Hihi,
So sorry for the delay,
I tried again with this new value. It's still giving me way too high values for the torque. I'm really stuck now. Everything in the equation looks fine but it's just the ## \sigma ## which drives up the torque numbers a lot.
Hi,
Sorry for the delay,
So, right after looking at what is an annulus, I decided that it is in fact an annulus. My values for inner and outer radii are: $$r_1 =0.6984 m ; r_2 =0.11984 m$$. Using these values and the density of aluminium as 2810 kg/m3, I get the value of the moment of inertia...
Ok, right,
So firstly, to find the moment of inertia of this semicircular region, I first find the total moment of inertia and then the moment of inertia of the inner ring, and then subtract to get the moment of inertia of the ring itself. This is then divided by two, to give me the moment for...
Right,
So, What I did is equate the torque to ## \sigma \cdot \omega \cdot B^2 \cdot I_c ##
So, what I did for finding ## I_c ##, was firstly approximate the total area of effect of the magnets, which is approximately in the shape of a ring around the disk, but only covering half of the...
The torque would be $$T= F \cdot r$$ which is equivalent to $$ \sigma \cdot \omega r \cdot (dA \cdot \tau) B^2 \cdot r$$ $$\sigma \cdot \omega r^2 \cdot (dA \cdot \tau) B^2$$ So the total torque is $$ \sigma \cdot \omega r^2 \cdot \tau \cdot B^2 \int dA $$ Or am I missing something?
Err.. Then it would become $$F= \sigma \cdot \omega r \cdot (dA \cdot \tau) B^2 $$
Where ## \tau ## represents the thickness of the disk.
If I'm right you now want me to integrate this whole equation with respect to ##dA## ?
Small v is velocity. Capital V is volume. This is because ##J \times B ## only gives force per unit volume. So, in order to get the total force, it has to be multiplied by the volume
The torque and the tangential speed would be irrespective of which element and depend only on the distance from the centre. Thus, respectively they would be ##T=Fd## and ##v=\omega r##
In my case, both would be the same.
It would act at a range of distances, however, not the whole disk as the magnetic field strength would drop off, after a distance. I feel like the answer would be to integrate. However, I'm not sure what to integrate, and over what range
Hi,
As far as I know, the velocity is defined as $$\omega r$$ and the torque would be defined as $$Fd$$ I'm sorry I'm not able to get at what you're suggesting..
Homework Statement: Finding the braking torque applied on a moving disk as a result of the Eddy Currents
Homework Equations: $$F = \sigma v V B^2$$
So right,
Basically my problem is how to find the torque exerted by an eddy current braking setup.
My setup consists of a disk rotating on an...