1Q20.50 Hinged Stick and Ball
Abstract
A plank, with one end hinged so that it acts like a pivot, undergoes angular acceleration when the free end
is released from rest at some inclination. For certain angles of inclination, the acceleration at the free end of the
plank is greater than the acceleration due to gravity. In this demonstration, a ball is balanced atop the free end.
When the end of the plank is allowed to fall from rest, the plank “falls away” from the ball, while the ball falls
with the usual acceleration due to gravity. The ball falls into a cup that is located at a different point on the
plank, showing that the end of the plank must have had a greater acceleration than the ball.
Picture
Safety Concerns
None.
Equipment
• Hinged board apparatus
• Supporting stick (Dowel)
• rubber ball
• metal ball
Procedure
First, adjust the foam pad inside the cup so that the foam is about a quarter of an inch (or a little less than
1 cm) above the bottom of the cup. This helps minimize the rebound of the falling balls. Next, prop the board
using the dowel. Note that the ends of the dowel are slanted, so that there is an obvious directional orientation
1
y
x̂
l
xcup
l/2
n̂
Mg
z
θ
Table
Figure 1: Diagram of the apparatus with the plank raised by an angle θ with the ball set and the co-ordinate
system labeled.
for the dowel between the two boards. There are small circular marks on the boards that show where the dowel
should be placed. Place one of the balls atop the golf tee.
Remove the prop with a quick motion of the hand. The ball will fall into the cup, though it may sometimes
bounce out. Either the rubber or the steel ball can be used with equal success.
Theory
Consider a plank that is able to pivot about one end, with the other end free to move. As in Figure 1, the free
end of this plank is raised a certain angle θ above a table, with a ball set on the raised end. The plank is then
released. All points on the plank (except those at the pivot) will follow circular trajectories, each with a different
radius. The linear velocity of each point, directed tangentially to the circular path of the point, is then dependent
upon its distance from the fixed end of the plank. The ball, however, falls independently of the plank.
It is of interest to find the acceleration of various points on the plank. First, consider the rotational dynamics
of the system while the plank is in motion. Once the plank is released from its rest position, the two forces acting
on it are gravity and a normal force. The latter is exerted by the table at the pivot, so it does not exert a torque
about the pivot. The gravitational force does exert a torque on the plank about the pivot, giving the net torque
τnet on the plank as
τnet =
1
(M gl cos θ) k̂,
2
(1)
where M is the mass of the plank, g is the acceleration due to gravity g = 9.81m/s2 , l is the plank’s length
and θ is the angle between the raised plank and the table. The appropriate unit vector is found by applying the
right-hand rule. Note that, as in Figure 1, the x-axis is fixed to the plank, so its direction changes in time. The
y- and z-axes are stationary with respect to the table.
The rotational equivalent of Newton’s second law applies to the plank’s motion, namely
τnet = Iα,
(2)
where I is the plank’s moment of inertia and α is the angular acceleration of the plank (equal, in a given instant,
at each point on the plank). Approximating the plank to be uniform and thin, the moment of inertia about the
pivot is
1
I = M l2 .
(3)
3
2
Next, the linear acceleration a (x, θ) of a point on the plank that is at a distance x from the pivot is given by
a = αxn̂,
(4)
where n̂ is a unit vector perpendicular to the x-axis, in the direction of the plank’s motion. By inspection, it is
possible to express Equation 4 as
a
α = ı̂ × ,
(5)
x
where the unit vector and cross-product are found by applying the right-hand rule. Observe that substituting
Equation 4 into Equation 5 will give the correct vector expression for α.
Substituting Equations 1, 3, and 5 into Equation 2 yields
1
1
a
2
,
(6)
(M gl cos θ) k̂ =
Ml
ı̂ ×
2
3
x
from which it is possible to solve for a, with the result being,
3gx
(cos θ) n̂.
(7)
2l
Finally, it is of interest to find the y-component ay of the acceleration given in Equation 7. This is given by,
a=
ay = |a| (cos θ) ̂
3gx
=
cos2 θ ̂.
(8)
2l
Consider Equation 8. Let x = l, so that the point of consideration is the moving end of the stick, which
coincides with the location of the ball at initial time t = 0. The end of the stick then has an acceleration of
greater magnitude than g (i.e. |a| > g) for certain angles that are less than some critical angle. The critical angle
is given by θ that satisfies
2
cos2 θ = ,
(9)
3
that is, θ ≈ 35.3◦ . The angle of inclination must be less than 35.3◦ in order for the end of the plank to fall with
an acceleration greater than g.
In this demonstration, the condition that must be met for the ball to fall into the cup is that the time taken for
the cup to fall to the table, tcup , must be less than the time taken by the ball, tball . A precise approach requires the
solving of elliptical integrals, as discussed by Theron, 1988. Applying this method to the demonstration apparatus,
a time of tcup = 0.280 s is calculated. This method takes into account the changing angular acceleration of the
rod. However, if we assume that the angular acceleration is approximately constant, then basic kinematics gives
a good approximation for tcup , differing from the precise value by only about two hundredths of a second. The
details of the kinematics approach follow.
Integrating Equation 8 twice with respect to time gives the vertical component of the displacement of a point
on the plank at a distance x from the pivot. Solving for time in this expression yields
2h
ay
s
4
hl
1
=
·
·
2
3 g cos θ x
tplank =
(10)
where h is the initial height of the point on the plank. In this demonstration, the length of the plank is l = 99.3
cm, and the angle of inclination is θ = 29.5◦ . Consider the point of interest, namely the cup into which the ball
will fall. The x-coordinate of its centre is xcup = 84.9 cm, and its initial height is h = 40.5 cm. Accounting for
the length of the golf tee, the ball starts at a height of hball = 53.0 cm. Using these values, we can compare the
falling times for each the cup and the ball. For the cup,
s
4 (0.405m) (0.993m)
tcup =
3 (9.81ms−2 ) (cos 29.5◦ ) (0.849m)
(11)
= 0.292s,
3
while the time it takes for the ball to fall a distance of 53 cm is
2hball
g
r
2 (0.530m)
=
9.81ms−2
= 0.329s.
tball =
The ball takes more time to fall than the cup does, and so it falls into the cup.
4
(12)
References
[1] Cenco Physics. Operating Instructions: Hinged Stick and Falling Ball Demonstrator, No. CP74874-00. SargentWelch, Buffalo Grove (IL), December 2000.
[2] G. D. Freier and F. J. Anderson. ”My-6. Falling Chimney”, A Demonstration Handbook for Physics. American
Association of Physics Teachers, 2002. pg M-64.
[3] W. A. Hilton. ”Free Fall Paradox”. The Physics Teacher, Vol 3, No. 7, October 1965. pg 323.
[4] R. M. Sutton. ”M-206. Falling Chimney - Free-fall Paradox”, Demonstration Experiments in Physics. McgrawHill Book Company Inc, New York and London, 1938. pg 89.
[5] W. F. D. Theron. ”The Faster than Gravity Demonstration Revisited”. American Journal of Physics, Vol 56,
No. 8, August 1988. pg 736.
[6] W. M. Young. ”Faster than Gravity!”. American Journal of Physics, Vol 52, No. 12, 1984. pg 1142.
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