Intro Pre-Lab for LAB#5 EQUATIONS & CONCEPTS mmmm Figure 1: Top view of a mass, m, undergoing horizontal uniform circular motion with radius R moving at a constant speed v. The source of the centripetal force keeping the mass in the circular motion is a stretched spring. In the vertical direction the weight of mass, m, is balanced by the tension force [not shown in the diagram]. The speed of the mass, m, can be calculated in terms of the circumference of the 2R circle, 2R and the time period of one revolution, T. Speed v = ....(1) The mass experiences a centripetal ("center-directed") acceleration, ac, due to a continuous change in the direction of its velocity. Magnitude of ac= v2 R ...(2) According to Newton's 2nd law of motion centripetal force, F = mxa...(3) The source of the centripetal force, Fe, keeping the mass in the circular motion in the above situation is the only force due to a stretched spring, Fs, such that Problem Fc=Fs (4) The magnitude of a force F needed to stretch a spring a distance Ax from its equilibrium length is described by the equation F = kAx, where k is the spring constant, or "stiffness" of the spring. The constant k may be found by applying a force to the spring, and measuring 4x. The SI units for k are N/m. If the end of the spring is not accelerating, the spring exerts a force that is equal in magnitude and in the opposite direction to Ax (labeled as Dx in figure on the next page). Draw In the figure below, draw a free body diagram for the forces acting on a mass m hanging from the bottom of a spring. m Dx = 5 cm mmm 7.cm 12 em Imm Equilibrium Point Stretching of a spring. 1. A spring stretches by Ax = 10.0 cm when a mass of 25.0 g is hung from it. What is the spring constant k of the spring? 2. The same spring stretches by Ax = 22.3 cm when a force F is applied to its end. What is the magnitude of the force F?