[Closed] Help with a physics question please.

What sort of help do you want?

Biscuit

You've got the kinetic energy of the container, which is your regular:

1/2 mv^2

Plus the rotational kinetic energy of the two wheels (or whatever it's called these days). Not sure what a radius of gyration is, must have been invented recently.

That gives you your first answer.

Second answer. probably easiest is to take the energy in a spring and find the extension at which it equals the kinetic energy above.

If you squint, it looks a bit like a Bender from Futurama.

What's the frictional resistance of the rails?

Frictional coefficiency of the wheel bearings?

All taking place at sea-level or in a vacuum?

has anyone calibrated Plank's constant recently? I'd want that checked out before I went any further.....

Thread to molgrips...Thread to molgrips.

Is it on a conveyor belt?

I'll post tomorrow 🙂 bedtime now.

Had to look up Radius of Gyration - new term since I last did A level Physics.

Radius of gyration or gyradius is the name of several related measures of the size of an object, a surface, or an ensemble of points. It is calculated as the root mean square distance of the objects' parts from either its center of gravity or a given axis.

I'm trying to help out a friend at work but this has me a bit baffled. I think I can work out the first part easily enough but the second one seems more difficult.

I thought that if I know initial and final velocity and displacement, I could work out deceleration.

Then, knowing deceleration and the mass, I would know the force, then just feed it into Hookes Law and out pops the sprink constant. But I can't hlp thinking I've missed something because I've not used the radius of gyration.

Makes me think my method if flawed but I can't see where.

radius of gyration and wheel mass lets you find moment of inertia.

Find W from W = v /r

you can then use E rot = 0.5 x I x Wsquared

double this for the two wheels, and then add in 1/2 m v sqaured for the whole system (m=130kg)

Total the rotational and linear kinetic energies.

Then use strain energy formula for springs E = 0.5 k x sqaured, where E is the total energy from earlier and x is the 0.3m spring compression.

Solve for k, the spring stiffness in N/m

The underside of the rail pictured above suggests it was made by Fox, and the regular recommended service intervals have not been adhered to.

I suggest the answer would involve some form of Kashima coating...

what he ^^ said.

once you've got the kinetic energy, you can work out the spring force.

Energy = force x distance.

in this case distance = 0.3metres

rearrange for force.

i think...

(i've had a migraine, i'm thinking a little screwy today)

edit: ignore me, i'm an idiot.

Radius of gyration is the point equivalent of the mass

African or European Swallow?

Radius of gyration can be used for calculating Inertia in a wheel/flywheel etc.

I=m*Ksquared

I=Inertia
M=mass of object (in this case the 15kg wheel)
K=Radius of Gyration (in M so 0.4)

gyration

IGMC

23kN/m

I think but it is late and I have had a little drink, well it is the wifes birthday.

has anyone calibrated Planck's constant recently? I'd want that checked out before I went any further...

I wouldn't get involved at all. That whole set up is a H&S nightmare and if you specify the spring, you'll be culpable...

WALK AWAY.

If you google the first sentence you will find various people discussing the same!

The reason the wheels are specified is that the spring has to absorb their rotational energy too AND their linear momentum don't forget..

You'll probably want to swap the standard spring out for a heavy.

I wouldn't add the wheel's rotational energy to the total kinetic energy, because at any moment as much of the wheel is going backwards as forwards. (And up as down, and up and right as down and left, and etc...)

(But I'm always too alert for a trick question!)

And if spring stiffness is measure in joules per meter (is it?) the whole thing becomes trivial.

Still not sure what rebound damping they should set on the fork....

I wouldn't add the wheel's rotational energy to the total kinetic energy, because at any moment as much of the wheel is going backwards as forwards.

So you're saying a rotating wheel has no net energy? How would flywheels work then?

So you're saying a rotating wheel has no net energy? How would flywheels work then?

No, I was thinking the rotation energy didn't contribute to the apparatus' linear energy. I'm not convinced though, and the most compelling reason to suspect it does is that it was included in the question. 😀

It is of course true that the wheels are turning, and once the apparatus stops they won't be, and there's only one thing absorbing that energy.

Unless it's a frictionless rail. Just list frictionless rail in your assumptions. (Frictionless [b]spherical[/b] rail, to be extra sure.)

It is of course true that the wheels are turning, and once the apparatus stops they won't be, and there's only one thing absorbing that energy.

Yes, and this is a powerful tactic for these physics problems. Take a step back and say what you can about the whole system and work from there. So you know it's got x energy (by adding up all the things with energy) and you know at some point it's going to end up with 0 kinetic energy (just before it starts to bounce off the spring). You know what distance it has to bring the whole lot to a stop, so there you go. Job's a goodun.

My last refuge for my suspected "trap" was that the energy the spring needed to absorb would be more than the kinetic energy [b]if[/b] spinning energy doesn't "count"; a quick brush-up on the semantics shows it definitely does.

How boringly straightforward. 😉

Engineer's answer: put adjustable dampening on the spring and do a few test runs 😀

Long spring, medium spring and short spring, fitted concentrically, sorted.

Slight issue, the object will only be stopped for a finite just non zero amount of time by the spring. Then the Spring will accelerate the object back the way it came! (No damping in the system.......) Doesn't sound very safe to me 😉