# What's the energy of a rotating bearing?

1. Aug 29, 2015

### Karol

1. The problem statement, all variables and given/known data
The outer ring is rotating, the inner is fixed. n is the number of balls of radius r and mass m. the outer shell has mass M. the bearing (the outer shell) rotates with angular velocity Ω. what's it's kinetic energy.

2. Relevant equations
Moment of inertia of a ball: $I=\frac{2}{5}mr^2$
Moment of inertia of a thin ring: $I=Mr^2$
Energy of a rigid body: $E=\frac{1}{2}I\omega^2$

3. The attempt at a solution

When a ball rotates an angle β and advances through angle α, it actually rotates $\theta=\beta+\alpha$
Relation between α and β: $s=\alpha R=\beta r\; \rightarrow\; \beta=\frac{R}{r}\alpha$
$$\dot \theta=\dot\alpha\left( 1+\frac{R}{r} \right)$$
I mark with ω the actual angular velocity of a ball and so:
$$\dot \theta=\omega,\; \dot\alpha=\Omega\; \rightarrow\; \omega=\Omega\left( 1+\frac{R}{r} \right)$$
$$E=\frac{1}{2}I_{ball}\omega^2+\frac{1}{2}I_{ring}\Omega^2$$
$$E=\frac{1}{2}\left[\left(1+ \frac{2}{5} \right) nmr^2\left( 1+\frac{R}{r} \right)^2\Omega^2+M(R+2r)^2\Omega^2 \right]$$
$$E=\frac{\Omega^2}{2}\left[\left(1+ \frac{2}{5} \right)nm\left( 1+\frac{R}{r} \right)^2r^2+M(R+2r)^2 \right]$$

#### Attached Files:

File size:
14.2 KB
Views:
117
• ###### Snap2.jpg
File size:
15.3 KB
Views:
115
Last edited: Aug 29, 2015
2. Aug 29, 2015

### SammyS

Staff Emeritus

It's so much easier to read and understand your post when you display the images.

3. Aug 29, 2015

### SammyS

Staff Emeritus
What is it that you're trying to determine?

4. Aug 29, 2015

### Karol

I think my solution is right but i ask to be sure

5. Aug 29, 2015

### TSny

Is it correct that $\dot{\alpha} = \Omega$?

Something to consider: If a wheel is rolling without slipping along the ground, how does the speed of a point at the top of the wheel compare to the speed of the center of the wheel?

6. Aug 30, 2015

### Karol

The speed is greater on the top but i am given Ω, the angular speed which is the same for the top and the center, so, why $\dot{\alpha} = \Omega$ isn't correct?
I understand Ω as the angular velocity of the outer rim

7. Aug 30, 2015

### TSny

I don't think the angular speed is the same for the outer shell and the center of a ball.

8. Aug 30, 2015

### Karol

The angle γ the outer rim travels if the ball would be in place and rotate at the combined angle θ=β+α:
$$\theta r=\gamma (R+2r)\;\rightarrow\;\gamma=\frac{r}{R+2r}\theta$$
But the ball also travels an angle α, and relative to the inner ring:
$$\theta r=\alpha R\;\rightarrow\;\alpha=\frac{r}{R}\theta$$
$$\Omega=\dot\gamma+\dot\alpha=\left( \frac{r}{R+2r}+\frac{r}{R} \right)\dot\theta$$
For the energy of the balls i use:
$$\dot\theta=\frac{\Omega}{\frac{r}{R+2r}+\frac{r}{R}}$$
$$E=\frac{1}{2}\left[\left(1+ \frac{2}{5} \right) nmr^2\left( 1+\frac{R}{r} \right)^2\dot\theta^2+M(R+2r)^2\Omega^2 \right]$$

9. Aug 30, 2015

### TSny

Note that you include the effect of $\alpha$ in the definition of $\theta$. But then you add the effect of $\alpha$ again when you write:$$\Omega=\dot\gamma+\dot\alpha$$ So I think you might be over counting the effect of $\alpha$.

Anyway, I obtain a different expression for $\Omega$ in terms $\dot{\theta}$. For me, it was easier to approach this in terms of linear speeds.

How does the linear speed of the outer rim compare to the linear speed of the center of a ball. (Think about a ball rolling without slipping on a horizontal surface and compare the linear speed of a point at the top of the sphere with the linear speed of the center of the sphere. The sphere can be thought of as instantaneously rotating about the point of contact with the surface.)

