The 3 kg collar B slides on the frictionless arm AA\'. The arm is attached to dr
ID: 2184416 • Letter: T
Question
The 3 kg collar B slides on the frictionless arm AA'. The arm is attached to drum D and rotates about O in a horizontal plane at the rate theta . = 0.75t, where theta. and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of 0.5 m/s. Knowing that at t = 0. r = 0. answer the following multiple choice questions. 12.70.1 The radial acceleration of the collar B has a magnitude of 0.00 m/s2. 0.28t3 m/s2. 0.38t m/s2. 0.38t2 m/s2. The direction of the radial acceleration of the collar B is toward the center of rotation O. away 'rom the center of rotation O. counterclockwise direction. clockwise direction. The transverse acceleration o' the collar B has a magnitude of 0.00 m/s2. 0.38t m/s2. 0.75t m/s2. 1.13t m/s2. What is the role of contact force N between the arm AA' and the collar B? Nothing since friction is not present. It balances the cord tension T. It causes the collar to rotate. It allows the collar to slide without friction. The Newton's second law equation in the radial direction reads -T = mB(r.. - r theta . 2). T = mB(r.. - r theta .2). -T - mg sin theta = mB(r.. - r theta .2). T + N = mB(r.. - r theta .2). The Newton's second law equation in the transverse direction reads N - mg cos theta = mB(2r. theta . + r theta ..). -N = mB(2r. theta . + r theta ..). N = mB(2r. theta . + r theta ..). N + mg = mB(2r. theta . + r theta ..). The time it takes for the tension in the cord to be equal to the contact force is 2.00 seconds. 4.00 seconds. 6.00 seconds. 8.00 seconds. How many contact forces between the collar B and the arm AA'. where each is perpendicular to each other, arc required so that the whole mechanism performs as specified? One. Two. Three. Zero.Explanation / Answer
d b d c a b d c
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