Answer Happy

Show the algebraic steps in deriving the equation (6) of the“Theory” section
using the equations (1) through (5).
Theory A free body diagram of the rubber stopper involving the tension, the weight, and the inertia force while it is moving in a horizontal circle is shown in Fig.2. Based on this diagram we can write the following force equation in the x-direction using Newton's second law. T cos a = Mar (1) where T'is the tension in the string in [N], a is the angle between the horizontal and the string in [deg], M is the mass of the rubber stopper in [kg], and a, is the centripetal acceleration in [m/s²]. S. Araci □ α r M S. Araci Fig. 2 - Free body diagram of the rubber stopper x Ma But from kinematics of circular motion we know that the centripetal acceleration is given by ar = 22 (2) where v is the speed of the object in uniform circular motion in [m/s] and r is the radius of the circular motion in [m]. The speed of the object in uniform circular motion can be calculated as 2πr t v= Mg (3) where t is the time in [s] the object takes to complete one revolution around the circular path. The radius r of the circular motion can be related to the string length using the simple trigonometric relationship in view of the diagram shown in Fig.2 and given by r = L cosa (4) where L is the strength length in [m] between the middle of rubber stopper and the top of the handle of the centrifugal force apparatus. The tension T' in the string is equal to the weight of the hanging mass in the centrifugal force apparatus and is given by T = mg (5) where m is the hanging mass in [kg] and g = 9.81 m/s² is the gravitational acceleration constant. g t² (47²) ML = Using equations (1) through (5), we can obtain the following relationship between the time for one revolution t, and the hanging mass m. m (6) 1 Note that in the above equation the units of M and m can be kept in [g] and the unit of L can be kept in [cm], if the gravitational constant is expressed as g = 981 cm/s². 2