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Arm AB has a constant angular velocity of 24 rad/s counterclockwise. At the instant when theta = 90 degree, determine the acceleration of collar D and the midpoint G of bar BD. (Round the final answers to the nearest whole number). The acceleration of collar D is _____ in./s^2. The acceleration of the midpoint G of bar BD is _____ in./s^2 _____.

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Without knowing the specific geometry and dimensions of the system involving arm AB, collar D, and midpoint G of bar BD, it is not possible to provide an exact solution to this problem. However, we can provide a general method for finding the accelerations of collar D and midpoint G using the equations for angular velocity, angular acceleration, linear velocity, and linear acceleration.

Let's assume that the arm AB is rotating counterclockwise about a fixed point at a constant angular velocity of 24 rad/s. At the instant when theta = 90 degrees, we can use the following equations to find the linear velocities and accelerations of collar D and midpoint G:

Angular velocity: w = d(theta)/dt = 24 rad/s (given)
Angular acceleration: alpha = dw/dt = 0, since the angular velocity is constant
Linear velocity: v = r*w, where r is the distance from the fixed point to the point of interest (collar D or midpoint G)
Linear acceleration: a = ralpha + vdw/dt, where alpha = 0
For collar D, let's assume that r is the distance from the fixed point to the point where the arm AB is attached to collar D. Let's also assume that the distance from collar D to the fixed point is L. Then, we can write:

r = L/2, since collar D is at the midpoint of bar BD
v = rw = (L/2)(24) = 12L
alpha = 0
a = ralpha + vdw/dt = (L/2)*0 + (12L)*0 = 0
Therefore, the acceleration of collar D is 0 in/s^2.

For midpoint G, let's assume that r is the distance from the fixed point to midpoint G, and that the distance from midpoint G to the fixed point is d. Then, we can write:

r = sqrt((L/2)^2 + d^2), since midpoint G is at the midpoint of the hypotenuse of right triangle ABD
v = r*w = sqrt((L/2)^2 + d^2)*24
alpha = 0
a = ralpha + vdw/dt = sqrt((L/2)^2 + d^2)0 + sqrt((L/2)^2 + d^2)dw/dt(-sin(theta)) = -24sqrt((L/2)^2 + d^2)*cos(theta)
At theta = 90 degrees, the acceleration of midpoint G is:

a = -24*sqrt((L/2)^2 + d^2)cos(90) = -24sqrt((L/2)^2 + d^2)

Note that the accelerations of collar D and midpoint G depend on the dimensions and geometry of the system, which are not specified in the problem statement. Therefore, we cannot provide numerical values for the accelerations without additional information.
User Hyung Ook An
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