a small bead can slide without friction on a circular hoop that is in a vertical plane and has a radius of 0.100 m. The hoop rotates at a constant rate of 3.00 rev/s about a vertical diameter. Find the angle beta at which the bead is in vertical equilibrium. what would happen if the hoop rotates at 1.00 rev/s?

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  1. The bead is accelerating on the horizontal plane in a circle with radius r=RsinB. This is derived from the angle that the bead makes with the vertical. Once you have the radius of the particles trajectory you need to find the forces acting on it. The two forces acting on the bead are the force of gravity and the normal force the hoop exerts on the bead. The normal force is simply the force that is pointing towards the centre of the vertical hoop. Once you draw your FBD you can see now that the Force that's causing acceleration on the horizontal axis is simply NsinB. Using Newton's Second Law of motion, (F=ma) NsinB=ma. Acceleration of a moving body in a uniform circle is a=v^2/r. Therefore, by substitution, NsinB=mv^2/R. Now to find N we go back to our FBD. NcosB=mg, thus, N=mg/cosB. Subbing it back into the original equation we arrive at, gtanB=v^2/r. For v, v=2πr/T. Subbing that into the equation we arrive at, gtanB=(4π^2(RsinB)/T^2). r=RsinB. Rearrange, then find that cosB=(gT^2)/(4π^2)R). T=1/3s, So B= 74 degrees. To find what happens if the hoop rates at 1.00 rev/s we take a look at the equation, tanB=(4π^2R/gT^2)sinB. We notice that if the term inside the brackets is above one we can receive a numerical value for B other than 0. However, if T=1s, we cant find a value, therefore the only way to equate this is to have B=0. Therefore is bead stays stationary at the bottom.

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