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8. Prove that the moment of inertia of a uniform rod, of mass m and length 2l, about an axis through its centre, perpendicular to its plane, is $\frac{1}{3} m l^{2}.$ (b) A uniform rod, of length one metre and with centre O, oscillates about a horizontal axis through P, which is a distance x from O - Leaving Cert Applied Maths - Question 8 - 2015

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8. Prove that the moment of inertia of a uniform rod, of mass m and length 2l, about an axis through its centre, perpendicular to its plane, is $\frac{1}{3} m l^{2}.... show full transcript

Worked Solution & Example Answer:8. Prove that the moment of inertia of a uniform rod, of mass m and length 2l, about an axis through its centre, perpendicular to its plane, is $\frac{1}{3} m l^{2}.$ (b) A uniform rod, of length one metre and with centre O, oscillates about a horizontal axis through P, which is a distance x from O - Leaving Cert Applied Maths - Question 8 - 2015

Step 1

Prove that the moment of inertia of a uniform rod is $\frac{1}{3} m l^{2}$

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Answer

To prove that the moment of inertia $I$ of a uniform rod of mass $m$ and length $2l$ about an axis through its center and perpendicular to its plane is given by the equation:

Define mass per unit length:
Let the mass per unit length be ( M = \frac{m}{2l} ) where ( m ) is the total mass of the rod.
Consider a differential element:
Taking a small element of length ( dx ) at a distance ( x ) from the center:
- The mass of the element is ( dm = M dx = \frac{m}{2l} dx ).
- The moment of inertia of the element is ( dI = dm \cdot x^2 = (\frac{m}{2l}dx)x^2 = \frac{m}{2l} x^{2}dx ).
Integrate over the length of the rod:
To find the moment of inertia of the whole rod, integrate from ( -l ) to ( l ):

$I = \int_{-l}^{l} \frac{m}{2l} x^2 dx = \frac{m}{2l} \left[ \frac{x^3}{3} \right]_{-l}^{l} = \frac{m}{2l} \left( \frac{l^3}{3} - \frac{(-l)^3}{3} \right)$

Simplifying gives
( I = \frac{m}{2l} \cdot \left( \frac{2l^3}{3} \right) = \frac{1}{3} m l^{2} ).

Thus, the moment of inertia $I$ of the uniform rod is indeed given by $\frac{1}{3} m l^{2}$ .

Step 2

(i) Find, in terms of x, the length of the equivalent simple pendulum.

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Answer

To find the length ( L ) of the equivalent simple pendulum:

Use the formula for moment of inertia: The moment of inertia for the rod is given as
$I = \frac{1}{3} m (1)^{2} + mx^{2} = \frac{1}{12} m + mx^{2}.$
Apply the formula for the period of oscillation: The formula for the oscillation period ( T ) is given by:
$T = 2\pi \sqrt{\frac{I}{M g h}},$
where ( h = \frac{L}{2} ) (since the center of mass is at the midpoint).
Derive the formula in terms of L: Setting ( I = \frac{1}{12} m + m x^{2}, ) we can substitute it into the period formula:

$T = 2\pi \sqrt{\frac{(\frac{1}{12} m + m x^{2})(\frac{L}{2})}{mg}} = 2\pi \sqrt{\frac{m(\frac{1}{12} + x^{2})}{2g}}.$

Thus, the length of the equivalent simple pendulum in terms of ( x ) is derived from this formula.

Step 3

(ii) Find the value of x for which the period of oscillation is a minimum.

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Answer

To find the value of ( x ) for which the period of oscillation is minimized, we will:

Differentiate the expression for T: We start by squaring ( T ) to simplify:

$T^2 = 4\pi^{2} \frac{m\left(\frac{1}{12} + x^{2}\right)}{2g} .$
Set up the differentiation: To find the value of ( x ) that minimizes ( T^2 ):

$\frac{dT^2}{dx} = 4\pi^{2} \frac{(2x)}{2g} = 0,$
leading to
( x = 0 \text{ or } \frac{1}{\sqrt{12}}. )

Thus, the value of ( x ) for minimum oscillation period is ( x = \frac{1}{\sqrt{12}}. )

Step 4

(iii) Find the minimum period of oscillation.

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Answer

To find the minimum period of oscillation based on the expression derived:

Calculate T using the value of x: Substitute ( x = \frac{1}{\sqrt{12}} ) into the period formula:

$T_{min} = 2\pi \sqrt{\frac{\frac{1}{12} m + m (\frac{1}{\sqrt{12}})^{2}}{mg}}.$
Simplifying gives: $T_{min} = 2\pi \sqrt{\frac{\frac{1}{12} m + \frac{m}{12}}{mg}} = 2\pi \sqrt{\frac{\frac{2}{12} m}{mg}}.$
1. Further simplification leads to the final equation: $T_{min} = 2\pi \sqrt{\frac{1}{6g}}.$ Using approximate values for g (approximately 9.81 m/s²), we can compute the value:
After calculation, we find that the minimum period of oscillation is approximately 1.525 seconds.

Prove that the moment of inertia of a uniform rod is $\frac{1}{3} m l^{2}$

(i) Find, in terms of x, the length of the equivalent simple pendulum.

(ii) Find the value of x for which the period of oscillation is a minimum.

(iii) Find the minimum period of oscillation.

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