Measurement of acceeration due to gravity (g) by a compound penduum Aim: (i) To determine the acceeration due to gravity (g) by means of a compound penduum. (ii) To determine radius of gyration about an axis through the center of gravity for the compound penduum. Apparatus and Accessories: (i) A bar penduum, (ii) a knife edge with a patform, (iii) a sprit eve, (iv) a precision stop watch, (v) a meter scae and (vi) a teescope. Theory: A simpe penduum consists of a sma body caed a bob (usuay a sphere) attached to the end of a string the ength of which is great compared with the dimensions of the bob and the mass of which is negigibe in comparison with that of the bob. Under these conditions the mass of the bob may be regarded as concentrated at its center of gravity, and the ength of the penduum is the distance of this point from the axis of suspension. When the dimensions of the suspended body are not negigibe in comparison with the distance from the axis of suspension to the center of gravity, the penduum is caed a compound, or physica, penduum. A rigid body mounted upon a horizonta axis so as to vibrate under the force of gravity is a compound penduum. In Fig.1 a body of irreguar shape is pivoted about a horizonta frictioness axis through P and is dispaced from its equiibrium position by an ange θ. In the equiibrium position the center of gravity G of the body is verticay beow P. The distance GP is and the mass of the body is m. The restoring torque for an anguar dispacement θ is τ = - mg sinθ (1) P For sma ampitudes (θ 0), I d2 θ dt 2 = mgθ, (2) where I is the moment of inertia of the body through the axis P. Eq. (2) represents a simpe harmonic motion and hence the time period of osciation is given by T = 2π Ι mg () Now I = I G + m 2, where IG is the moment of inertia of the body about an axis parae with axis of osciation and passing through the center of gravity G. IG = mk 2 (4) where K is the radius of gyration about the axis passing through G. Thus, L G θ O Fig. 1 1
T = 2π mk2 +m 2 mg = 2π K2 + g The time period of a simpe penduum of ength L, is given by (5) Comparing with Eq. (5) we get T = 2π L g L = + K2 (6) (7) This is the ength of equivaent simpe penduum. If a the mass of the body were concentrated at a point O (See Fig.1) such that OP = k2 +, we woud have a simpe penduum with the same time period. The point O is caed the Centre of Osciation. Now from Eq. (7) 2 L + K 2 = 0.(8) i.e. a quadratic equation in. Equation 6 has two roots 1 and 2 such that 1 + 2 = L and 1 2 = K 2 (9) Thus both 1 and 2 are positive. This means that on one side of C.G there are two positions of the centre of suspension about which the time periods are the same. Simiary, there wi be a pair of positions of the centre of suspension on the other side of the C.G about which the time periods wi be the same. Thus there are four positions of the centers of suspension, two on either side of the C.G, about which the time periods of the penduum woud be the same. The distance between two such positions of the centers of suspension, asymmetricay ocated on either side of C.G, is the ength L of the simpe equivaent penduum. Thus, if the body was supported on a parae axis through the point O (see Fig. 1), it woud osciate with the same time period T as when supported at P. Now it is evident that on either side of G, there are infinite numbers of such pair of points satisfying Eq. (9). If the body is supported by an axis through G, the time period of osciation woud be infinite. From any other axis in the body the time period is given by Eq. (5). From Eq.(6) and (9), the vaue of g and K are given by g = 4π 2 L T 2 K = 1 2 (10) (11) By determining L, 1 and 2 graphicay for a particuar vaue of T, the acceeration due to gravity g at that pace and the radius of gyration K of the compound penduum can be determined. 2
Description: The bar penduum consists of a metaic bar of about one meter ong. A series of circuar hoes each of approximatey 5 mm in diameter are made aong the ength of the bar. The bar is suspended from a horizonta knife-edge passing through any of the hoes (Fig. 2). The knifeedge, in turn, is fixed in a patform provided with the screws. By adjusting the rear screw the patform can be made horizonta. Fig. 2 Fig. Procedure: (i) Suspend the bar using the knife edge of the hook through a hoe nearest to one end of the bar. With the bar at rest, focus a teescope so that the vertica cross-wire of the teescope is coincident with the vertica mark on the bar. (ii) Aow the bar to osciate in a vertica pane with sma ampitude (within 4 0 of arc). (iii) Note the time for 20 osciations by a precision stop-watch by observing the transits of the vertica ine on the bar through the teescope. Make this observation three times and find the mean time t for 20 osciations. Determine the time period T. (iv) Measure the distance d of the axis of the suspension, i.e. the hoe from one of the edges of the bar by a meter scae. (v) Repeat operation (i) to (iv) for the other hoes ti C.G of the bar is approached where the time period becomes very arge.
(vi) Invert the bar and repeat operations (i) to (v) for each hoe starting from the extreme top. (vii) Draw a graph with the distance d of the hoes as abscissa and the time period T as ordinate. The nature of graph wi be as shown in Fig.. Draw the horizonta ine ABCDE parae to the X-axis. Here A, B, D and E represent the point of intersections of the ine with the curves. Note that the curves are symmetrica about a vertica ine which meets the X-axis at the point G, which gives the position of the C.G of the bar. This vertica ine intersects with the ine ABCDE at C. Determine the ength AD and BE and find the ength L of the equivaent simpe penduum from L = AD+BE = L x. 2 2 Find aso the time period T corresponding to the ine ABCDE and then compute the vaue of g. Draw severa horizonta ines parae to X-axis and adopting the above procedure find the vaue of g for each horizonta ine. Cacuate the mean vaue of g. Aternativey, for each horizonta ine obtain the vaues of L and T and draw a graph with T 2 as abscissa and L as ordinate. The graph woud be a straight ine. By taking a convenient point on the graph, g may be cacuated. Simiary, to cacuate the vaue of K, determine the ength AC, BC or CD, CE of the ine ABCDE and compute ACΧBC or CDΧCE. Repeat the procedure for each horizonta ine. Find the mean of a K. Observations: Tabe 1-Data for the T versus d graph Seria no of hoes from one end One side of C.G Other side of C.G Distance d of the hoe from one end 1 2 Time for 20 osciations (sec). 1 2. Mean time t for 20 osciations (sec) Time period T = t/20 (sec) 4
TABLE 2- The vaue of g and K from T vs. d graph No. of obs. 1. ABCDE 2.. L (AD+BE)/2 T (sec) g = 4π 2 L T 2 (cm/sec 2 ) Mean g (cm/sec 2 ) K ACΧBC or CDΧCE Mean K Computation of proportiona error: We have from Eq. (10) g = 4π 2 (L x/2) ( t 20 )2 (12) Since L = Lx/2 (Lx= AD+BE) and T = t/20, therefore, we can cacuate the maximum proportiona error in the measurement of g as foows δg g = ( δl x L x ) 2 + (2 δt t )2 (1) δl x = the vaue of the smaest division of the meter scae δt = the vaue of the smaest division of the stop-watch Precautions and Discussions: (i) (ii) (iii) (iv) Ensure that the penduum osciates in a vertica pane and that there is no rotationa motion of the penduum. The ampitude of osciation shoud remain within 4 0 of arc. Use a precision stop-watch and note the time accuratey as far as possibe. Make sure that there is no air current in the vicinity of the penduum. References: 1. Fundamentas of Physics: Resnick & Haiday 2. Practica physics: R.K. Shuka, Ancha Srivatsava, New Age Internationa (P) Ltd, New Dehi. Eric J. Irons, American Journa of Physics, Vo. 15, Issue 5, pp.426 (1947) 5