A machine part consists of a thin, uniform m1 = 3.85-kg bar that is 1.50 m long,

Question

A machine part consists of a thin, uniform m1 = 3.85-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass m2 = 3.10 kg and length 1.80 m. The longer bar has a small but dense m3 = 1.70-kg ball at one end (Figure 1)

By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90 to make the entire part horizontal?
Find the magnitude of horizontal displacement.

Find the direction of horizontal displacement.

Find the magnitude of vertical displacement.

Find the direction of vertical displacement.

Explanation / Answer

Let's set the origin at the left end of the horizontal bar.
Now, we need to Weight each mass by its horizontal or vertical distance from
the origin and then divide by the total mass.
NOw,
total mass M = (3.85+3.1+1.7) = 8.65 kg
Now,
When it is Bent:
x_cm = (3.85*1.50/2 + 3.1*1.50 + 1.7*1.50)kg·m / 8.65 = 1.17 m
y_cm = (0 - 3.1*1.80/2 - 1.7*1.80)kg·m / 8.65 = -0.68 m
Now,
When it is Straight:
x'_cm = (3.85*1.50/2 + 3.1*(1.50 + 1.80/2) + 1.7*(1.50 + 1.80))kg·m / 8.65
x'_cm = 1.84 m
y'_cm = 0
Hense,

x = (1.84 - 1.17) m = 0.67 m to the right
y = 0.68 m upward

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