Determining the Acceleration Due to Gravity through Experimentation, Study notes of Physics

Download Determining the Acceleration Due to Gravity through Experimentation and more Study notes Physics in PDF only on Docsity! 1 The Acceleration Due to Gravity Introduction: Acceleration is defined as the rate at which the velocity of a moving object changes with time. Accelerations are always caused by forces. In this laboratory we will investigate the acceleration due to the force of gravity. Theory: In its simplest form, Newton's law of force relates the amount of force on an object to its mass and acceleration. F = m a (1) or force = mass times acceleration. Therefore, to impart an acceleration to an object, one must impart a force. One of the most obvious (and the weakest) of all forces in nature is the gravitational force. Newton's Universal Law of Gravitation describes the gravitational force (Fg) as follows: Fg = Gmm' r 2 (2) This equation states that the force between the two masses m and m' is equal to the product of their masses (mm' ) multiplied by a constant (G ) and divided by the distance between them squared (r 2 ). The constant (G ) is called the gravitational constant. To compute the gravitational force between the Earth and an any object, we substitute the mass of the Earth (ME) and the distance from the object to the center of the Earth (r ). When the objects are on or near the Earth's surface, this distance can be approximated by the value for the radius of the Earth* (RE) so that Equation (2) becomes: Fg = GmME RE2 (3) * We can approximate this because in the scale of the size of the Earth (many hundreds of kilometers) the value for r at the top of our lab table is virtually equal to the value of r at the floor. Algebraically this is shown by: RE ≅ RE + ∆R 2 in which we see that the force only depends on the mass of the object, because G, ME, and RE are all constants . This force (measured at the Earth's surface) is called the weight of the object. Now looking at Equation (1) and equating F to the gravitational force (Fg), we see that: ma = GmME RE2 = mg , (4) where g = GME RE2 . In this last equation, we see that g, the gravitational acceleration, is itself a constant because it depends on quantities which do not change with time. This result was first demonstrated by Galileo when he dropped cannonballs of different masses (weights) from the Leaning Tower of Pisa to show that although they had different masses, when dropped together, they landed together. This happened in this manner because they both experienced the same acceleration. A similar experiment may also be performed by dropping a coin and a feather. When dropped in air, the coin always lands first, but when they are dropped in a vacuum, an environment where there is no air, they land together! In the coin and feather case, the different velocities are due to another force called air friction. Our Equation (4) equates the total force to the gravitational force, and therefore neglects the effects of air friction. We must now try to discover a quantitative method for determining the gravitational acceleration, g. We first look to the equation for one-dimensional motion (motion in one direction) under constant acceleration. x(t) = xo + vo t + (1/2) at 2 (5) In Equation (5), the motion is described by the function x(t) which is position as a function of time. On the right, we have the parameters xo , vo , and a. These are, respectively, the initial position, the initial velocity and the acceleration. In this experiment, like Galileo, we will be dropping an object from rest, so that vo will be zero. If we call the place from which we drop the object to be x = 0, then xo will also be zero. with ∆R = height of the lab table. 5 The Acceleration Due to Gravity Name:___________________________ Abstract: Data: tennis ball 1 trial X t t 2 g σ = |gave-g | 1 2 3 4 mass of the ball σave Data: tennis ball 2 trial X t t 2 g σ = |gave-g | 1 2 3 4 mass of the ball σave 6 Data: tennis ball 3 trial X t t 2 g σ = |gave-g | 1 2 3 4 mass of the ball σave Data: tennis ball 4 trial X t t 2 g σ = |gave-g | 1 2 3 4 mass of the ball σave Data: ping-pong ball 1 trial X t t 2 g σ = |gave-g | 1 2 3 4 mass of the ball σave 7 Data: ping-pong ball 2 trial X t t 2 g σ = |gave-g | 1 2 3 4 mass of the ball σave average measured value for g: _______________ accepted value for g:_______ % error:_______ Calculations: Conclusions: Error Analysis and Discussion: