Just one sec...
Hey! Sign in to get help with your study questionsNew here? Join for free to post

Circular motion question

Announcements Posted on
Take our short survey, £100 of Amazon vouchers to be won! 23-09-2016
    • Thread Starter
    Offline

    0
    ReputationRep:
    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.
    Offline

    3
    ReputationRep:
    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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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
    Offline

    3
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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.
 
 
 
Write a reply…

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: September 20, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Who is going to win Bake Off?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.