A uniform rod of length L and mass M is free to rotate about a frictionless pivo
ID: 2014038 • Letter: A
Question
A uniform rod of length L and mass M is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. (Use Torque Methods)a.) What is the initial angular acceleration of the rod?
b.) What is the initial linear acceleration of the right end of the rod?
c.) What is the angular velocity of the rod when it reaches its vertical position (lowest point)?
d.) What is the linear velocity of the center of mass in the vertical position?
e.) What is the linear velocity of the lowest point on the rod in the vertical position?
Explanation / Answer
Length of the rod = L Mass of the rod = M (a) Initial angular acceleration = Torque/Moment of inertia = /I The (initial) torque due to gravity = M*g*L/2 (the center of gravity is at the middle of the rod). Moment of inertia I = ML^2/3 So Initial angular acceleration = (M*g*L/2)/( ML^2/3) = (3/2)(g/L) (b) Initial linear acceleration of the right end of the rod a = L = L (3/2)(g/L) = (3/2)g (c) Let be the angular speed of the rod when it reaches the vertical position. From conservation of energy Mg(L/2) = (1/2)I^2 + (1/2)Mv^2 MgL = ( ML^2/3)^2 + M[(L/2)]^2 g = (L/3)^2 + (L/4)^2 g = [(L/3) + (L/4)]^2 g = (7L/12)^2 ^2 = 12g/7L = [12g/7L] (d) Linear velocity of the center of mass in the vertical position = (L/2) = (3gL/7) (e) Linear velocity of the lowest point on the rod in the vertical position = L = (12gL/7)Related Questions
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