MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC-W08D2-10 Bead and Ring Solution
A ring of mass mr hangs from a thread, and two identical beads of mass mb slide on it
without friction. The beads are released simultaneously from the top of the ring and slide
down opposite sides.
a) Draw free body force diagrams for the ring and the beads. What direction is the
force of the bead on the ring pointing? Does it change has the bead moves. Can
you still proceed with an analysis using Newton’s second Law if you are not sure
which way this force points? Try to find a physical explanation for the direction
of this force.
b) Show that the ring will start to rise if mb > (3 / 2)mr , and find the angle θ with
respect to the vertical direction that this occurs.
Solution: Given that there is no friction, the contact force between the ring and either
bead must be normal to the ring. Denote the radial normal force that the ring exerts on
either bead as N1 and the radial normal force that either bead exerts on the ring as N 2 .
The forces on the bead and the ring are shown in the figure below.
By Newton’s Third Law, the magnitude of the normal forces are equal
N 2 = N1
(1)
Initially, with the beads at rest or moving very slowly at the top of the ring, the ring
exerts an upward force on each bead and the beads exert downward forces on the ring;
then the direction of N1 and N 2 are opposite the directions shown in the figure above.
We expect that as the beads gather speed, there must be an inward force to maintain
motion along the ring. If mb = mr , so that the rising of the ring is not an issue, we know
that at the bottom of the ring there must be an inward force on the beads, the directions of
N1 and N 2 are as shown in the figure above; thus we should expect that the direction of
the force, and hence the directions of N1 and N 2 will change as the beads move.
If used properly, the algebraic sign of any component of force in Newton’s Second Law
should only reflect a direction of the force. We may have to adjust our interpretation of
the physical situation; our algebraic solution should give us some insight.
b) With the above convention, the radial component of the force on either bead will be
Fradial = −mb g cos θ − N1 ,
(2)
with the angle θ that shown in the above figure. The radial component of the force must
be equal to the product of the radial component of the acceleration and the bead’s mass,
−mb g cos θ − N1 = −mb
v2
,
R
(3)
where v is the speed of the bead (which will of course vary as the bead moves along the
ring) and R is a reasonable choice for the radius of the ring. Thus
N1 = mb
v2
− mb g cosθ
R
(4)
In Equation (3), the speed of the bead may be found from conservation of energy. Choose
U g = 0 at the top of the ring. Then the initial mechanical energy is
Ei = 0 .
(5)
At an angle θ , the bead has fallen a distance Δy = R (1 − cos θ ) , so that the mechanical
energy is given by
E(θ ) =
1
mbv 2 − mb gR(1− cosθ ) .
2
(6)
There are no non-conservative forces,
0 = E(θ ) − Ei =
1
m v 2 − mb gR(1− cosθ ) .
2 b
(7)
The change in kinetic energy is the decrease in potential energy,
1
mbv 2 = mb gR (1− cosθ ) .
2
(8)
This implies that
mb v 2
= 2mb g (1 − cos θ )
R
After some minor algebra, the radial force equation becomes
v2
N1 = mb − mb g cosθ
R
2
mbv
= 2mb g(1− cosθ )
.
R
N1 = 2mb g(1− cosθ ) − mb g cosθ
(9)
N1 = mb g(2 − 3cosθ )
Indeed, this expression shows that N1 will be positive for angles in which cos θ < 2 / 3 .
and negative for angles in which cos θ > 2 / 3 . The change in sign reflects the fact that the
direction of the normal force on the bead is pointing initially outward N1 < 0 and
eventually inward N1 > 0 .
When the ring just starts to rise T = 0 and az = 0 . Therefore the net upward force on the
ring will be equal to zero
2 N 2 cosθ − mr g = 0 ,
(10)
where the factor of 2 reflects the presence of two beads. Using N1 = mb g(2 − 3cosθ )
along with N 2 = N1 leads to
2 mb g(2 − 3cosθ )cosθ = mr g
6 mb cos 2 θ − 4 mb cosθ + mr = 0.
The second equation in (11) is a quadratic in cosθ , with solutions (obtained after a bit of
minor algebra)
(11)
1⎛
3 mr
cos θ = ⎜⎜1 ± 1 −
3⎝
2 mb
⎞
⎟⎟ .
⎠
Thus, for the ring to rise, the expression in the square root must be positive,
2mbead > 3mring .
To find the angle, there’s a bit of subtlety. In Equation (9), we are looking for the smaller
angle so the argument of the cosine must be larger, hence we need to take the positive
sign in Equation (12), so that the ring will rise at the angle
1⎛
3 mring
cos θ = ⎜1 + 1 −
3 ⎜⎝
2 mbead
⎞
⎟.
⎟
⎠
(12)