Rotational Motion & Center of Mass Solved Example

ROTATIONAL MOTION & CENTER OF MASS SOLVED EXAMPLE Ex.1 Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long. Ans. With the X and Y axes chosen as shown in Fig., the coordinates of points O, A and B forming the equilateral Ex.3 Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 kg. Ans. Choosing the X and Y axes as shown in Fig. we have the coordinates of the vertices of the L-shaped lamina as given in the figure. We can think of the Lshape to consist of 3 squares each of length 1m. The mass of each square is 1kg, since the lamina is uniform. The centres of mass C1, C2 and C3 of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/2,1/2), (1/2,3/2) respectively. We take the masses of the squares to be concentrated at these points. The centre of mass of the whole L shape (X, Y) is the centre of mass of these mass points. triangle are respectively (0,0), (0.5,0), (0.25,0.25 3 ). Let the masses 100 g, 150g and 200g be located at O, A and B be respectively. Then, X = = 100(0)  150(0.5)  200(0.25)gm (100  150  200) g 75  50 125 5 m m m 450 450 18 Y= = m1x1  m2x2  m3x3 m1  m2  m3 100(0)  150(0)  200(0.25 3) gm 450 g 50 3 3 1 m m m 450 9 3 3 Ex.2 Find the centre of mass of a triangular lamina. Ans. The lamina (LMN) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig. 7.10 Hence X 1(1 / 2)  1(3 / 2)  1(1 / 2)k g m  5 m Y 1(1 / 2)  1(1 / 2)  1(1 / 2)k g m  5 m (1  1  1) kg (1  1  1) kg 6 6 The centre of mass of the L-shape lies on the line OD. Ex.4 Find the scalar and vector products of two  ^ ^ ^ ^ ^  ^ vectors. a =  3 i  4 j  5 k  and b =   2 i  j  3 k      ^ ^ ^ ^ ^ ^ Ans. aib = (3 i - 4 j+5 k)i(-2 i + j- 3 k) = –6 – 4 –15 = –25 ^ a×b= ^ i j 3 4 2 ^ ^ k ^ ^ ^ 5  7 i  j 5 k 1 3 ^ ^ Note a × b = – 7 i  j  5 k By symmetry each strip has its centre of mass at its midpoint. If we join the midpoint ofall the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle. Ex.5 Show that moment of a couple does not depend on the point about which you take the moments. Ans. Consider a couple as shown in Fig. acting on a rigid body. The forces F and -F act respectively at points B and A. These points have position vectors r1 and r2 with respect to origin O. Let us take the moments of ROTATIONAL MOTION & CENTER OF MASS the forces about the origin. The moment of the couple = sum of the moments of the two forces making the couple = r1 × (–F) + r2 × F = r2 × F – r1 × F = (r2–r1) × F But r1 + AB = r2, and hence AB = r2 – r1. The moment of the couple, therefore, is AB × F. Clearly this is independent of the origin, the point about which we took the moments of the forces. Ex.6 A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg weight is suspended at 30 cm from one end. Find the reactions at the knifeedges. (Assume the bar to be of uniform cross section and homogeneous.) Ans. Ans. The ladder AB is 3 m long, its foot A is at distance AC = 1 m from the wall. From Pythagoras theorem, BC = 2 2 m. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces F1 and F2 of the wall and the floor respectively. Force F1 is perpendicular to the wall, since the wall is frictionless. Force F2 is resolved into two components, the normal reaction N and the force of friction F. that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall. For translational equilibrium, taking the forces in the vertical direction, N – W = 0 (i) Taking the forces in the horizontal direction, F – F1 = 0 (ii) For rotational equilibrium, taking the moments of the forces about A, 2 2 F1 – (1/2) W = 0 (iii) Now W = 20 g = 20 × 9.8 N = 196.0 N From (i) N = 196.0 From (iii) F1 =W 4 2 = 196.0/4 2 = 34.6N From (ii) 1 F = F = 34.6 N Figure shows the rod AB, the positions of the knife edges K1 and K2 , the centre of gravity of the rod at G and the suspended weight at P. The weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP = 30 cm, PG = 5 cm, AK1= BK2 = 10 cm and K1G = K2G = 25 cm. Also, W= weight of the rod = 4.00 kg and W1= suspended weight = 6.00 kg; R1 and R2 are the normal reactions of the support at the knife edges. For translational equilibrium of the rod, R1+R2 –W1 –W = 0 (i) W1 and W act vertically down and R1 and R2 act vertically up. For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of R2 and W1 are anticlockwise (+ve), whereas the moment of R1 is clockwise (-ve). For rotational equilibrium, –R1 (K1G) + W1 (PG) + R2 (K2G) = 0 (ii) It is given that W = 4.00g N and W1 = 6.00g N, where g = acceleration due to gravity. We take g = 9.8 m/s2. With numerical values inserted, from (i) R1 + R2 – 4.00g – 6.00g = 0 or R1 + R2 = 10.00g N (iii) = 98.00 N From (ii) – 0.25 R1 + 0.05 W1 + 0.25 R2 = 0 or R2 – R1 = 1.2g N = 11.76 N (iv) From (iii) and (iv), R1 = 54.88 N, R2 = 43.12 N Thus the reactions of the support are about 55 N at K1 and 43 N at K2. Ex.7 A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig. Find the reaction forces of the wall and the floor. F2 = F 2  N2 = 199.0 N The force F2 makes an angle ? with the horizontal, tan = N F = 4 2  = tan–1(4 2 )  80º Ex.8 What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end? Ans. For the rod of mass M and length l, I = M2/12. Using the parallel axes theorem, I’= I + Ma2 with a = /2 we get, 2 I'  M 2 M 2   M   12 3 2 We can check this independently since I is half the moment of inertia of a rod of mass 2M and length 2l about its midpoint, I'  2M. 42 1 M2   12 2 3 Ex.9 What is the moment of inertia of a ring about a tangent to the circle of the ring? Ans. The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes theorem, Itan gent  Idia  MR 2  MR 2 3  MR 2  MR 2 2 2