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Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its centre perpendicular to its plane is \( \frac{1}{2} mr^{2} \) - Leaving Cert Applied Maths - Question 8 - 2012
Question 8
Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its centre perpendicular to its plane is \( \frac{1}{2} mr... show full transcript
Worked Solution & Example Answer:Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its centre perpendicular to its plane is \( \frac{1}{2} mr^{2} \) - Leaving Cert Applied Maths - Question 8 - 2012
Step 1
Prove that the moment of inertia of a uniform circular disc is \( \frac{1}{2}mr^2 \)
Answer
Let M = mass per unit area.
The mass of an element = ( M \cdot 2\pi r \cdot dx ).
Moment of inertia of the element = ( M \cdot 2\pi r \cdot dx \cdot x^2 ).
Integrating, the moment of inertia of the disc is given by:
[ I = 2\pi M \int_{0}^{r} x^2 , dx = 2\pi M \cdot \left[ \frac{x^3}{3} \right] _{0}^{r} = \frac{2\pi M r^{3}}{3} ].
Since mass M is ( m = \pi r^2 ), replacing yields:
[ I = \frac{mr^2}{2} ]
Thus proved.
Step 2
(i) Find, in terms of I, m and r, the tension in the string.
Answer
Applying conservation of energy and dynamics:
[ \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 = mgh ].
[ \frac{1}{2} \left( \frac{mgr^2}{I + mr^2} \right) (2h) = mgh ].
By rearranging, we find that [ T = \frac{mg r^2}{I + mr^2} ].
Step 3
(ii) If the acceleration of the particle is \( \frac{g}{5} \), find the mass of the pulley wheel in terms of m.
Answer
From the equation of motion:
[ mg - T = ma ]
[ T = mg - ma = mg - m \frac{g}{5} ].
Substituting into the tension equation yields:
[ mg - \frac{mgr^2}{I + mr^2} = m\frac{g}{5} ].
Solving for M gives us:
[ M = 8m ].