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    Hi all need a bit of help please

    A small fluffy toy hangs on a string in the rear window of a car. The car rounds a bend at 30ms-1 and the string hangs 20 degrees from the vertical.

    There is a diagram just of the string hanging 20 degrees from vertical.

    a) mark the forces acting on the toy

    b) Calculate the radius of the curve in the road.

    I have no idea so any help would be good.

    Thanks
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    (Original post by TheStrokess)
    Hi all need a bit of help please

    A small fluffy toy hangs on a string in the rear window of a car. The car rounds a bend at 30ms-1 and the string hangs 20 degrees from the vertical.

    There is a diagram just of the string hanging 20 degrees from vertical.

    a) mark the forces acting on the toy

    b) Calculate the radius of the curve in the road.

    I have no idea so any help would be good.

    Thanks
    For a) i got the tenson in the string, weight acting down on the toy, and centripetal force towards the center.

    for b) Well i"m pretty weak at this topic myself, but I got 252.1m, seems a lot. I did it by using tan(20)=30^2/9.81xR R=252.1m. ?

    Anyone confirm this or am i totally wrong?
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    I'm poor at it also. I always thought the car was experiencing the centripetal force and the bodies inside the car kept 'trying to move in a straight line' according to newtons law. Thanks for the help so far.
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    (Original post by Im_a_cyborg)
    For a) i got the tenson in the string, weight acting down on the toy, and centripetal force towards the center.

    for b) Well i"m pretty weak at this topic myself, but I got 252.1m, seems a lot. I did it by using tan(20)=30^2/9.81xR R=252.1m. ?

    Anyone confirm this or am i totally wrong?
    I just checked the MS and 252 is correct. Can you explain what you've done here in more detail? I would appreciate it.

    Thanks
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    Yes, no problem, from the diagram of the toy, sin(20) gives the centripetal force (towards the centre) so sin(20)=mv^2/R and
    cos(20)= mg (the weight)

    You cannot use sin(20)=mv^2/R since you do not have the mass of the car, so you combine mv^2/R with mg. 'm" cancels leaving v^2/gr which is equal to tan(20) and get 'r' from that

    Its hard to explain but if you remember;
    sin(angle)=mv^2/R
    cos(angle)= mg
    tan(angle)=v^2/gr

    You can't go wrong.
    Hope that helps.
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    (Original post by Im_a_cyborg)
    Yes, no problem, from the diagram of the toy, sin(20) gives the centripetal force (towards the centre) so sin(20)=mv^2/R and
    cos(20)= mg (the weight)

    You cannot use sin(20)=mv^2/R since you do not have the mass of the car, so you combine mv^2/R with mg. 'm" cancels leaving v^2/gr which is equal to tan(20) and get 'r' from that

    Its hard to explain but if you remember;
    sin(angle)=mv^2/R
    cos(angle)= mg
    tan(angle)=v^2/gr

    You can't go wrong.
    Hope that helps.
    Yeah i know sin/cos is tan. Brilliant! Just didnt see where the dividing by g came from but of course now i do.
    Thanks. much appreciated.
    So for the first part is it safe to assume for these q's anything in the car experiences the same centripetal force as the car itself?
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    (Original post by TheStrokess)
    Yeah i know sin/cos is tan. Brilliant! Just didnt see where the dividing by g came from but of course now i do.
    Thanks. much appreciated.
    So for the first part is it safe to assume for these q's anything in the car experiences the same centripetal force as the car itself?
    Yes I think so, as everything in the car is traveling at the same speed and direction.

    No problem.
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    I got 205 for this answer, can anyone tell me what i've done wrong, i think the mark scheme may be incorrect

    v^2/g x tan(20)

    30^2 = 900
    g=9.81
    tan(20) = 2.24

    r = 900 x 2.24 / 9.81 = 205.2m
 
 
 
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