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#Katers_pendulum For More: https://sites.google.com/view/cselabb... Course Title: Physics Lab I Course Code: PHY 102 Experiment No.: 12 Name of the Experiment: To determine the value of g, acceleration due to gravity, by means of Kater’s Pendulum Objectives of the Experiment: • To study the motion of Katter’s Pendulum, • To study simple harmonic motion, • To determine the acceleration due to gravity using the theory, results, and analysis of this experiment. Theory: Kater’s pendulum, shown in Fig. 1, is a physical pendulum composed of a metal rod 1.20 m in length, upon which are mounted a sliding metal weight W1, a sliding wooden weight W2, a small sliding metal cylinder w, and two sliding knife edges K1 and K2 that face each other. Each of the sliding objects can be clamped in place on the rod. The pendulum can be suspended and set swinging by resting either knife edge on a flat, level surface. The wooden weight W2 is the same size and shape as the metal weight W1. Its function is to provide as near equal air resistance to swinging as possible in either suspension, which happens if W1 and W2, and separately K1 and K2, are constrained to be equidistant from the ends of the metal rod. The Centre of mass G can be located by balancing the pendulum on an external knife edge. Due to the difference in mass between the metal and wooden weights W1 and W2, G is not at the Centre of the rod, and the distances h1 and h2 from G to the suspension points O1 and O2 at the knife edges K1 and K2 are not equal. Fine adjustments in the position of G, and thus in h1 and h2, can be made by moving the small metal cylinder w. In Fig. 1, we consider the force of gravity to be acting at G. If hi is the distance to G from the suspension point Oi at the knife edge Ki, the equation of motion of the pendulum is I_i θ ̈=-Mgh_i sinθ And the equation of motion of the simple pendulum x ̈=-g/l_i x We see that the two equations of motion are the same if we take (Mgh_i)/I_i =g/l_i ……… (1) It is convenient to define the radius of gyration of a compound pendulum such that if all its mass M were at a distance from Oi, the moment of inertia about Oi would be Ii , which we do by writing I_i=Mk_i^2 Inserting this definition into equation (1) shows that k_i^2=h_i l_i ………….. (2) If IG is the moment of inertia of the pendulum about its Centre of mass G, we can also define the radius of gyration about the Centre of mass by writing l_G=Mk_G^2 The parallel axis theorem gives us k_i^2= h_i^2+ k_G^2 o that, using (2), we have l_i= (h_i^2+ k_G^2)/h_i The period of the pendulum from either suspension point is then T_i= 2π√(l_i/g) T_i= 2π√((h_i^2+ k_G^2)/(gh_i ))…………(3) Squaring (3), one can show that 〖h_1 T〗_1^2-〖h_2 T〗_2^2= (4π^2)/g (h_1^2-h_2^2) and in turn, (4π^2)/g=(〖h_1 T〗_1^2-〖h_2 T〗_2^2)/((h_1^2-h_2^2))=(〖h_1 T〗_1^2-〖h_2 T〗_2^2)/((h_1+ h_2)(h_1-h_2))=(T_1^2+T_2^2)/(2(h_1+ h_2))+(T_1^2-T_2^2)/(2(h_1-h_2)) ∴g=8π^2 [(T_1^2+T_2^2)/((h_1+ h_2))+(T_1^2-T_2^2)/((h_1-h_2))]^(-1) Apparatus: A Kater’s Pendulum A small metallic wedge A holder in fixed point Stop watch Experimental Procedure: Shift the weight W1 to one end of katers pendulum and fix it. Fix the knife edge K1 just below it. Keep the knife edge K2 at the other end and fix the wooden weight W2 symmetrical to other end. Keep the small weight 'w' near to Centre. Suspend the pendulum about the knife edge 1 and take the time for about 10 oscillations. Note down the time t1 using a stopwatch and calculate its time period using equation T1=t1/10. Now suspend about knife edge K2 by inverting the pendulum and note the time t2 for 10 oscillations. Calculate T2 also. If T_2≠T_1adjust the position of knife edge K2 so that T_2≈T_1 Balance the pendulum on a sharp wedge and mark the position of its Centre of gravity. Measure the distance of the knife-edge K1 as h1 and that of K2 as h2 from the Centre of gravity. Data Collection: Distance between K1 and CG (h_1)=……..cm Distance between K2 and CG (h_2)=……..cm Table for time period T_1 (oscillation about K1) Table for time period T_2 (oscillation about K2) Calculation: g=8π^2 [(T_1^2+T_2^2)/((h_1+ h_2))+(T_1^2-T_2^2)/((h_1-h_2))]^(-1) Results and Discussions: The acceleration due to gravity, g = …….. cm/s2