How Long Does It Take for a Flywheel to Stop with Applied Tangential Force?

Kalus
Messages
36
Reaction score
0

Homework Statement


You have a flywheel rotating at 1000rpm. It has a moment of inertia of 0.5 Kgm^2 and a diameter of 0.5m.

It is unclear if you know the mass. You know the moment of inertia, which is calculated using mass, but whilst the flywheel is circular I'm not sure that the surface area is that of a circle (as it may have ridges or something cut into it)

A tangential friction force is then applied to the rim of the flywheel (in the opposite direction of rotation to the flywheel), the force is 1000N.

Find the time it takes in minutes for the flywheel to come to rest.


Homework Equations



image006.gif


Not sure if any other equations are needed...

I've tried a few attempts, but i don't want to list them incase they influence anyones thinking

Many thanks, Kalus.
 

Attachments

  • image006.gif
    image006.gif
    4 KB · Views: 14,263
Physics news on Phys.org
You have, in that array of equations that you splatter up there,

Tau = I alpha
and if you think about it, the torque will be Tau = f*r = friction force * radius, both known

Thus alpha = (f*r)/I where everything on the right is known.
This is motion under a constant (angular) acceleration. Can you take it from there?
 
As you suggested, i did that previously...

if you plugin the values:

alpha = (f*r)/I

alpha = (1000*0.25)/0.5

which means alpha= 500

Then if you use the equation omega = omega_0 + alpha*t from down below...

you know omega (the rotational velocity, which is 1000rpm * 2pi /60)

but when you plug the numbers in you get 0.2 seconds... it doesn't seem right at all, especially given the question asks for the time in minutes...
 
It is entirely possible to express 0.2 seconds in minutes.

The way the problem was expressed, it indicated that the friction force was 1000 N. If the intent was that the normal force was 1000 N, then there needs to be included a coefficient of friction that will multiply this value to give a somewhat lower actual friction force and hence a lower angular acceleration. This will take longer to slow down. But, with what is given, this is what we get.
 
I know it is of course possible to express it in minutes,but i thought that an answer of 0.0033 minutes seemed a bit off, especially as they asked it for the time in minutes (leading me to expect +60s times). Unfortunatey because I've never seen a flywheel of this size and don't know the mass, i have no idea if the answer seem sensible.

The way the question is worded the force is tangetial, and no coefficeient of friction is given.

Can you think of any other possible approaches or where this could be going wrong?

Many thanks for your help Dr. D

Kalus
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top