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Bead Sliding on Wire
A vertical hoop supports a wire which is attached from the top of the hoop to any other point. Illustrate that the time required for a frictionless bead to slide down the wire is the same for any destination point. A relevant kinematics equation is
x = v0t +1/2at2.
The acceleration is g cos θ. Since v0 is zero we can write
x =1/2(cosθgt2).
But x = 2Rcos θ from geometry. So
2Rcosθ =1/2(cosθgt2)
2R =1/2(gt2)
And,
t = 2√R/g,
Which is independent of the angle θ, and depends only on the radius of the hoop and the acceleration due to gravity.
Note that the problem and its solution is unchanged if one end of the wire is connected to the bottom of the hoop instead of the top.
Sir James Jeans, in his remarkable book "An Elementary Treatise on The-oretical Mechanics" noted that the solution suggests an interesting mini-mization problem. Namely, where to place a wire from a fixed point to aninclined plane such that the time for a bead to slide from the point to the plane is a minimum?
The practical form of the solution is to configure a vertical hoop in a plane at a 90 degree angle to the ramp with its top at the fixed point P and to adjust its diameter until it just touches the
ramp at point T. A wire from the top to the point of tangency will provide the least time. Why? Because every other path from the point will touch the hoop at the same time, but the wire chosen is the only one which will have reached the plane in this time.
It is the difference among potential at two points of a conductor. Please note that by potential(at a point A) ,we actually mean potential difference among point A and another
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