You can then relate the linear speeds to the angular speeds $\Omega$ and $\dot{\alpha}$.

10. Aug 30, 2015

### OldEngr63

It is necessary to get the kinematics correct, and this is often difficult. Does the ball roll without slip, or is there slipping? It makes a difference in the way the kinetic energy is to be calculated.

Also, most rolling element bearings involve a "cage," a spacer element to keep the balls properly spaced. This needs to be included in the kinetic energy calculation as well.

11. Aug 30, 2015

### OldEngr63

Is the shaft center stationary or is it moving? This also affect the energy calculation.

12. Aug 31, 2015

### Karol

The shaft is stationary and there is no cage. the balls rotate without slipping.
$$v=\omega r,\;V=2r\omega=(R+2r)\Omega$$
The distance s the ball makes on the inner ring:
$$s=\omega r=\dot\alpha R\;\rightarrow\dot\alpha=\frac{r}{R}\omega$$
$$\dot\theta=\omega+\dot\alpha=\omega+\frac{r}{R}\omega=\left( 1+\frac{r}{R} \right)\omega\;\rightarrow\;\omega=\frac{\dot\theta}{1+\frac{r}{R}}$$
Combining:
$$\dot\theta=\left[ \frac{(R+2r)(R+r)}{2Rr} \right]\Omega$$

13. Aug 31, 2015

### TSny

OK. Here, $\omega = \dot{\theta}$ where $\theta$ is the total angle of rotation of the sphere around its center.
Your expression $s = \omega r$ is dimensionally incorrect.
Also, recall from your first post that $s = \beta r$ where $\beta = \frac{R}{r} \alpha$ and $\theta = \alpha + \beta$.

14. Aug 31, 2015

### Karol

$$2r\dot\theta=(R+2r)\Omega \; \rightarrow \; \dot\theta=\frac{(R+2r)\Omega}{2r}$$
And this value of $\dot\theta$ i use for the energy calculation of the ball. there is no need to separate θ into α and β, and so the relation $\beta = \frac{R}{r} \alpha$ is meaningless

15. Aug 31, 2015

### TSny

Yes, that's correct. Although I wouldn't necessarily say that $\beta = \frac{R}{r} \alpha$ is meaningless. It's just not needed if you derive the relation between $\Omega$ and $\dot{\theta}$ this way.

16. Sep 1, 2015

### Karol

Thank you TSny, i want to reach the same result from angular movements, but i can't. if i take:
$$\gamma R=\theta r+\alpha(R+2r),\;\theta=\alpha+\beta=\left( 1+\frac{R}{r} \right)\alpha$$
$$\rightarrow\;\dot\theta=\frac{R(R+r)}{r(2R+3r)}$$

17. Sep 1, 2015

### TSny

I don't see where this comes from.

Imagine the ball remaining in place and rotating CW through angle $\beta$ as shown below. The inner ring will rotate CCW by angle $\alpha$ while the outer ring will rotate CW through angle $\gamma$. Since the two red arc lengths are equal, we have $\alpha R = \gamma (R+2r)$. So, $\gamma = \frac{R}{R+2r} \alpha$.

Now imagine the whole figure rotated clockwise by angle $\alpha$ so that the inner ring returns to its initial position. Then the outer ring will now be rotated from its initial position by $\gamma + \alpha = 2\frac{R+r}{R+2r}\alpha$.

So, the angular speed of the outer ring is $\Omega = 2\frac{R+r}{R+2r} \dot{\alpha}$.

When the figure was rotated as a whole through angle $\alpha$, the angle of rotation of the ball was increased by $\alpha$. So, the total angle of rotation of the ball is $\theta = \beta + \alpha = (\frac{R}{ r}+ 1)\alpha = \frac{R+r}{r} \alpha$. The rate of rotation of the ball is then $\dot{\theta} = \frac{R+r}{r} \dot{\alpha}$.

So, we find $\dot{\theta} = \frac{R+2r}{2r} \Omega$.

#### Attached Files:

• ###### Bearing.png
File size:
4.8 KB
Views:
26
Last edited: Sep 1, 2015
18. Sep 1, 2015

### Karol

Thank you very much TSny!