This is what I get:

$$\frac{L}{2\cos\theta}+\Delta \sin\theta+R\cos\theta$$

Looks good?

You could do that, but it would be simpler to sum the two green line segments, both easily found.

 

[voko]

If you meant to say $\vec u = (\sin \theta, \cos \theta)$, yes. Pretty much by the unit circle trig definitions.

 

[Saitama]

You could do that, but it would be simpler to sum the two green line segments, both easily found.

Oops, you are right, it is simply $(R+L/2)\cos\theta + R\theta \sin \theta$. So the potential energy of block is $mg((R+L/2)\cos\theta + R\theta \sin \theta)$. Should I use the small angle approximation and then differentiate?

 

[gneill]

Oops, you are right, it is simply $(R+L/2)\cos\theta + R\theta \sin \theta$. So the potential energy of block is $mg((R+L/2)\cos\theta + R\theta \sin \theta)$. Should I use the small angle approximation and then differentiate?

Differentiate first. Isolate what you want to solve for, then apply any approximations required. You can also use limits.

 

[voko]

Is there any need to approximate anything? We just need to know the sign of the second derivative at $\theta = 0$.

 

[Saitama]

Differentiate first. Isolate what you want to solve for, then apply any approximations required. You can also use limits.

Let
$$U(\theta)=mg((R+L/2)\cos\theta+R\theta\sin\theta)$$
Differentiating wrt $\theta$
$$U'(\theta)=mg((R+L/2)(-\sin\theta)+R\sin\theta+R\theta\cos\theta)$$
Clearly, U(0)=0
$$U''(\theta)=mg((R+L/2)(-\cos\theta)+R\cos\theta+R\cos\theta-R\theta\sin\theta)$$

Since $U''(0)>0$,
$$-R-\frac{L}{2}+2R>0 \Rightarrow L>2R$$

Is this correct?

 

[voko]

Yes, that is correct. For an extra credit you could try and figure out what happens when $L = 2R$.

 

[Saitama]

Yes, that is correct. For an extra credit you could try and figure out what happens when $L = 2R$.

Many thanks voko & gneill!

I am not sure but I think it won't return to its initial position, it would stay in the same configuration. Right?

Also, does the block perform oscillatory or simple harmonic motion? Is it possible to find the time period of this motion?

 

[voko]

I am not sure but I think it won't return to its initial position, it would stay in the same configuration. Right?

That's only possible if the function were completely flat in some region around zero. Which is not the case here.

When examining extrema, vanishing of the second derivative is a special case, which requires examining higher order derivatives.

Also, does the block perform oscillatory or simple harmonic motion? Is it possible to find the time period of this motion?

Now that you have PE, you can write down conservation of energy and, using small-angle apprximations, get the SHM equation.

 

[Saitama]

When examining extrema, vanishing of the second derivative is a special case, which requires examining higher order derivatives.

I substituted $L=2R$. For this value of L, $U''(\theta)=-mgR\sin\theta \Rightarrow U'''(\theta)=-mgR\cos\theta$ For $\theta=0$, $U'''(0)=-mgR$. How do I interpret this?

Now that you have PE, you can write down conservation of energy and, using small-angle approximations, get the SHM equation.

After taking the approximations,
$$U(\theta)=mg\left(R+\frac{L}{2}+\frac{R\theta^2}{2}-\frac{L\theta^2}{4}\right)$$

At any instant of time, energy of block is
$$\frac{1}{2}I\omega^2+U(\theta)$$
where I is moment of inertia about CM and $\omega$ is angular velocity of block about CM, am I doing it right?

 

[voko]

I substituted $L=2R$. For this value of L, $U''(\theta)=-mgR\sin\theta$.

No. It is $-mg R \theta \sin \theta$.

How do I interpret this?

The principle is the same as with the second-order analysis. If the third derivative is not zero, that is not an extremum (why?). If it is, then everything depends on the sign of the fourth derivative, in exactly the same way (why?).

At any instant of time, energy of block is
$$\frac{1}{2}I\omega^2+U(\theta)$$
where I is moment of inertia about CM and $\omega$ is angular velocity of block about CM, am I doing it right?

Right, except that you want the moment with respect to the center of the drum, so that the block undergoes pure rotation.

 

[Saitama]

No. It is $-mg R \theta \sin \theta$.

Yes, very sorry. So $U'''(0)=-2mgR$.

The principle is the same as with the second-order analysis. If the third derivative is not zero, that is not an extremum (why?). If it is, then everything depends on the sign of the fourth derivative, in exactly the same way (why?).

From what I remember, if the third derivative is non-zero, then the point is an inflextion point. I don't understand how to interpret that in the given case.

Right, except that you want the moment with respect to the center of the drum, so that the block undergoes pure rotation.

voko, I don't know why but "pure rotation" about centre of drum does not make sense to me. I can't visualize how it is a pure rotation. The block tips to the right, comes back, tips left, comes back and the cycle goes on, how is it pure rotation about centre of drum? Sorry for asking so many dumb questions.

Differentiating the energy equation wrt time and setting the derivative to zero.

$$I\omega\alpha+U'(\theta)=0=I\omega\alpha+mg\left(\frac{R}{2}-\frac{L}{4}\right)(2\theta\omega)$$
$$\Rightarrow \alpha=\frac{mg}{I}\left(\frac{L}{2}-R\right)\theta$$

I don't think the above is correct, for SHM we must have $\alpha \propto -\theta$ but that doesn't seem to be the case here. :(

 

[voko]

Yes, very sorry. So $U'''(0)=-2mgR$.

