A non-uniform 2.0-kg rod is 2.0 m long. The rod is mounted to rotate freely about a horizontal axis perpendicular to the rod that passes through one end of the rod. The moment of inertia of the rod about this axis is 4.0 kg.m². The centre of mass of the rod is 1.2 m from the axis. If the rod is released from rest in the horizontal position, what is its angular speed as it swings through the vertical position? A)3.4 rad/s B)4.4 rad/s C)4.3 rad/s D)5.8 rad/s E)6.8 rad/s

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Quizplus · Accepted Answer

Explanation:The moment of inertia about the axis of rotation is 4.0 kg.m2. The potential energy of the rod when it is released from rest in the horizontal position is:$$U_i = mgl(1 - cosθ_i)$$$$= (2.0 kg)(9.8 m/s^2)(1.2 m)(1 - cos 0°) = 23.5 J$$The kinetic energy of the rod as it swings through the vertical position is:$$K_f = (1/2)Iω^2$$$$= (1/2)(4.0 kg.m^2)ω^2 = 2.0ω^2 J$$By conservation of energy, we have:$$U_i = K_f$$$$23.5 J = 2.0ω^2 J$$$$ω^2 = 11.75 rad^2/s^2$$$$ω = 3.4 rad/s$$

A Non-Uniform 2