# Wheel attached to a rope falls

1. Sep 19, 2015

### Karol

1. The problem statement, all variables and given/known data
A wheel is made of two hoops, one of radius R and mass $\frac{2M}{3}$ and the inner is of radius R/2 and mass M/3.
1. the wheel rotates around it's axis with angular velocity ω but remains in place. what is it's kinetic energy
2. the wheel rotates without friction at velocity V, what's it's kinetic energy
3. a rope is wound around the inner hoop. the rope's end is attached to the ceiling and the wheel falls. what is it's angular velocity if the velocity of the center is u.
4. what is the linear velocity of the center if the wheel descended the height h
5. what is the linear acceleration of the center

2. Relevant equations
Moment of inertia of a thin hoop: I=mr2
Torque and angular acceleration of a rigid body: T=Iα

3. The attempt at a solution
The moment of inertia of the entire wheel:
$$I=\left( \frac{2M}{3}+\frac{M}{3} \right)R^2=\frac{3}{4}MR^2$$
Total mass=M
1. When the wheel rotates at ω:
$$E=I\omega^2=\frac{3}{4}MR^2\omega^2$$
2. $$V=\omega R,\quad E=MV^2+I\omega^2$$
$$E=MV^2+\frac{3}{4}MV^2=\frac{7}{4}MV^2$$
3.
$$u=\omega \frac{R}{2}$$
4. The energy calculated in paragraph 2 equals the loss in potential energy:
$$\frac{7}{4}MV^2=Mgh\quad\rightarrow\quad V=\frac{4}{9}gh$$
5. I take the moment of inertia round the small radius.
$$Mg\frac{R}{2}=\left( \frac{3}{4}MR^2+M\frac{R^2}{4} \right)\alpha \quad\rightarrow\quad a=\frac{g}{2}$$

2. Sep 19, 2015

### haruspex

Where you obtain the moment of inertia of the system, you get the right answer but there appears to be a typo along the way. I assume this is an error introduced in making the post.

In the energy formulae, you've left out a constant factor.

In the statement of q2, I am guessing it should say the wheel descends at speed V. But the rope is not mentioned until q3, so maybe that's wrong. Perhaps it is supposed to be rotating about a fixed centre here, with V being the tangential velocity.

There is a subtlety in the last two parts. Strictly speaking, the string would not remain vertical. To some extent, the whole system would behave like a pendulum. However, I'm sure you are not expected to get into that complication, so I agree with your answer to q5.

3. Sep 22, 2015

### Karol

The moment of inertia of the entire wheel about the center:
$$I=\frac{2M}{3}R^2+\frac{M}{3}\frac{R^2}{4}=\frac{3}{4}MR^2$$
When the wheel rotates at ω:
$$E=\frac{1}{2}I\omega^2=\frac{3}{8}MR^2\omega^2$$
$$V=\omega R,\quad E=\frac{1}{2}MV^2+\frac{1}{2}I\omega^2$$
$$E=\frac{1}{2}MV^2+\frac{1}{2}\frac{3}{4}MV^2=\frac{7}{8}MV^2$$
The energy calculated in paragraph 2 equals the loss in potential energy:
$$\frac{7}{8}MV^2=Mgh\quad\rightarrow\quad V=\frac{2}{9}gh$$
Moments around A: the weight Mg creates torque T:
$$T=-I\alpha\quad\rightarrow\quad Mgx=I\alpha$$
The rope unwinds and the distance y increases, so the moment of inertia increases. $I_c$ is the moment of inertia round the center.
$$Mgx=-(I_c+My^2)\alpha$$
Trigonometry:
$$y\alpha=\ddot x\quad\rightarrow\quad \alpha=\frac{\ddot x}{y}$$
$$\rightarrow Mgx=-(I_c+My^2)=\frac{\ddot x}{y}$$
Is this the equation for the oscillation?

4. Sep 22, 2015

### TSny

I agree with your answers to parts 1 and 3.

I don't understand part 2. Could it be asking for the KE when the wheel is rolling without slipping along a surface?

I don't agree with your answers to parts 4 and 5. Did you use the result of part 3?

I also don't see why there would be any pendulum motion. What force has a horizontal component that would move the center of mass of the wheel horizontally?

5. Sep 23, 2015

### Karol

Yes
When the wheel rotates at ω:
$$E=\frac{1}{2}I\omega^2=\frac{3}{8}MR^2\omega^2$$
$$E=\frac{1}{2}MV^2+\frac{1}{2}I\omega^2=\frac{1}{2}MV^2+\frac{3}{8}MR^2\omega^2$$
When the velocity of the center is V the angular velocity is $V=\frac{R}{2}\omega\rightarrow \omega=\frac{2V}{R}$
$$E=\frac{1}{2}MV^2+\frac{3}{8}MR^2\frac{4V^2}{R^2}=2MV^2$$
When a weight is hang not along the center of mass it rotates:
You mean the wheel will descend right down, and the rope will change gradually it's angle, like in the drawing:

6. Sep 23, 2015

### haruspex

Yes, I had second thoughts about my comment there. It seemed intuitively true, but I cannot justify it.

