A uniform rod, of length 2 m and relative density \( \frac{7}{9} \), is pivoted at one end \( P \) and is free to move about a horizontal axis through \( P \) - Leaving Cert Applied Maths - Question 9 - 2008
Question 9
A uniform rod, of length 2 m and relative density \( \frac{7}{9} \), is pivoted at one end \( P \) and is free to move about a horizontal axis through \( P \). The o... show full transcript
Worked Solution & Example Answer:A uniform rod, of length 2 m and relative density \( \frac{7}{9} \), is pivoted at one end \( P \) and is free to move about a horizontal axis through \( P \) - Leaving Cert Applied Maths - Question 9 - 2008
Step 1
Find the length of the immersed part of the rod.
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Answer
Let the length of the immersed part be ( B ).
Using the balance of moments about point ( P ):
[
\frac{\ell}{2} \cdot W \cdot \sin(\theta) = B \cdot \frac{W}{9} \cdot \left(\frac{\ell}{2} - \frac{B}{2}\right) \sin(\theta)
]
Solving this gives:
[
B = \frac{2W}{7}
]
Assuming ( W = 14 ), we have:
[
9B = 36\ell + 28 =0
]
[\ell = 1.06 \text{ m} \text{ (length of the immersed part)}]
Step 2
Find (i) the relative density of the body.
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Answer
Let the cross-sectional area of the cylinder = ( A ).
Using the equilibrium condition:
[
B = W \times \text{g}
]
Substituting for the body:
[B = \frac{0.04}{\rho} (1000) = \frac{0.05}{0.8}(1000)]
Thus,
[s = 0.8\text{ (relative density)}]
Step 3
Find (ii) the tension in the string.
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Answer
Using the equilibrium formula:
[B = T + W]
Substituting values:
[0.06g = T + 0.06g]
This results in:
[T = 0.015\text{ N or } 0.147 N]
Step 4
Find (iii) the radius of the cylinder.
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Answer
Using the formula for buoyancy:
[W = 0.06g]
From the relation between volume and area:
[\rho A = 0.06g]
Substituting measures:
[ \pi r^2 = 0.0015]
Thus,
[r = 0.0218\text{ m or } 2.18 cm]
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