A single bead can slide with negligible friction on a stiff wire that has been b

Question

A single bead can slide with negligible friction on a stiff wire that has been bent into a circular loop of radius 11.3 cm, as in the figure below. The circle is always in a vertical plane and rotates steadily about its vertical diameter with a period of 0.475 s. The position of the bead is described by the angle ? that the radial line, from the center of the loop to the bead, makes with the vertical.
(a) At what angle(s) up from the bottom of the circle can the bead stay motionless relative to the turning circle? (Select all that apply.) 90

Explanation / Answer

The other answer posted in response to this question is wrong, except in a minor technicality that I will explain at the end. Here is the right 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. With your numbers in part a, omega=14.0 rad/s. theta=arccos[(9.8 m/s^2)/(0.15 m)/(14.0 /s)^2] =arccos(0.33) =70.5 degrees. Now for a fairly pedantic point: It is true that a bead in the 6 o'clock position on a wire experiences no forces to move it away from the bottom; however, this is not a stable state. If the bead moves even a tiny bit away from exactly the bottom, the bead will be accelerated by the rotation of the loop. The bead's innertia will carry it outward and the normal force of the loop on the bead will push it upward, until it reaches the point where the centrepital acceleration is just the right amount to be supplied by the horizontal component normal force. This will be at some angle theta above the bottom, with theta between zero and 90 degrees

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