Physics Friday 142

Consider a small bead of mass m sliding frictionlessly down a wire in the shape of a helix, with a vertical distance d=2πc between coils and a radius R, aligned along a vertical axis. If the bead is released from rest at t=0, find, as functions of time, the distance s(t) travelled along the helix by the bead, the speed v(t) of the bead, and the magnitude F(t) of the (normal) force exerted on the bead by the wire.

Only two forces act on the bead, its weight and the force exerted by the wire. Only the former has a component tangential to the wire. Now, the wire makes a constant angle θ from the horizontal, with . Then the component of the net force tangent to the wire is , for a tangential acceleration of . This is the time derivative of our speed, so integrating with v(0)=0, we have ,
and integrating again,
.
Note that these are the same as for a straight wire inclined at an angle .

Now, we need only find the normal force. Here, we need to work in three dimensions. For the bead at a given moment, we have three mutually orthogonal directions, the Frenet frame, given by the Frenet-Serret formulas:
We have the tangent vector T, which points along the wire; the normal vector N, which points inward to the helix, and has no vertical component; and the binormal vector , perpendicular to both. Choosing a downward tangent, we then see that B is downward normal to the wire.
The weight has no component along the normal vector; its other components are
(as found earlier), and .
Similarly, our force of constraint has no tangent component, but we do have both
FN and FB.
The net acceleration is a combination of the tangential (aT) and centripetal (aN) accelerations.
Thus, the net force has no binormal component, and so
.
The centripetal acceleration is
, where v is the speed, and is the radius of curvature (not the radius R of the helix). As the curvature of a helix is , the radius of curvature is
,
and we have

Thus, the normal component of the net force is this times the mass, and since the weight has no normal component, we have
.
Thus, the magnitude of the force the wire exerts is
,
and plugging in for ,
we get
.

Note that if you take the limit as , then , and , since the helix approaches a horizontal wire in that limit.
Similarly, taking the limit gives and , since in that limit, we approach the bead falling down a vertical wire. Lastly, taking and while holding the ratio (and thus the angle ) constant, we get
, which is the result for a straight wire at an angle θ.

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4 Responses to “Physics Friday 142”

  1. Dr.D Says:

    Interesting solution for what is basically an elementary dynamics problem with the complication of three dimensions. Well done!

  2. Physics Friday 143 « Twisted One 151's Weblog Says:

    […] decelerate, coming to a stop somewhere along the wire. Now, let us take the bead on a helix from last week, and add a friction force, with a coefficient of kinetic friction μ. Here, our constraint force […]

  3. Shouvik Says:

    Hi….
    none of the formulae in your post are appearing.
    Is it a problem with my browser? I am using Chrome.
    Do I need some add-on for them to be displayed properly?

    • twistedone151 Says:

      It’s not you. The webpage/server that I used for generating images from LaTeX is no longer up. I’m in the process of slowly swiching the old posts over to WordPress’s LaTeX renderer, but certain bits (mainly arrays) are handled differently, so it’s not a straightforward process. Eventually, they’ll be back.

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