Teacher Notes
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Galileo’s Paradox—Hinged Stick vs. Falling BallDemonstration Kit![]() IntroductionIn the absence of air resistance, all free-falling objects, regardless of their mass will accelerate toward the Earth at the same rate. This “law of motion” was proposed by Galileo Galilei (1564–1642) and later confirmed by Isaac Newton (1642–1727). However, in this discrepant event, a falling stick appears to break the laws of motion as it “falls” faster than a ball. Concepts
Safety PrecautionsAlthough this activity is considered non-hazardous, please follow normal laboratory safety guidelines. Prelab Preparation
Procedure
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Correlation to Next Generation Science Standards (NGSS)†Science & Engineering PracticesAsking questions and defining problemsDeveloping and using models Planning and carrying out investigations Analyzing and interpreting data Obtaining, evaluation, and communicating information Disciplinary Core IdeasMS-PS2.A: Forces and MotionMS-PS3.A: Definitions of Energy HS-PS2.A: Forces and Motion HS-PS2.B: Types of Interactions HS-PS3.A: Definitions of Energy HS-PS3.B: Conservation of Energy and Energy Transfer HS-PS3.C: Relationship between Energy and Forces Crosscutting ConceptsSystems and system modelsEnergy and matter Structure and function Stability and change Performance ExpectationsMS-PS4-2: Develop and use a model to describe that waves are reflected, absorbed, or transmitted through various materials. DiscussionThis demonstration reveals an apparent contradiction to Galileo's original observations—all masses fall with the same acceleration. However, a careful examination of the apparatus shows that the stick and ball are not falling in the same manner. The falling stick is not truly in "free-fall" motion. The stick is attached at one end. {12591_Discussion_Figure_6}
However, since the stick is attached to a support stand by a hinge, the stick rotates about its attached end as it falls, instead of falling straight down as the ball does. Therefore, the stick falls with an angular acceleration (αs) instead of linear acceleration. Angular acceleration is the acceleration of a rotating object. Angular acceleration (αs) and tangential linear acceleration (at) are proportional according to Equation 1 (see Figure 7).
{12591_Discussion_Figure_7}
{12591_Discussion_Equation_1}
at = linear acceleration {12591_Discussion_Equation_2}
αs = angular acceleration of the stick {12591_Discussion_Equation_3}
Equation 3 shows that any point on the board beyond the center of mass (r > rcm) will have a linear downward acceleration that is faster than the center of mass (faster than the acceleration due to gravity, g). Points on the hinge-side of the center of mass (r < rcm) have a slower downward acceleration than the center of mass. It is important to remember, however, that all the points on the stick still have the same, constant, angular acceleration, αs. Therefore a cup placed beyond the center of mass of the stick falls with a faster acceleration than a free-falling ball. When both the cup and the ball are released from the same height, the cup will beat the ball to the benchtop.There is a limit to the angle at which the stick can be released and still beat the ball. Unfortunately, this requires a thorough investigation of torque and angular acceleration which is explained in Further Mathematical Discussion. Simply stated, for initial angles between 0 and 35º, the end of the board will always beat the ball down when released from the same height. At angles greater than 35º, the ball will hit the ground first (as long as it does not hit the board as it falls). Further Mathematical Discussion To determine the “break-even” angle, the torque on the rotating stick produced by the pull of gravity must be analyzed. Torque, τ, is the “force-equivalent” for a rotating system and is equal to the moment of inertia of the system, I, multiplied by the angular acceleration, α. {12591_Discussion_Equation_4}
This is similar to the linear equation of force, F = ma. A torque produced by a force acting on a rotating lever arm is equal to the force multiplied by the distance between the pivot point and the position where the force acts, multiplied by the sine of the angle between the applied force and the lever arm (Equation 5). See Figure 8.
{12591_Discussion_Equation_5}
{12591_Discussion_Figure_8}
For a rotating stick pivoting at one end, the moment of inertia equal to (mL2)/2 – 43, where L is the length of the stick. The torque about the center of mass of the stick (L/2) is, therefore, given by Equation 6.
{12591_Discussion_Equation_6}
Solving for the angular acceleration gives
{12591_Discussion_Equation_7}
Referring to Equation 1, the tangential linear acceleration at the end of the stick is equal to the distance from the pivot point (the length of the stick, L) multiplied by the angular acceleration.
{12591_Discussion_Equation_8}
The vertical component of the tangential acceleration (avert) of the end of the board is equal to at multiplied by cos ϴ (see Figure 9).
{12591_Discussion_Figure_9}
{12591_Discussion_Equation_9}
Note: sin δ is equal to sin (90 – ϴ), which also equals cos ϴ.For the special case when the end of the stick has the same vertical acceleration as the falling ball, avert = g, the initial angle of the board is equal to: {12591_Discussion_Equation_10}
As the stick's release angle increases, it can be seen that the initial movement (rotation) of the board is slower compared to the quickness at which the ball falls. For small initial angles, the faster acceleration of the end of the stick is large enough to overcome its slow start. This is not the case for larger angles. The “break-even” initial angle at which the end of the stick falls with the same vertical acceleration as the ball is 35°. The end of the stick beats the ball when the initial angle is smaller than 35°. If the initial angle of the stick is larger than 35°, the end of the stick will not beat the ball to the benchtop. During the demonstration, it may be observed that the ball will not lose contact with the board when the initial angle is greater than 35°, so it may appear that the ball and board fall with the same acceleration. The ball is actually forcing the stick to fall with the same acceleration as the ball rides above it.
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