No, still not correct. $U''(\theta) = -mg R \theta \sin \theta \to U'''(\theta ) = - mg R \sin \theta - mg R \theta \cos \theta \to U'''(0) = 0$

voko, I don't know why but "pure rotation" about centre of drum does not make sense to me. I can't visualize how it is a pure rotation. The block tips to the right, comes back, tips left, comes back and the cycle goes on, how is it pure rotation about centre of drum? Sorry for asking so many dumb questions.

Those are not dumb questions. The truth is, I was wrong and you are right doubting me. You could instead obtain the KE of CoM motion and the KE of block's rotation about CoM, just like you intended originally.

 

[Saitama]

No, still not correct. $U''(\theta) = -mg R \theta \sin \theta \to U'''(\theta ) = - mg R \sin \theta - mg R \theta \cos \theta \to U'''(0) = 0$

Very sorry again, I accidentally posted the value of the fourth derivative at $\theta=0$.

The fourth derivative is non-zero and negative, so do I say that equilibrium is stable in the case when $L=2R$?

Those are not dumb questions. The truth is, I was wrong and you are right doubting me. You could instead obtain the KE of CoM motion and the KE of block's rotation about CoM, just like you intended originally.

So the energy of block is
$$\frac{1}{2}mv^2+\frac{1}{2}I_{CM}\omega^2+U(\theta)$$

How do I relate v with $\omega$? Is $v=(L/\sqrt{2})\omega$?

 

[voko]

Very sorry again, I accidentally posted the value of the fourth derivative at $\theta=0$.

The fourth derivative is non-zero and negative, so do I say that equilibrium is stable in the case when $L=2R$?

The treatment of the fourth-order derivative is analogous to the second-order case, for largely the same reasons.

How do I relate v with $\omega$? Is $v=(L/\sqrt{2})\omega$?

You have already found the location of CoM in terms of the angle. Finding the velocity from that should be straight forward.

 

[Saitama]

You have already found the location of CoM in terms of the angle. Finding the velocity from that should be straight forward.

$$v_y=\frac{dh}{dt}$$

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

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?

 

[voko]

What is $h$?

I was referring to the following formula of the CoM position: $(R + L/2) \vec u - R \theta \vec t$, where $\vec u$ is the unit radial vector, and $\vec t$ is the unit tangent vector, as explained earlier. The CoM velocity, then, is $v_c = (R + L/2) \dot {\vec u} - R \dot \theta \vec t - R \theta \dot {\vec t} = (R + L/2) \vec t \dot \theta - R \dot \theta \vec t + R \theta \vec u \dot \theta = (L/2) \vec t \dot \theta + R \theta \vec u \dot \theta$, and $v_c^2 = (L^2/4 + R^2 \theta^2) \dot \theta^2$

 

[Saitama]

What is $h$?

h is the height of CoM as shown in gneill's sketch in post #25.

I was referring to the following formula of the CoM position: $(R + L/2) \vec u - R \theta \vec t$, where $\vec u$ is the unit radial vector, and $\vec t$ is the unit tangent vector, as explained earlier. The CoM velocity, then, is $v_c = (R + L/2) \dot {\vec u} - R \dot \theta \vec t - R \theta \dot {\vec t} = (R + L/2) \vec t \dot \theta - R \dot \theta \vec t + R \theta \vec u \dot \theta = (L/2) \vec t \dot \theta + R \theta \vec u \dot \theta$, and $v_c^2 = (L^2/4 + R^2 \theta^2) \dot \theta^2$

I love the way you solve these problems using vectors. I wonder when I will get efficient in using them. :)

But I don't get the same answer as yours using my definition of $v_x$ and $v_y$.

Here's what I did:
$$v_x=\frac{L}{2}\dot{\theta}$$

Expression for h:
$$h=\left(R+\frac{L}{2}\right)\cos \theta+R\theta\sin\theta \approx \left(R+\frac{L}{2}\right)\left(1-\frac{\theta^2}{2}\right)+R\theta^2$$
Differentiating wrt time and simplifying, I get
$$v_y=\frac{dh}{dt}=\left(R-\frac{L}{2}\right)\theta\dot{\theta}$$
Also $v=\sqrt{v_y^2+v_x^2}$. But substituting the values doesn't give the same answer as yours. :(

 

[voko]

I suggest that you do the small angle approximation after you are done with finding the square of velocity. Then you will find the sine and cosine terms disappear and you do not have to approximate anything at all.

As for getting the hang of using vectors, you do that by using them :)

 

[voko]

$$v_y=\frac{dh}{dt}$$

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

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).

 

[Saitama]

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?

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

 

[voko]

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?

 

[Saitama]

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. :(

 

[voko]

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: $K(0)\ddot \theta + U''(0) \theta = 0$.

 

[Saitama]

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!

 

[BruceW]

Is $K(0)$ like an 'effective moment of inertia', evaluated at $\theta=0$ ?

 

[voko]

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 $K(0)$.

 

[Saitama]

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?

 

[voko]

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 canceled 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 :)