What's the energy of a rotating bearing?

[Karol]

Homework Statement

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.

Homework 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:
$$\dot \theta=\omega,\; \dot\alpha=\Omega\; \rightarrow\; \omega=\Omega\left( 1+\frac{R}{r} \right)$$
$$E=\frac{1}{2}I_{ball}\omega^2+\frac{1}{2}I_{ring}\Omega^2$$
$$E=\frac{1}{2}\left[\left(1+ \frac{2}{5} \right) nmr^2\left( 1+\frac{R}{r} \right)^2\Omega^2+M(R+2r)^2\Omega^2 \right]$$
$$E=\frac{\Omega^2}{2}\left[\left(1+ \frac{2}{5} \right)nm\left( 1+\frac{R}{r} \right)^2r^2+M(R+2r)^2 \right]$$

 

[SammyS]

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

 

[SammyS]

Homework Statement

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.

Homework 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:
$$\dot \theta=\omega,\; \dot\alpha=\Omega\; \rightarrow\; \omega=\Omega\left( 1+\frac{R}{r} \right)$$
$$E=\frac{1}{2}I_{ball}\omega^2+\frac{1}{2}I_{ring}\Omega^2$$
$$E=\frac{1}{2}\left[\left(1+ \frac{2}{5} \right) nmr^2\left( 1+\frac{R}{r} \right)^2\Omega^2+M(R+2r)^2\Omega^2 \right]$$
$$E=\frac{\Omega^2}{2}\left[\left(1+ \frac{2}{5} \right)nm\left( 1+\frac{R}{r} \right)^2r^2+M(R+2r)^2 \right]$$

What is it that you're trying to determine?

 

[Karol]

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

 

[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?

 

[Karol]

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?

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

 

[TSny]

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,...

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

 

[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:
$$\dot\theta=\frac{\Omega}{\frac{r}{R+2r}+\frac{r}{R}}$$
$$E=\frac{1}{2}\left[\left(1+ \frac{2}{5} \right) nmr^2\left( 1+\frac{R}{r} \right)^2\dot\theta^2+M(R+2r)^2\Omega^2 \right]$$

 

[TSny]

The angle γ the outer rim travels if the ball would be in place and rotate at the combined angle θ=β+α:
$$\theta r=\gamma (R+2r)\;\rightarrow\;\gamma=\frac{r}{R+2r}\theta$$

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 $\dot{\alpha}$.

 

[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.

 

[OldEngr63]

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

 

[Karol]

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

The shaft is stationary and there is no cage. the balls rotate without slipping.

For me, it was easier to approach this in terms of linear speeds.

$$v=\omega r,\;V=2r\omega=(R+2r)\Omega$$
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:
$$\dot\theta=\left[ \frac{(R+2r)(R+r)}{2Rr} \right]\Omega$$

 

[TSny]

The shaft is stationary and there is no cage. the balls rotate without slipping.

View attachment 88033 $$v=\omega r,\;V=2r\omega=(R+2r)\Omega$$

OK. Here, $\omega = \dot{\theta}$ where $\theta$ is the total angle of rotation of the sphere around its center.

The distance s the ball makes on the inner ring:
$$s=\omega r=\dot\alpha R\;\rightarrow\dot\alpha=\frac{r}{R}\omega$$

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 $\theta = \alpha + \beta$.

 

[Karol]

OK. Here, ω=θ˙\omega = \dot{\theta} where θ\theta is the total angle of rotation of the sphere around its center.

$$2r\dot\theta=(R+2r)\Omega \; \rightarrow \; \dot\theta=\frac{(R+2r)\Omega}{2r}$$
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

 

[TSny]

$$2r\dot\theta=(R+2r)\Omega \; \rightarrow \; \dot\theta=\frac{(R+2r)\Omega}{2r}$$
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

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.

 

[Karol]

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

 

[TSny]

if i take:
$$\gamma R=\theta r+\alpha(R+2r)$$

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 $\dot{\theta} = \frac{R+2r}{2r} \Omega$.

 

[Karol]

Thank you very much TSny!