Ramp Angle and Acceleration ```Name: Kyler Status: student Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: If you were to take a long ramp, about 30 feet long, angle it at a 45 degrees angle to make a right triangle with 45, 45, and 90 degree angles, and roll the ball down it, would it have half the acceleration of a ball free falling at an initial rate of 9.8 meters per second per second? Replies: Acceleration is a funny thing and difficult to understand at first. The first thing I want to point out is the flaw in the statement "an Initial rate of 9.8 m/s^2". The acceleration of a ball being freely dropped is ALWAYS 9.8 m/s^2 (given the only force is due to gravity, no wind or on the moon etc). That means when the ball is first dropped it has a speed of 0 m/s, but a downward acceleration of 9.8 m/s^2. After 5 seconds it will have a higher velocity (speed) but still have an acceleration of 9.8m/s^2. A common physics question is "A ball is shot directly up with an initial velocity of 10 m/s. When will the acceleration be 0 m/s^2? At the beginning, at its highest point or at the end?" The answer is never. The ball will always have a downward acceleration of 9.8 m/s^2. OK...On to the second part with the ramp. The ball will have a constant acceleration along the ramp (the vector of the acceleration is parallel to the ramp) of Sin(45)*g =.707*9.81 m/s^2. If you want to change the angle of ramp you need to use Sin (angle at the bottom of the ramp)*gravity. Try what happens when you have the ramp at 30° You need to be careful when dealing with rolling balls. If you are doing an experiment you probably care more about speed or time it takes for the ball to roll. When something is rolling there are more factors involved such as "Potential energy", "Kinetic Energy", "Acceleration", "Friction" and "Moments of inertia". If you are doing an experiment actually with a rolling ball I suggest you look into these other topics but unfortunately I do not have time to get into all of them. Best of luck and remember to keep wondering about the world around you, it is what makes life interesting. Kevin Hardin Hello Kyler, This is a great example of a question that's deeper than it appears on the surface. You are correct in thinking that a ball going down an incline will accelerate more slowly than one in free-fall, but there are 2 subtle points that must be considered to get the correct answer. First, let us compare a ball in free-fall to one that is sliding down a ramp. In the second case, the acceleration is muted to an extent due to the presence of the ramp. However, for a ramp of 45deg, it is not half. The relation is that the acceleration of the ball on the ramp is equal to the sin function of the angle. The sin(45deg) is about 0.707, so the ball feels an acceleration of 0.707 * g. To get a ball that to experience only 1/2 the acceleration, you need to have sin(angle) = 0.5. That is an angle of 30degrees. So you see for a sliding ball that experiences no friction you do not need as steep of an angle as you were expecting. Now onto the second subtle point. I said sliding above for a reason. Sliding and rolling in fact are two very different things and that adds a level of difficulty to your question! Just as an object moving in a straight line have forward momentum, a rotating object has angular momentum. That is, in the absence of any forces, if an object is rotating, it will continue to rotate at the same rate. If the ball rolls down the incline, then there will a frictional force present which will act to reduce the linear acceleration the ball experiences (though it's now gaining rotational acceleration too). You can think of it as if some of the acceleration due to gravity is being transferred into rotating the ball. Because of that, it is linear acceleration will be less than if it were just sliding down the ramp. Another way to think of it compared to the first example is that above it only took an angle of 30 degrees, but now you will have to raise the ramp a bit to get the ball to experience the same linear acceleration as before because now it is also rotating. The equation describing the acceleration of the ball (including the rotation, assuming it is a uniformly solid sphere) : a = sin(theta) * g * 5 / 7 "a" being the acceleration of the object, "g" being free-fall acceleration due to gravity here on earth, and "theta" being the angle of the ramp. I did my homework and derived the equation for you. So now I will leave it to you to calculate what the angle theta must be in order for the acceleration to be exactly 1/2! You will be in for a surprise as the answer is close (but not the same as) your guess for the original angle. Michael Pierce Kyler, Gravitational acceleration is straight downward, directly toward the floor. Like any quantity that has direction (i.e. a vector), it can always be split into two perpendicular parts. In this case use parallel to the ramp and perpendicular to the ramp. To find how big each part is will require that you draw a right triangle with the hypotenuse (the "total" side) vertical, pointing toward the ground. One of the shorter sides will be parallel to the ramp and one will be perpendicular to the ramp. If the hypotenuse is 9.8 centimeters long, then the length of the parallel side will tell you the acceleration along the ramp. Ratios work well. Measure (acceleration along the ramp)/(total acceleration) equals (length of parallel side)/(9.8 m/s^2). If you prefer using trigonometry, the acceleration triangle 45-45-90. A sine function or cosine function should work quite well to find the correct ratio of accelerations. Dr. Ken Mellendorf Physics Instructor Illinois Central College No, it would not, for three reasons. 1. The vertical component of a 45-degree angle is the sine of 45 degrees, which is about 0.71, not 0.5. What this means is that while gravity pushes down on the ball, the ramp pushes the ball away from the ramp (because the ball cannot penetrate the ramp). For every 1 meter the ball rolls down the ramp, the ball loses only 0.71 m of elevation. 2. Some of the kinetic energy of the ball goes into its rotation because it rolls rather than slides. This will make the ball's center of mass move more slowly than if all its kinetic energy went into sliding without rolling. 3. There is some rolling friction which reduces the rolling acceleration of the ball. Note that the first of these effects makes the ball accelerate faster than you expected, while the other make it accelerate more slowly than you expected. The effects may exactly cancel, but that depends on specifically how much rolling friction your ball encounters on the ramp. Richard Barrans, Ph.D., M.Ed. Department of Physics and Astronomy University of Wyoming That is an interesting question Kyler. I worked with actual numbers so for the ramp you describe the vertical height would be 21.213 ft. To determine the falling time for a freely dropping ball (not on the ramp) use the equation dist = 1/2 X accel X time squared. So with dist known and with accel at 32.2 ft/s** you can solve for t and find it to be 1.146 sec. With velocity = accel X time the final velocity would be 36.9 ft / sec. For the ball on the ramp you have to consider that in addition to the vertical force of gravity acting on the ball there is also a force normal (perpendicular) to the ramp pushing upward on the ball where it contacts the ramp. I am assuming you are ruling out any rolling or friction resistance which would be another force on the ball. The two primary forces (gravity and the normal force) can be resolved geometrically with the 45 degree ramp angle to determine a resultant force that causes the ball to roll down the ramp. For the vertical force associated with the 32.3 ft/s** acceleration both of the other force components would have values of 32.2 / sq rt of 2. That would be a force resulting in an acceleration of 22.77 ft/s** along the direction of the ramp. Now though the ball must travel 30 ft. Using the same distance equation above with the 30 ft and the 22.77 ft/s**acceleration results in a time of 1.62 sec for the ball to travel the ramp. That is about 1/2 second longer. Interestingly though, when you check final velocity (accel x time) you find it to be 36.9 ft / sec (along the direction of the ramp). I think I have looked at this problem correctly but I will be interested in the other answers you may receive. Carlton Schroeder Click here to return to the Physics Archives

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