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oldhw22sol - rote(cmr2495 oldhomework 22 Turner(58120 This...

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rote (cmr2495) – oldhomework 22 – Turner – (58120) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uniform rod of mass 1 . 8 kg is 20 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 20 m from the center of mass of the rod. Initially the rod makes an angle of 60 with the horizontal. The rod is released from rest at an angle of 60 with the horizontal, as shown in the figure below 20 m 20 m 1 . 8 kg O 60 What is the angular acceleration of the rod at the instant the rod is in a horizontal posi- tion? The acceleration of gravity is 9 . 8 m / s 2 and the moment of inertia of the rod about its center of mass is 1 12 m ℓ 2 . Correct answer: 0 . 452308 rad / s 2 . Explanation: Let : = 20 m , d = = 20 m , θ = 60 , and m = 1 . 8 kg . I = I CM + m d 2 = 1 12 m ℓ 2 + m ℓ 2 = 13 12 m ℓ 2 . Since the rod is uniform, its center of mass is located at ℓ . Recalling that the weight m g acts at the center of mass, the magnitude of the torque at the horizontal position is τ = I α r F = parenleftbigg 13 12 m ℓ 2 parenrightbigg α ( m g ) = 13 12 m ℓ 2 α α = 12 g 13 = 12 (9 . 8 m / s 2 ) 13 (20 m) = 0 . 452308 rad / s 2 . 002 (part 1 of 2) 10.0 points A uniform flat plate of metal is situated in the reference frame shown in the figure below. 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 x ( m ) y ( m ) Calculate the x coordinate of the center of mass of the metal plate. Correct answer: 3 . 33333 m. Explanation: Let : ( x 1 , y 1 ) = (0 m , 0 m) ( x 2 , y 2 ) = (10 m , 0 m) ( x 3 , y 3 ) = (0 m , 2 m) . The equation for the hypotenuse is y - y 2 = s ( x - x 2 ) = y 3 x 2 ( x - x 2 ) . Since dm = ρ y dx , for the areal density, the x -coordinate of the center of mass is x cm = integraldisplay x dm integraldisplay dm = σ integraldisplay x 2 0 x y dx σ integraldisplay x 2 0 y dx
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rote (cmr2495) – oldhomework 22 – Turner – (58120) 2 = integraldisplay x 2 0 x s ( x - x 2 ) dx integraldisplay x 2 0 s ( x - x 2 ) dx = integraldisplay x 2 0 x ( x - x 2 ) dx integraldisplay x 2 0 ( x - x 2 ) dx = 1 3 x 3 - 1 2 ( x 2 ) x 2 vextendsingle vextendsingle vextendsingle vextendsingle x 2 0 1 2 x 2 - ( x 2 ) x vextendsingle vextendsingle vextendsingle vextendsingle x 2 0 = 1 3 x 3 2 - 1 2 ( x 2 ) x 2 2 1 2 x 2 2 - ( x 2 ) x 2 = x 3 2 3 x 2 2 = 1 3 x 2 = 1 3 (10 m) = 3 . 33333 m . 003 (part 2 of 2) 10.0 points If the mass of the plate is 4 kg, find the moment of inertia of the triangle with the y -axis as the axis of rotation. Correct answer: 66 . 6667 kg · m 2 . Explanation: The moment of inertia of the triangle about the x -coordinate is I x =0 y integraldisplay x 2 0 x 2 dm = σ integraldisplay x 2 0 x 2 y dx = σ s integraldisplay x 2 0 x 2 ( x - x 2 ) dx = σ s parenleftbigg 1 4 x 4 - 1 3 x 2 x 3 parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle x 2 0 = σ s parenleftbigg 1 4 x 4 2 - 1 3 x 4 2 parenrightbigg = parenleftbigg 2 m x 2 y 3 parenrightbigg parenleftbigg - y 3 x 2
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