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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC_W07D1-3 Bead on a Track Solution
A small bead of mass m is constrained
to move along a frictionless track. At the
top of the circular portion of the track of
radius R , the bead is pushed with an
unknown speed v0 . The bead comes
momentarily to rest after compressing a
spring (spring constant k ) a distance
x f . The magnitude of the acceleration
due to the gravitational force is g . What
is the direction and magnitude of the
normal force of the track on the bead at
the point A, a height R above the base
of the track? Express your answer in
terms of m , k , R , x f , and g .
Solution:
We will use conservation of energy to find the kinetic energy of the bead at point A. We
will then use Newton’s Second Law, to derive the equation of motion for the bead when
it is at point A. Specifically, we will find the speed v1 in terms of the gravitational
constant g and the track radius R . We will then combine these results to find direction
and magnitude of the normal force of the track on the bead at the point A.
Choose for the initial state the instant when the bead is at point A. Choose for the final
state the instant the bead is momentarily at rest compressing the spring at the end of the
track. Choice of Zero for Potential Energy: choose the gravitational potential energy to be
zero at the bottom of track.
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Initial Energy: The initial kinetic energy is K1  (1/ 2)mv12 . The initial potential energy is
non-zero, U1  mgR . The initial mechanical energy is then
E1  K1  U1  K1  (1 / 2)mv12  mgR .
(1)
Final Energy: The final kinetic energy is K f  0 . The final potential energy is non-zero,
U f  (1 / 2)k x 2f . The final mechanical energy is then
E f  K f  U f  U f  (1 / 2)k x 2f .
(2)
Non-conservative Work: Since we are assuming the track is frictionless, there is no nonconservative work.
Change in Mechanical Energy: The change in mechanical energy is therefore zero,
0  Wnc  Emech  E f  E0 .
(3)
Mechanical energy is conserved, E f  E0 , or
(1 / 2)k x 2f  (1 / 2)mv12  mgR .
(4)
From Equation (4), the kinetic energy at point A is
(1 / 2)mv12  (1 / 2)k x 2f  mgR
(5)
At point A, the forces on the bead are the gravitational force of magnitude mg pointing
downward and the normal force of magnitude N pointing inward. Note that N must
point inward because that is the force that is responsible for the inward acceleration.
Newton’s Second Law in the radial direction, is
2
N  
mv12
R
.
(6)
We can rewrite Equation (6) in terms of the kinetic energy as
(1 / 2)NR  (1 / 2)m v12 .
(7)
Combing Equations (5) and (7) yields
(1 / 2)NR  (1 / 2)k x 2f  mgR .
(8)
Thus the magnitude of the normal force at point A is
N  k x 2f / R  2mg .
(9)
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