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ID: 2144577 • Letter: 1
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A disk with mass m = 9.4 kg and radius R = 0.39 m begins at rest and accelerates uniformly for t = 17.8 s, to a final angular speed of ? = 27 rad/s. What is the angular acceleration of the disk? You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. What is the angular displacement over the 17.8 s? What is the moment of inertia of the disk? What is the change in rotational energy of the disk? What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? What is the final speed of a point on the disk half-way between the center of the disk and the rim? What is the total distance a point on the rim of the disk travels during the 17.8 seconds?Explanation / Answer
1). angular accelleration = (27-0)/17.8 = 1.517 rad/s^2 2). angular displacement = w0*t +0.5*angular acceleration*t^2 = 0 + 0.5*1.517*(17.8)^2 =240.32 rad 3). I = m*r^2/2 = (9.4*0.39^2)/2 = 0.715 kg-m^2 4). initial R.E. = 0 final R.E. = 0.5*I*w^2 = 0.5*0.715*27^2 = 260.62 J change = 260.62-0 = 260.62 J 5). angular acceleration = 1.517 rad/s^2 tangential acceleration = 1.517*0.39 = 0.592 m/s^2 6). radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed = w^2*r = (27/2)^2*0.39 = 71.08 m/s^2 7). v = w*r/2 = 27*(0.39/2) = 5.265 m/s 8). angular displacement*r = 240.32*0.39 = 93.72 m
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