Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point C such that it can rotate freely around C in the XY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative x direction in the XY plane with speed

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Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point C such that it can rotate freely around C in the XY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative x direction in the XY plane with speed 100 ms^ -1 , hits the circumference of the disk at a point P. After collision the particle moves along negative y direction at a speed of 90 ms^ -1 . [Given: the acceleration due to gravity (g) = -10 j ms^ -2 ] Amount of energy loss (in J) in the collision is:

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Correct answer: A. A

Let the center of the uniform circular disk be the origin O(0,0). The disk is pivoted at its top point C(0, R). Radius of disk R = 0.2 m, Mass M = 1 kg. Mass of particle m = 20 g = 0.02 kg. Initial velocity of particle v _i = -100 i ms^ -1 . Final velocity of particle v _f = -90 j ms^ -1 . The particle hits the disk at point P. From the figure, the line OP makes an angle of 45^ with the negative y-axis. Coordinates of P with respect to O are (R 45^ , -R 45^ ). Position vector of P with respect to the pivot C is: r _ P/C = (R 45^ - 0) i + (-R 45^ - R) j = R 2 i - R (1 + 1 2 ) j Angular momentum is conserved about the pivot C during the collision. Initial angular momentum of the system about C is only due to the particle: L _i = r _ P/C (m v _i) = [ R 2 i - R (1 + 1 2 ) j ] (-100m i ) L _i = -100mR (1 + 1 2 )( j i ) = -100mR (1 + 1 2 ) k Substituting m = 0.02 kg and R = 0.2 m: L _i = -100(0.02)(0.2) (1 + 1 2 ) k = -0.4 (1 + 1 2 ) k = - (0.4 + 0.2 2 ) k kg m^2s^ -1 Final angular momentum of the particle about C: L _ f,p = r _ P/C (m v _f) = [ R 2 i - R (1 + 1 2 ) j ] (-90m j ) L _ f,p = -90m R 2 ( i j ) = - 90mR 2 k L _ f,p = - 90(0.02)(0.2) 2 k = - 0.36 2 k = -0.18 2 k kg m^2s^ -1 Let be the angular velocity of the disk after collision. Its moment of inertia about C is: I_C = I_ cm + MR^2 = 1 2 MR^2 + MR^2 = 3 2 MR^2 = 3 2 (1)(0.2)^2 = 0.06 kg m^2 Final angular momentum of the disk is L _ f,d = I_C k = 0.06 k By conservation of angular momentum about C: L _i = L _ f,p + L _ f,d -(0.4 + 0.2 2 ) = -0.18 2 + 0.06 0.06 = -0.4 - 0.02 2 = - 40 + 2 2 6 = - 20 + 2 3 rad s^ -1 Initial kinetic energy of the system: E_i = 1 2 mv_i^2 = 1 2 (0.02)(100)^2 = 100 J Final kinetic energy of the system: E_f = 1 2 mv_f^2 + 1 2 I_C ^2 = 1 2 (0.02)(90)^2 + 1 2 (0.06) ( 20 + 2 3 )^2 E_f = 81 + 0.03 ( 400 + 2 + 40 2 9 ) = 81 + 402 + 40 2 300 = 81 + 201 + 20 2 150 Amount of energy loss: E = E_i - E_f = 100 - (81 + 201 + 20 2 150 ) = 19 - 201 + 20 2 150 E = 2850 - 201 - 20 2 150 = 2649 - 20 2 150 Using 2 1.414, E 2649 - 28.28 150 = 2620.72 150 17.47 J

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