A thin uniform rod (length = 1.2 m,mass = 2.0 kg)is pivoted about a horizontal,frictionless pin through one end of the rod.(The moment of inertia of the rod about this axis is ML²/3. )The rod is released when it makes an angle of 37$\degree$ with the horizontal.What is the angular acceleration of the rod at the instant it is released? A)9.8 rad/s² B)7.4 rad/s² C)8.4 rad/s² D)5.9 rad/s² E)6.5 rad/s²

Question

Quizplus · Accepted Answer

Explanation:The torque about the pivot is:$$	au = I\alpha = \frac{1}{3}ML^2\alpha$$The net force on the rod is:$$F = mg = 2.0 	ext{ kg} 	imes 9.8 	ext{ m/s}^2 = 19.6 	ext{ N}$$The distance from the pivot to the center of mass is:$$d = \frac{1}{2}L = 0.6 	ext{ m}$$The torque due to gravity is:$$	au = Fd = 19.6 	ext{ N} 	imes 0.6 	ext{ m} = 11.76 	ext{ Nm}$$Therefore, the angular acceleration is:$$\alpha = \frac{	au}{I} = \frac{11.76 	ext{ Nm}}{\frac{1}{3} 	imes 2.0 	ext{ kg} 	imes (1.2 	ext{ m})^2} = 9.8 	ext{ rad/s}^2$$

A Thin Uniform Rod (Length = 1 °\degree° With the Horizontal

A Thin Uniform Rod (Length = 1 $\degree$ With the Horizontal