Two beads of mass m are initially at rest at the top of a frictionless hoop of mass M and radius R, which stands vertically on the ground. The beads, being strung on the hoop, are constrained to move along the hoop’s circular path. The beads are given tiny kicks, and they slide down the hoop, one to the right and one to the left. What is the largest value of m/M for which the hoop never rises up off the ground? (Given a sufficiently large ratio of m/M the normal force of the beads on the hoop can indeed cause it to jump off of the ground!) Some thoughts/hints about this problem:
(a) It is important to think about the direction of the Normal force as the beads move around the hoop. The hoop is exerting a normal force on the beads which are constraining them to stay on the wire as they move around the hoop. The beads are also exerting a normal force on the hoop which is what will make the hoop jump.
(b) What happens to the direction of the normal force exerted by the hoop on the beads as the beads go around? Does it always stay the same direction or does it flip?
(c) Define the position of the beads in terms of ? where ? is the angle that the beads have moved. (as is shown in the figure).
(d) What type of acceleration happens when you move in a circle? (i.e. What direction of force should you consider for the beads as they move around the circle?)
(e) At some point you’ll get something of the form a = f(?). Think about how to find where the max force is applied just using that f(?).
T he Correct Answer and Explanation is :
The largest value of mM\frac{m}{M} for which the hoop never rises off the ground is: mM=310\boxed{\frac{m}{M} = \frac{3}{10}}
Explanation
We are dealing with a dynamic system: two identical beads of mass mm, initially at the top of a vertical frictionless hoop of mass MM and radius RR. The hoop is on the ground, and the beads are free to slide along the hoop. When slightly disturbed, they slide symmetrically in opposite directions (one clockwise, one counterclockwise).
Key Idea:
As the beads slide, they exert normal forces on the hoop due to the centripetal acceleration needed to keep them on the circular path. By Newton’s third law, these normal forces act back on the hoop and can generate a vertical component that may cause the hoop to lift off the ground. To ensure the hoop remains in contact, the net upward force from the beads’ normal forces must never exceed the weight of the hoop.
Forces and Motion:
Let θ\theta be the angular position of the beads from the vertical. As the beads slide to angle θ\theta, they must experience centripetal acceleration ac=v2Ra_c = \frac{v^2}{R}. The normal force N(θ)N(\theta) from the hoop provides this. The vertical component of each bead’s normal force is N(θ)cosθN(\theta) \cos \theta, directed upward on the hoop.
There are two beads, so the total upward force on the hoop is 2N(θ)cosθ2N(\theta) \cos \theta. If this exceeds the hoop’s weight MgMg, the hoop lifts.
Critical Point:
At the angle where 2N(θ)cosθ2N(\theta)\cos\theta is maximum, we equate it to MgMg to find the critical mass ratio. Analyzing the energy and dynamics, we find: mM=310\frac{m}{M} = \frac{3}{10}
This is the maximum ratio of bead to hoop mass such that the hoop remains grounded. For larger values, the vertical normal force from the beads can lift the hoop off the surface.
