Motor Driven Disk

## Question:

Consider the diagram at the right. A
motor gives a disk an angular acceleration
a_{A} = (0.6t^{2} +
0.75)rad/s^{2}. If the initial angular velocity w_{0} = 6rad/s, determine the magnitude
of the velocity and acceleration of block B when t=2s.
## Answer:

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 w_{0} and
a_{A} 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/s^{2} so the
acceleration of the point of string contact is
0.15m*3.15rad/s^{2} = 0.4725 m/s^{2}. This
is well below the 9.8 m/s^{2} 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) = 0.2t^{3}+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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