Steps-Part aStep 1) In order to calculate the spring constant, the spring is positioned vertically, and a small mass is attached to its lower hook.Step 2) Visit the site:https://phet.colorado.edu/sims/html/masses-and-springs/latest/masses-and-springs_en.htmlStep 3) Select the Lab icon.Step 4) Make sure the damping is set to none. Using the ruler, measure the displacement of the spring and calculate the spring constant.Step 5) The mass is removed and securely attached to an inelastic cord. On other end of the cord is attached to the spring. Another segment of cord is attached to the other side of the spring. The opposite end of this cord is left free. This is the end in which the system will be rotated. The setup is shown in Fig. 1. The ball is rotated in a horizontal circle.Step 6) The radius of the system is measured before the spring is displaced.Step 7) The system is now spun around until uniform circular motion is achieved and the spring displacement is measured.Steps-Part bStep 1) On a circle, sketch the components of the total acceleration from Eq. (4)Step 2) Sketch the components of the total centripetal force acting on the ball from Eq. (5). Spring Constant DataSuspended Mass: 0.100 kgSpring DisplacementmSpring constantN/mUniform Circular Mation DataMass of Ball: 10.0 gUnstretched Spring Length: 10.0 cmTotal Cord Length: 0.5 mStretched Spring Length: 30.0 cmRadius of Circle Before DisplacementmRadius of Circle After DisplacementmCentripetal Accelerationm/sTangential Velocitym'sCentripetal ForceNTension in the CordN43When the velocity of the ball is no longer constant, the circular motion is no longer uniform. The totalacceleration on the ball is now the sum of the tangential component and the radial component,a = a, +a,(4)The total centripetal force acting on the ball is,EF = F, + F,(5)Steps-Part bStep 1) On a circle, sketch the components of the total acceleration from Eq. (4)Step 2) Sketch the components of the total centripetal force acting on the ball from Eq. (5). IntroductionIn this laburatory unifum circular uotiuu (UCM) will be studicd and Newton's Sccund Law focentripetal acceleration and centripetal force will be verified. A plastic ball will be attached to a springand a cord and rotated in UCM. Since the spring constant and displacement of the spring can casily bedetermined, the centripetal force can be calculated. From this, the tension in the cord can be determined.When an object rotates in a circle with constant velocity, this type of motion is called uniformcircular motion. Consider a ball, which can be approximated as a particle, on a string rotating in a circle.Newton's Second Law of motion still holds; however, the acceleration of the ball is given by theexpression as shown in Eq. 1. Where v, is the tangential velocity of the ball and r is the radius of thecircle. Substituting this back into Newton's Second Law gives us Eq. 2. Since direct measurements ofthe tension of the string would be difficult a spring can be used to detemine the centripetal force actingon the string. Hooke's law is given by Eq. 3.From Eq. 3, k is the spring constant and x is the displacement of the spring. This is the amountthe spring has stretched from its equilibrium position. The negative sign is the restoring force of thespring. Note that if the spring is displaced in the y-direction, just replace x with y.a. =(1)EF = ma. =(2)E= -ki(3)A plastic ball will be attached to a spring and a cord and rotated in UCM, as shown in Fig. 1. Thespring constant of the spring is not known but can be detemined by measuring the displacement of thespring and applying Hooke's law.CordCordwwwwiwiAxFigure 1. Diagram for the ball-spring system indicating the displacement of the spring.

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