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Acceleration of a bead sliding down a Helix

  1. Mar 4, 2005 #1
    Can a bead desecnd down a helix have zero vertical component for its acceleration??? (i.e. If I found my result to only has coefficients for its horizontal radial component unit vector (e_r), but zero for that of the vertical unit vector (e_z) !! ... That means I got zero vertical component for its acceleration right? Otherwise, please define the unit vectors <e_r, e_phi, e_z> for me please...?)
    Is it possible for the bead to still have its descend down a vertical helix if it has zero component for its e_z unit vector's direction? Note that it is an HELIX... so perhaps the radial component alone, something like centripetal accel, or something like that, can pull it down to allow its downward descend still?? or no...? Why???

    (See below for the question & how I tried to approach it) :


    A ball is constrained to slide down a frictionless helix path, with the helix axis to be vertical. The radius of the helix is R and each of the successive circular motion down the helix (distance of vertical drop after going round one full circular motion), is, 2pi*b.

    The helix can be described by cylindrical coordinates: r=R, z=b*phi.

    a) Write out the position vector and velocity vector.

    b) Assuming that the bead is released from rest and allowed to determine freely along the helix, using convervation of energy to determine the time dependence of the angle theta and the vertical component of accelration.


    I wrote out what represents the position and velocity vector, then used the conservation of energy equation, and integration to find angle phi, as shown below:

    First, note unit vectors = <e_r, e_phi , e_s>, and d(e_r)/dt=e_phi*d(e_phi)/dt, and also d(e_phi)/dt = - [e_r*d(e_phi)/dt].

    We found velocity vector to be v= R*(d-phi/dt)*e_r + b*(d-phi/dt)*e_z.

    Thus, we also know that: acceleration vector = -R*(d-phi/d)*e_r + b*(d-phi^2/dt^2)*e_z

    Now, using the conservation of energy equation of a ball falling from rest,
    change of KE = change of PE, we get:

    Substituting the speed found from velocity vector: v=(d-phi/dt)*(R^2+b^2)^(1/2),
    I get d-phi/dt= a bunch of constants.

    So separating variables and integrating the resulting expression, we get phi:

    phi=(sq.root[(4*pi*g*b/(R^2+b^2) ]*t (:confused:)

    Taking the derivatives of phi twice with respect to time t, (which fits into the acceleration vector e_z coefficient component), however,
    gives us zero.
    i.e. d^2(phi)/dt^2 = 0 :eek: :eek: :confused:

    Can this be right??

    i.e. The acceleration vector= a= -R*((d-phi/dr)^2)*e_r + b*0*e_z = -R*((d-phi/dr)^2)*e_r ... only!! :eek:

    i.e. the accel vector only have its coefficient for the radial unit vector (e_r), but ZERO coefficient for its vertical unit vector (e_z). :eek: :eek:

    Is this possible??

    Does my result mean the vertical component of acceleration is zero? (What are the definitions for the unit vector e_r, e_z etc?) Is there something wrong with my set up? How then should I do it?

    Please comment on the approach i used, whether i have missed some concepts, and suggest an alternative better approach. Thank you!!
    Last edited by a moderator: Mar 4, 2005
  2. jcsd
  3. Mar 14, 2005 #2
    You shouldn't be too confused by the helix shape. All that matters is the inclination of the helix thread i.e. the 'pitch angle'. As the latter is constant along the helix it should therefore be sufficient to consider a bead constricted to a straight line with the same inclination angle.

    The pitch angle of the helix is

    alpha = arctan[b/(2*PI*r)] ,

    where b is the pitch (vertical drop between full circles) and r the radius of the helix.

    The acceleration of the bead along the helix is thus

    a = g*sin[alpha] ,

    with g the gravitational acceleration.

    Projecting this again back onto the vertical gives the vertical acceleration

    av = g*sin[alpha] *sin[alpha] = g * sin^2[alpha]


    z= b*phi(t) = 1/2*av*t^2

    you have therefore the time dependence of the angle

    phi(t) = 1/2/b *av *t^2 .
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