Lab 8

Physics 220 Name: Tayler Holcomb Date: October 15, 2023 Centripetal Force Purpose: By setting up an experiment, and conducting it, we will be able to see that an object in a circular motion is accelerating towards the center of the object. We will be able to prove that the force that keeps an object in a circular path is called centripetal force and is equal to the product of the mass of the object by the acceleration. Procedure: Assemble the experiment by following the picture or the demo given by the instructor. Connect the mass to the T with a string and a spring. Place at one of the three holes. Once the setup is complete, you may begin the experiment. Pull the string enough to make it start to spin, and also try to keep everything horizontal. It may fly off, so wear safety goggles and move all electronics. Record the time to take for it to make 10 revolutions. Repeat this for 5 trials. To find T, take the time it took for 10 revolutions and divide by 10 to get the time for 1. To measure R, pull it so that everything is horizontal and measure from the middle to the hole for the radius. Calculate the velocity. Find the elastic force by using a force tool. Measure the mass of the canister, and calculate the centripetal force. Data: Hole 1 Trial m(kg) r(m) T(s) v= 2pi r/T Fc= mv^2/r Fe 1 0.375 0.406 0.684 3.73 12.85 5.3 2 0.375 0.406 0.626 4.08 15.38 5.3 3 0.375 0.406 0.657 3.88 13.9 5.3 4 0.375 0.406 0.79 3.23 9.64 5.3 5 0.375 0.406 0.665 3.84 13.62 5.3

Hole 2 Trial m(kg) r(m) T(s) v= 2pi r/T Fc= mv^2/r Fe 1 0.37 0.381 0.61 3.92 14.92 5.1 2 0.37 0.381 0.623 3.84 14.32 5.1 3 0.37 0.381 0.586 4.09 16.25 5.1 4 0.37 0.381 0.623 3.84 14.32 5.1 5 0.37 0.381 0.624 3.84 14.32 5.1 Hole 3 Trial m(kg) r(m) T(s) v= 2pi r/T Fc= mv^2/r Fe 1 0.28 0.342 0.606 3.55 10.32 4.3 2 0.28 0.342 0.613 3.51 10.09 4.3 3 0.28 0.342 0.607 3.54 10.26 4.3 4 0.28 0.342 0.611 3.52 10.14 4.3 5 0.28 0.342 0.605 3.55 10.32 4.3 Sample Calculations: v= 2pi r/T Fc= mv^2/r Discussion