7. Sep 23, 2015

### TSny

That looks good.

Yes, if you release the block when it is in the position shown on the right in your figure, then as it "falls" the block will rotate about the point of attachment with the string and the string will not stay vertical. The string will initially swing to the left a little. At least that's what I think will happen.
In this case, I think the string will stay vertical as the wheel descends.

8. Sep 24, 2015

### Karol

Interesting why do you think it's so, can you explain your intuition? i will also think about that

9. Sep 24, 2015

### TSny

In the case of the block, the string has a fixed length. As the center of mass of the block falls the block must rotate. The point of attachment of the string to the block must then move to the left. The string is thus pulled away from vertical to the left. The resultant horizontal component of the tension in the string will then cause the center of mass of the block to start swinging to the right. So, you get some swinging motion in the string as the block pivots about the point of attachment.

In the case of the wheel, the string unwraps and increases in length. As the center of mass of the wheel falls the wheel must rotate. But now the point where the string unwraps from the wheel always stays at the same point relative to the center of mass of the wheel. The string is not pulled to the left as in the case of the block. My main reason for believing that the string remains vertical for the wheel is that you get a solution that is consistent with Newton's laws when you assume the string remains vertical.

10. Sep 25, 2015

### Karol

I understand the power of such a reasoning but i don't see which laws, specifically, are satisfied and how could any other solution not be based on those laws.
I know the equation $T=I\alpha$ that we used is parallel to Newton's law F=ma, but how does it assure you that the rope must stay vertical?

11. Sep 25, 2015

### TSny

The mathematical proof that the string must stay vertical for this problem would come from invoking a uniqueness theorem for a system of ordinary differential equations. The degrees of freedom for the system can be taken to be the angle θ that the string makes to the vertical and the length s of string that is unwrapped from the wheel. You can derive the coupled differential equations for θ and s as functions of time from the Lagrangian of the system. The differential equations will satisfy the conditions necessary for the uniqueness theorem to hold. So, if you find any solution that satisfies the differential equations for initial conditions s(0) = so, s'(0) = 0, θ(0) = 0, and θ'(0) = 0, then the theorem guarantees that the solution is the only solution. [Edited to replace α by θ.]

Last edited: Sep 25, 2015
12. Sep 25, 2015

### Karol

I don't know about lagrangian and the rest about differential equations and the theorem, and i won't question you about it, but if the initial conditions s'(0) = 0, θ(0) = 0 are satisfied, does it say θ won't obtain other values? the rope also starts from rest and s'(0) = 0, θ(0) = 0 are fulfilled.
I also don't see the differential equation that includes θ, besides the one i tried to derive, is it true? in the analysis we made so far there was no differential equation involved

13. Sep 25, 2015

### TSny

You don't need to use the Lagrangian approach. You can just apply Newton's laws for the general configuration shown below. However, the torque equation should be written $\tau = \frac{dL}{dt}$ where $L$ is the total angular momentum of the system. The angular momentum can be broken up into two parts: angular momentum due to motion of CM and angular momentum due to rotation about the CM.

The torque equation will give a differential equation involving $s, \dot{s}, \ddot{s}, \theta, \dot{\theta}$ and $\ddot{\theta}$

A second differential equation can be obtained by applying $F= ma$ in the direction perpendicular to the string. (Choose this direction to avoid bringing in the tension in the string.) This differential equation will involve $s, \dot{s}, \theta, \dot{\theta}$ and $\ddot{\theta}$.

So you get two coupled second-order differential equations for $s$ and $\theta$. The equations are too complicated to solve in closed form for general initial conditions. But, you can solve them for the special initial conditions $s(0) = s_0$, $\dot{s}(0) = \theta(0) = \dot{\theta}(0) = 0$. The solution keeps $\theta = 0$ as the wheel descends.

#### Attached Files:

• ###### Wheel unwrapping string.png
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14. Sep 29, 2015

### Karol

I change the markings for clarity: m is the total mass, r is the radius of the small rim, IC is the moment of inertia round the center and I is the moment of inertia at the unwrapping point.
The component of the wight parallel to the rope:
$$L=I\omega=(I_c+mr^2)\omega=\frac{I_c+mr^2}{r}\dot s$$
$$L'=\frac{I_c+mr^2}{r}\ddot s=mg\cdot\cos\theta$$
This equation doesn't contain $\dot{s}, \dot{\theta}, \ddot{\theta}$
The component of the gravity vertical to the rope:
$$F=ma:\quad mg\sin\theta=m\dot v=m(\theta\sqrt{s^2-r^2})'$$
The derivative is complicated and it won't contain $\ddot s, \dot{\theta}, \ddot{\theta}$

15. Sep 29, 2015

### TSny

This expression for $L$ is incomplete. Note that if you imagine $s$ as held constant as $\theta$ changes, you would have nonzero angular momentum about the point where the string attaches to the support. But your expression gives $L = 0$ if $s$ is held constant. You can fix your expression by adding appropriate terms that include the effect of changing $\theta$. These terms will involve $s$ and $\dot{\theta}$. So, when you take the time derivative of $L$ you will get something involving $\dot{s}, \dot{\theta}$, and $\ddot{\theta}$.

