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Unformatted text preview: rote (cmr2495) – oldhomework 22 – Turner – (58120) 1 This printout should have 12 questions. Multiplechoice 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 mℓ 2 . Correct answer: 0 . 452308 rad / s 2 . Explanation: Let : ℓ = 20 m , d = ℓ = 20 m , θ = 60 ◦ , and m = 1 . 8 kg . I = I CM + md 2 = 1 12 mℓ 2 + mℓ 2 = 13 12 mℓ 2 . Since the rod is uniform, its center of mass is located at ℓ. Recalling that the weight mg acts at the center of mass, the magnitude of the torque at the horizontal position is τ = I α r F = parenleftbigg 13 12 mℓ 2 parenrightbigg α ℓ ( mg ) = 13 12 mℓ 2 α α = 12 g 13 ℓ = 12 (9 . 8 m / s 2 ) 13 (20 m) = . 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. b b b b 1 2 3 4 5 6 7 8 9 10 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 = ρy dx , for the areal density, the xcoordinate of the center of mass is x cm = integraldisplay xdm integraldisplay dm = σ integraldisplay x 2 xy dx σ integraldisplay x 2 y dx rote (cmr2495) – oldhomework 22 – Turner – (58120) 2 = integraldisplay x 2 xs ( x x 2 ) dx integraldisplay x 2 s ( x x 2 ) dx = integraldisplay x 2 x ( x x 2 ) dx integraldisplay x 2 ( x x 2 ) dx = 1 3 x 3 1 2 ( x 2 ) x 2 vextendsingle vextendsingle vextendsingle vextendsingle x 2 1 2 x 2 ( x 2 ) x vextendsingle vextendsingle vextendsingle vextendsingle x 2 = 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 yaxis as the axis of rotation....
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This note was uploaded on 04/22/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.
 Spring '10
 McCord

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