- #1

Muffintoast

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## Homework Statement

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)?

## Homework Equations

Centre of mass equation:

x(centre of mass) = Σm1+x1/Σm

y(centre of mass) = Σm2+y2/Σm

## 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. :(