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Circular motion question Watch

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    I created a thread about this a few weeks ago but I didn't type the question clearly so I'll try again.


    A train on a fairground is initially stationary before it descends through a height of 45m into a dip that has a radius of curvature of 78m, as shown in Figure 5. (Basically, descends downwards before levelling off at the bottom).

    a. Calculate the speed of the train at the bottom of the dip, assuming air resistance and friction are negligible

    ANSWER: 29.7 ms^-1. I'm fine with this part.

    b.i. Calculate the centripetal acceleration of the train at the bottom of the dip.

    ANSWER: 11.3 ms^-2. Fine with this as well.

    ii. Calculate the extra support force on a person of weight 600 N in the train.

    ANSWER: 690 N. I don't understand how to get to this answer.

    S - mg = mv^2/r

    S = mg + mv^2/r

    r = 78, v = 29.7, g = 9.8, m = 600/9.8

    subbing the numbers in:

    S = 600 + 689 = 1289

    These working don't account for the mass of the carriage which isn't given either. Just can't get the answer. Any help would be really appreciated.

    Thanks.
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    A person who weighs 600N static has a mass of (600/9.81) kg
    =61.2 kg

    accelerating that person by 11.3 ms-2 requires a force of (61.2 * 11.3) N
    =692 N (690 to 2 s.f.)

    afaict the question is just asking about the force due to the centripetal acceleration of the rider only - not the grand total or the force on the carriage.
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    (Original post by Joinedup)
    A person who weighs 600N static has a mass of (600/9.81) kg
    =61.2 kg

    accelerating that person by 11.3 ms-2 requires a force of (61.2 * 11.3) N
    =692 N (690 to 2 s.f.)

    afaict the question is just asking about the force due to the centripetal acceleration of the rider only - not the grand total or the force on the carriage.
    I know your right but I don't understand. Wouldn't that get you the centripetal force not the support force?

    Thanks
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    (Original post by mtig)
    I know your right but I don't understand. Wouldn't that get you the centripetal force not the support force?

    Thanks
    It asks about extra force which I presumed it means the centripetal force which you'd then add to the weight of the passenger to get a total force.

    but it doesn't ask for the total force - so I think you're trying to anticipate too much.
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    (Original post by Joinedup)
    It asks about extra force which I presumed it means the centripetal force which you'd then add to the weight of the passenger to get a total force.

    but it doesn't ask for the total force - so I think you're trying to anticipate too much.
    Ok thank you very much I understand it now.
 
 
 
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