How does the mass of the barbell affect the centre of gravity?

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SUMMARY

The discussion focuses on the impact of mass distribution on the center of gravity of a barbell system. A 1.7m barbell with a 20kg weight on one end and a 35kg weight on the other has its center of gravity calculated at 1.07m from the left end when ignoring the bar's mass. When accounting for the barbell's own mass of 8.0kg, the center of gravity shifts due to the uneven weight distribution. Additionally, torque calculations related to an athlete holding a 3.0kg steel ball while considering arm length and angle are discussed, highlighting the importance of accurate angle measurements in torque calculations.

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bigsaucy
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Hello all, just a little centre of gravity question:

A 1.7m long barbell has a 20kg weight attached on its left end and a 35kg weight on its right end.

a) If you ignore the weight of the bar itself, how far would from the left of the barbell is the centre of gravity

- I had no trouble with this part and found the centre of gravity to be 1.07m from the left end

b) Where is the centre of gravity if the 8.0kg mass of the barbell itself is taken into account?

- I was stumped at this part: How could the mass of the barbell possibly affect the centre of gravity!? please help!
 
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bigsaucy said:
- I was stumped at this part: How could the mass of the barbell possibly affect the centre of gravity!? please help!
Hint: You can treat the bar just like any other mass. Where is its center of gravity?
 
the centre of mass would be spot on in the middle of the bar, since that is the case, why does it affect the centre of gravity?
 
bigsaucy said:
the centre of mass would be spot on in the middle of the bar,
Right. The center of mass of the bar would be right in the middle. But where's the center of mass of the entire object, including the weights on each end?
since that is the case, why does it affect the centre of gravity?
Because the center of mass of the two weights is not in the middle.

(If the weights on each end of the bar were equal, then you'd be right--the bar would not affect the center of mass of the system.)
 
Ok cool, took me a while but I get it now. Since you're here I might as well ask for help with another question.

An athlete at the gym holds a 3.0kg steel ball in his hand. His arm is 70cm long and has a mass of 4.0kg. What is the magnitude of the torque about his shoulder if he holds his arm:

a) Straight but 45 degrees below the horizontal

I worked out the centre of gravity to be 0.5m from the shoulder
Then i calculated the perpendicular component of the force (68.6N) and found that to be 52.55N I then multiplied that by 0.5 (distance from shoulder) and got 26.28 N . m as the torque. The back of the book says the answer is 24 N . m
 
bigsaucy said:
I worked out the centre of gravity to be 0.5m from the shoulder
OK, assuming the arm is uniform.
Then i calculated the perpendicular component of the force (68.6N) and found that to be 52.55N
Show how you did that calculation.
 
I ASSUMED that the angle was 40 degrees from the similar triangles rules

i then used cos (40) = A/H
therefore A = 68.6 * cos 40

which provided me with the 52.55N
 
bigsaucy said:
I ASSUMED that the angle was 40 degrees from the similar triangles rules
:confused: You were told that the angle of the arm was 45 degrees below horizontal.
 
oh God. an error really that simple and I missed it? I don't deserve to be human.
 
  • #10
bigsaucy said:
oh God. an error really that simple and I missed it? I don't deserve to be human.
:smile: Don't worry about that. I estimate that I've made thousands of errors over the years, some pretty silly. That's what being human is all about.
 
  • #11
bigsaucy said:
oh God. an error really that simple and I missed it? I don't deserve to be human.
No, if you were perfect you wouldn't deserve to be human.
 

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