A long, uniform rod of length L and mass M is pivoted about ahorizontal, frictio

Question

A long, uniform rod of length L and mass M is pivoted about ahorizontal, frictionless pin at a distance of L/4 from oneend. The rod is initially vertical and a very small bumpcauses it to rotate about the pin as shown. The moment ofinertia of the rod about either end is ML^2/3 At the instant the rod is horizontal determine thefollowing: 3. Angular speed about the pin 4. Translational speed of the center of mass 5. Torque about the pin 6. Angular acceleration about the pin 7. Horizontal componet of the accerleration of the center ofmass 8. Vertical componet of the acceleration of the center ofmass. A long, uniform rod of length L and mass M is pivoted about ahorizontal, frictionless pin at a distance of L/4 from oneend. The rod is initially vertical and a very small bumpcauses it to rotate about the pin as shown. The moment ofinertia of the rod about either end is ML^2/3 At the instant the rod is horizontal determine thefollowing: 3. Angular speed about the pin 4. Translational speed of the center of mass 5. Torque about the pin 6. Angular acceleration about the pin 7. Horizontal componet of the accerleration of the center ofmass 8. Vertical componet of the acceleration of the center ofmass.

Explanation / Answer

1.the distance moved by the rod when it is horizontal is S = L1 * L1 = (L - (L/4)) = (3L/4) and = (/2) radian = (3.14/2) radian = 1.57radian or S = (3L/4) * 1.57 = 1.177 * L we know that v2 - u2 = 2gS u = 0 and g = 9.8 m/s2 or v2 = 2 * 9.8 * 1.177 * L or v = 4.8 * L m/s the angular speed about the pin is v = L1 * w or w = (v/L1) = (4.8 * L/(3L/4)) =(6.4/L) rad/s 2.the translational speed of the center of mass is v = 4.8 *L m/s 3.Torque about the pin = F * r * sin F = M * g,r = (3L/4) and = 90o or = M * g * (3L/4) * sin(90o) or = (3/4) * 9.8 * ML = 7.35 * ML 4.Torque about the pin is also given by = I * I is the moment of inertia and is the angularacceleration or = (/I) = (7.35 * ML/(ML2/3)) =(22.05/L) rad/s2 5.Horizontal componet of the accerleration of the center ofmass ax = (v2/L1) here,v = 0 therefore ax = ((0)2/L1) = 0 5.Horizontal componet of the accerleration of the center ofmass ax = (v2/L1) here,v = 0 therefore ax = ((0)2/L1) = 0 6.Vertical componet of the accerleration of the center ofmass ay = (v2/L1) = ((4.8 *L)2/(3L/4)) = 30.72 m/s2

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