OK, $F_{\bot} = mg\sin\theta$. But I don't understand your expression for $a_{\bot}$. I don't think it's right.

16. Sep 29, 2015

### Karol

$$L=I_1\omega+I_2\dot\theta=(I_c+mr^2)\omega+[I_c+m(s^2-r^2)]\dot\theta=\frac{I_c+mr^2}{r}\dot s+[I_c+m(s^2-r^2)]\dot\theta$$
$$L'=\frac{I_c+mr^2}{r}\ddot s=mg\cdot\cos\theta+2ms\dot s\dot\theta[I_c+m(s^2-r^2)]\ddot\theta$$
But which force and torque arms to take? for I1 i take the tangential component and for I2-the vertical. can i take mg as a whole? to where to stretch the arm?
for $a_{\bot}:\quad v=\dot \theta\sqrt{s^2-r^2}$
$$mg\cdot\sin\theta=m\left[ \dot \theta\sqrt{s^2-r^2} \right]'$$

17. Sep 29, 2015

### TSny

Your final expression for $L$ almost agrees with what I got using a different approach. But shouldn't your $(s^2 - r^2)$ be replaced by $(s^2+r^2)$?
$s$ and $r$ are perpendicular.

It looks like your expression for $\dot{L}$ has some typographical errors and I'm not sure how you got the $mg\cos \theta$ term.

Your method of getting the component of acceleration of the CM that's perpendicular to the string does not look correct.

-----------------------------------------

For me, the least confusing way to approach the problem was to express the location of the center of mass of the wheel in Cartesian components where I chose the origin of Cartesian axes at the point where the string attaches to the support. I chose the positive $x$ direction to be to the right and the positive $y$ direction to be upward.

Let $x_c$ and $y_c$ denote the Cartesian coordinates of the center of mass of the wheel. These coordinates can be expressed in terms of $s$, $r$, and $\theta$. The part of the angular momentum that is due to motion of the center of mass can be expressed as $L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)$. It is then straightforward but a little tedious to crank out $\dot{L}_c$.

The only torque about the origin is due to the force of gravity and you can see that the lever arm is just $x_c$.

When applying F = ma perpendicular to the string, you can first find the acceleration vector from $x_c$ and and $y_c$ and then project the acceleration along the direction perpendicular to the string.

18. Sep 30, 2015

### Karol

$$L=I_1\omega+I_2\dot\theta=(I_c+mr^2)\omega+[I_c+m(s^2+r^2)]\dot\theta=\frac{I_c+mr^2}{r}\dot s+[I_c+m(s^2+r^2)]\dot\theta$$
$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2-r^2)]\ddot\theta=mg\sqrt{s^2+r^2}\sin\theta$$
For $a_{\bot}$ i tend to say $v=\dot \theta s\rightarrow a=\ddot\theta s+\dot\theta \dot s$ but it's not true since s changes and i have to get an expression with $\ddot s$
I will try to work on the solution with Cartesian coordinates as soon as i finish this one.

19. Sep 30, 2015

### TSny

OK
The factor in front of $\ddot{\theta}$ should have $s^2 + r^2$ instead of $s^2 - r^2$. The expression for the torque on the right is not quite right. The force $mg$ does not make an angle $\theta$ to the line connecting the point of support to the center of the wheel.

20. Oct 1, 2015

### Karol

$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg[s\cdot \sin\theta-r\sin(90-\theta)]$$
$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg(s\cdot \sin\theta-r\cos\theta)$$
$$F=ma:\quad mg\sin\theta=m\dot v=m(\dot\theta s)'=m(\dot\theta^2\dot s+s\ddot\theta)$$
Do i need the F=ma equation? the information about the CM motion is already included in the member $I_2\dot\theta$ of $L=I_1\omega+I_2\dot\theta$
I understand i need 2 equations for s and θ but doesn't the information in F=ma repeat itself?
For the approach $L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)$ i take y directed downward:
$$x_c=s\cdot \sin\theta-r\cos\theta$$
$$y_c=s\cdot \cos\theta+r\sin\theta$$
$$\dot{x_c}=(s\cos\theta+r\sin\theta)\dot\theta$$
$$\dot{y_c}=(r\cos\theta-s\cdot \sin\theta)\dot\theta$$
$$L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)=m\dot\theta[(s\cdot \sin\theta-r\cos\theta)(r\cos\theta-s\cdot \sin\theta)+(s\cdot \cos\theta+r\sin\theta)(s\cos\theta+r\sin\theta)]$$
$$L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)=m\dot\theta[2sr\sin\theta\cos\theta+(s^2-r^2)(\cos^2\theta-\sin^2\theta)]$$
$$L=I_1\omega+m\dot\theta[2sr\sin\theta\cos\theta+(s^2-r^2)(\cos^2\theta-\sin^2\theta)]$$
$$L=\frac{I_c+mr^2}{r}\dot s+m\dot\theta[2sr\sin\theta\cos\theta+(s^2-r^2)(\cos^2\theta-\sin^2\theta)]$$

Last edited: Oct 1, 2015