Motor Driven Disk


motor driven discConsider the diagram at the right. A motor gives a disk an angular acceleration aA = (0.6t2 + 0.75)rad/s2. If the initial angular velocity w0 = 6rad/s, determine the magnitude of the velocity and acceleration of block B when t=2s.


The velocity and acceleration of block B will be the same as the velocity and acceleration of the point on the disk where the string contacts the edge of the disk provided that the string remains tight and does not stretch. Since in the statement of the problem the sign of w0 and aA are the same and in a direction to relieve tension in the string, we need to check that it stays tight. That will be the case as long as the calculated downward acceleration of B is less than the acceleration due to gravity.

The acceleration of a point on the disk edge is the angular acceleration times the radius of the disk. At time t=2s the angular acceleration is 0.6*4+0.75 = 3.15rad/s2 so the acceleration of the point of string contact is 0.15m*3.15rad/s2 = 0.4725 m/s2. This is well below the 9.8 m/s2 due to gravity so there is no risk of the string going slack.

The change in velocity of the disk from time t=0 to time t=2 is found by integrating the time dependent acceleration. Integrating the acceleration expression we get w(t) = 0.2t3+0.75t+c. Evaluating this expression a t=2 and t=0 and subtracting the values gives us the change in angular velocity = 0.2*8+0.75*2 = 3.1rad/s. To this we must add the initial angular velocity to get 9.1rad/s. Multiplying by the radius gives us the velocity of B = 1.365m/s

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