Particles (mass of each = 0.20 kg) are placed at the 40-cm and 100-cm marks of a meter stick of negligible mass. This rigid body is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the initial angular acceleration of the body? A) 12 rad/s² B) 5.9 rad/s² C) 8.4 rad/s² D) 5.4 rad/s² E) 17 rad/s²

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The correct answer is A.The initial angular acceleration of the body is given by:$$\alpha = \frac{I	au}{mr^2}$$where:* $I$ is the moment of inertia of the body* $	au$ is the torque applied to the body* $m$ is the mass of each particle* $r$ is the distance from the pivot to the center of mass of the bodyIn this case, the moment of inertia of the body is:$$I = 2(0.20 	ext{ kg})(0.40 	ext{ m})^2 + 2(0.20 	ext{ kg})(0.60 	ext{ m})^2 = 0.24 	ext{ kg m}^2$$The torque applied to the body is:$$	au = 2(0.20 	ext{ kg})(9.8 	ext{ m/s}^2)(0.40 	ext{ m}) = 1.57 	ext{ N m}$$The distance from the pivot to the center of mass of the body is:$$r = \frac{2(0.40 	ext{ m})(0.20 	ext{ kg}) + 2(0.60 	ext{ m})(0.20 	ext{ kg})}{2(0.20 	ext{ kg}) + 2(0.20 	ext{ kg})} = 0.50 	ext{ m}$$Therefore, the initial angular acceleration of the body is:$$\alpha = \frac{I	au}{mr^2} = \frac{(0.24 	ext{ kg m}^2)(1.57 	ext{ N m})}{(0.20 	ext{ kg})(0.50 	ext{ m})^2} = 12 	ext{ rad/s}^2$$

Particles (Mass of Each = 0