Help with a physics question please.

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  • Help with a physics question please.
  • Premier Icon aracer
    Subscriber

    What sort of help do you want?

    oldnpastit
    Member

    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.

    Jujuuk68
    Member

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

    tymbian
    Member

    What’s the frictional resistance of the rails?

    Frictional coefficiency of the wheel bearings?

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

    Premier Icon footflaps
    Subscriber

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

    Premier Icon SaxonRider
    Subscriber

    Thread to molgrips…Thread to molgrips.

    johndoh
    Member

    Is it on a conveyor belt?

    Premier Icon molgrips
    Subscriber

    I’ll post tomorrow 🙂 bedtime now.

    Premier Icon footflaps
    Subscriber

    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.

    Premier Icon Onzadog
    Subscriber

    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

    crikey
    Member

    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…

    Premier Icon ahwiles
    Subscriber

    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

    Premier Icon muddy@rseguy
    Subscriber

    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)

    Premier Icon mikewsmith
    Subscriber

    gyration

    sniggers

    IGMC

    mark90
    Member

    23kN/m

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

    CountZero
    Member

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

    Mrs Planck certainly does… 😉

    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.

    poly
    Member

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

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

    Premier Icon BigJohn
    Subscriber

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

    klumpy
    Member

    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.

    Premier Icon footflaps
    Subscriber

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

    Premier Icon molgrips
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    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?

    klumpy
    Member

    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 spherical rail, to be extra sure.)

    Premier Icon molgrips
    Subscriber

    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.

    klumpy
    Member

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

    How boringly straightforward. 😉

    Premier Icon GrahamS
    Subscriber

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

    slowoldgit
    Member

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

    maxtorque
    Member

    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 😉

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