A bead can slide without friction on a wire that is bent into a circular loop of
ID: 2140732 • Letter: A
Question
A bead can slide without friction on a wire that is bent into a circular loop of radius R. Typically the wire loop is in a vertical plane and is rotated steadily about its vertical diameter with a period T; however this time this activity is done while standing on an incline banked at an angle x relative to the horizon. If the position of the bead is described by the angle v that the radial line from the center of the loop to the bead makes relative to the vertical axis of the loop-- i.e., the axis is aligned with the handle, is there at least one angle v for which the bead can remain motionless relative to the wire loop.
Answer using equations to prove.
Explanation / Answer
When the bead is at an angle theta from the bottom of the loop, there are two forces acting on the bead. One is ---the force of gravity, which points downward with magnitude mg. The other is the normal force of the wire, N, which points toward the center of the loop.
When the bead is at its equilibrium position, it has no net vertical acceleration (ay=0). Its horizontal acceleration is the acceleration due to motion in a circle about the axis of the loop's rotation: ax=r*omega^2, but we must be careful defining r. Lets define R as the radius of the loop of wire. Then r=R*sin(theta) is the radius of the circular motion of the bead. The bead is always a distance R from the center of the loop, but the radius that matters here is the distance from the axis of rotation to the bead, r.
The way you can see this is to draw a picture of the loop from the side. Pick a point for the bead, and label the angle theta from the bottom of the loop to the bead. You can form a right triangle by drawing a line from the center of the circle to the bead of length R, then drawing a line from the bead horiontally over to the axis of rotation, then drawing a line straight up back to the center of the loop. The length of the horizontal side of the triangle is r=R*sin(theta). If you now picture the diagram from the top, you'll see this is also the radius of the bead's rotation about the axis.
Now we use Newton's second law: the sum of the forces in a given direction is equal to the mass times the net accelertion in that direction. F=ma.
In equilibrium, the vertical component of N minus the downward pull of gravity equals the vertical acceleration, which is zero:
Ny-m*g=m*ay=0
so Ny=m*g.
The horizontal component of N is equal to the centrepital "force":
Nx=m*ax=m*r*omega^2
It is also true that
Nx=N*sin(theta)
and Ny= N*cos(theta)
since these are just the horizontal and vertical components of the normal force.
If you plug these in for Nx and Ny, and plug in r=R*sin(theta). You get that
N*cos(theta)=mg
N*sin(theta)=m*R*sin(theta)*omega^2
Solve the first equation for N and plug it back into the second equation. After simplifying, you get that
cos(theta)=g/R/omega^2
so theta=arccos(g/R/omega^2).
Omega is the angular speed of the loop.
omega=2*pi/T, where T is the period.
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