Particles (mass of each = 0.20 kg) are placed at the 40-cm and 100-cm marks of a metre 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

Question

Quizplus · Accepted Answer

Explanation:The torque about the pivot is:$$	au = (0.20 	ext{ kg})(9.8 	ext{ m/s}^2)(0.40 	ext{ m}) + (0.20 	ext{ kg})(9.8 	ext{ m/s}^2)(1.00 	ext{ m}) = 27.44 	ext{ N$\cdot$m}$$The moment of inertia of the system about the pivot is:$$I = (0.20 	ext{ kg})(0.40 	ext{ m})^2 + (0.20 	ext{ kg})(1.00 	ext{ m})^2 = 0.28 	ext{ kg$\cdot$m}^2$$Therefore, the initial angular acceleration is:$$\alpha = \frac{	au}{I} = \frac{27.44 	ext{ N$\cdot$m}}{0.28 	ext{ kg$\cdot$m}^2} = 97.96 	ext{ rad/s}^2$$

Particles (Mass of Each = 0