Part B Now at the same instant determine X Y and Z the i j a

Part B

Now, at the same instant, determine ?X, ?Y, and ?Z, the i, j, and k components of ?, the boom\'s angular acceleration in the fixed reference frame X, Y, Z. At this instant, the translating-rotating reference frame x, y, z, lines up with the fixed reference frame.

Express the i, j, and k components of ? numerically in radians per second squared to four significant figures separated by commas.

?X, ?Y, ?Z =

Part C

At the same instant, determine aX, aY, and aZ, the i, j, and k components of aA, the acceleration of point A in the fixed reference frame X, Y, Z. The boom\'s length is a = 75.0 ft . At this instant, the translating-rotating reference frame x, y, z, lines up with the fixed reference frame.

Express the i, j, and k components of aA numerically in feet per second squared to three significant figures separated by commas.

Solution

Paart A
wz = w1 = 0.1 rad/s
wz\' = w1\' = 0.05 rad/s/s
wx = w2 = 0.03 rad/s
wx\' = w2\' = 0.01 rad/s/s
when rotating frame lines up with fixed frame, let time be t
[2*pi - theta] = wx + wx\'*t
t = (2*pi -31*pi/180 - 0.03)/0.01 = 571.2133 s

at this instant
w = wx + wy + wz = (wx + wx\'t)i + (wy + wy\'t)j + (wz + wz\'t)k = (0.03 + 0.01*t)i + (0.1 + 0.05*t)k = 5.742i + 28.66k rad/s

Part B

w\' = wx\' i + wz\' k = 0.1i + 0.05k rad/s/s

Part C

Considering the linear acceleration of point A

The tangential accelerations are,
az = a*wx\' k = 75*0.01 = 0.75 k ft/s/s
ax = -a*wz\' i = 75*0.05 = -3.75 i ft/s/s

The radial accelerations are
ay = -wx^2*a - wz^2*a = -5.742^2*75 - 28.66^2*75 = -64077.4623 j ft/s/s

Net Acc = -3.75 i -64077.4623 j + 0.75 k ft/s/s