- #1
Oz Alikhan
- 12
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1. The Question from Text:
"A forest playground has a tyre hanging from a tree branch. The tyre behaves like a pendulum, with a rope of 4.0 metres length, and the tyres mass 15 kg. A child of mass 45 kg swings on the tyre by pulling it 3.0 metres to one side and leaping on. What is the maximum height that the tyre will reach above its equilibrium position?" 3. The Solutions:
Through Energy Conservation
T = 2∏ \sqrt{l/g} ≈4.0s
ω = 2∏/T ≈1.6 rad/s
Therefore, V_max = Aω = 3.0 x 1.6 = 4.7 (Using unrounded T & ω)
Max K.E = 0.5*(15+45)*(4.7)^2 = 660 J
Max P.E = 660 J = mgh
h = 660/mg = 1.1 mThrough Trigonometry
The right angled triangle: Cosθ = 3/4
If I use Pythagoras Theorem, I get \sqrt{4^{2} - 3^{2}} = \sqrt{7}
Then, 4 - \sqrt{7} = 1.35 mWhy is there a significant difference in my answers? (Quick side note: Why doesn't my MathTex not work?)
"A forest playground has a tyre hanging from a tree branch. The tyre behaves like a pendulum, with a rope of 4.0 metres length, and the tyres mass 15 kg. A child of mass 45 kg swings on the tyre by pulling it 3.0 metres to one side and leaping on. What is the maximum height that the tyre will reach above its equilibrium position?" 3. The Solutions:
Through Energy Conservation
T = 2∏ \sqrt{l/g} ≈4.0s
ω = 2∏/T ≈1.6 rad/s
Therefore, V_max = Aω = 3.0 x 1.6 = 4.7 (Using unrounded T & ω)
Max K.E = 0.5*(15+45)*(4.7)^2 = 660 J
Max P.E = 660 J = mgh
h = 660/mg = 1.1 mThrough Trigonometry
The right angled triangle: Cosθ = 3/4
If I use Pythagoras Theorem, I get \sqrt{4^{2} - 3^{2}} = \sqrt{7}
Then, 4 - \sqrt{7} = 1.35 mWhy is there a significant difference in my answers? (Quick side note: Why doesn't my MathTex not work?)
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