A machine part consists of a thin, uniform 4.00 kg (A) bar that is 1.50 m (X) long, hinged
perpendicular to a similar bar, placed vertically downward, of mass 3.00 kg (B) and length 1.80 m (Y).
The longer bar has a small but dense 2.00 kg (C) ball at one end (the ball can be considered as a
point mass). By what distance will the center of mass of this part move horizontally and vertically if
the vertical bar is pivoted 90° counterclockwise to make the entire part horizontal (8.51)?
2. Relevant equations
Centre of mass equation:
x(centre of mass) = Σm1+x1/Σm
y(centre of mass) = Σm2+y2/Σm
3. The attempt at a solution
So I tried finding the centre of mass before machine rotated, so the [mass of A x the (x) coordinate of A] + [mass of A x the (x) coordinate of B]/all masses together so [(4 x 1.4) + (4 x 0)]/(2 + 3 + 4) = 2/3
Then, [mass of B x the (y) coordinate of A] + [mass of B x the (y) coordinate of B]/all masses together so again, [(3 x 1/8) + (3 x 0)]/ ( 2 + 3 + 4) = 5.4/9
Then, I tried finding the centre of mass after the machine rotates so that the (y) coordinate of B becomes 0. We then know that y is 0, then to find x, it is [mass of A x (x) coordinate of A] + [mass of A x (x) coordinate of B]/ all the masses added together = [(4 x 1.5) + 94 x 1.800]/(2 + 3 + 0) = 1.4666
So to find the displacement, it is the centre of mass before the rotation – the centre of mass after the rotation = 2/3 – 1.46666 = 4/5 then, 5.4/9 – 0 = 5.4/9
So the answer is x(4/5), y(5.4/9)
This answer is completely wrong, I have the correct answer which is supposed to be Δx = 0.700m, Δy = 0.700m. Please tell me where I went wrong, this is my first week of uni and i’m just so confused. 🙁