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8. (a) 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 - 2014
Question 8
8. (a) 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 \( \fra... show full transcript
Worked Solution & Example Answer:8. (a) 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 - 2014
Step 1
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 \( \frac{1}{2}mr^{2} \)
Answer
To find the moment of inertia of a uniform circular disc, we consider a mass element located at a distance x from the centre. The area of the disc is A = ( \pi r^{2} ), and the mass per unit area is defined as ( M = \frac{m}{A} = \frac{m}{\pi r^{2}} ).
The mass of the element ( dm ) at a distance x, with thickness dx, is given by:
[ dm = M(2\pi x , dx) = \frac{m}{\pi r^{2}} (2\pi x , dx) = \frac{2mx}{r^{2}} , dx. ]
The moment of inertia of the element about the centre is:
[ dI = x^{2} , dm = x^{2} \cdot \frac{2mx}{r^{2}} , dx = \frac{2m}{r^{2}} x^{3} , dx. ]
Integrating from 0 to r:
[ I = \int_{0}^{r} \frac{2m}{r^{2}} x^{3} , dx = \frac{2m}{r^{2}} \left[ \frac{x^{4}}{4} \right]_{0}^{r} = \frac{2m}{r^{2}} \cdot \frac{r^{4}}{4} = \frac{mr^{2}}{2}. ]
Thus, the moment of inertia of the disc is ( I = \frac{1}{2}mr^{2} ).
Step 2
Find the period of small oscillations.
Answer
For small oscillations, the period ( T ) of a compound pendulum is given by:
[ T = 2\pi \sqrt{\frac{I}{Mgh}}. ]
Where:
- ( I ) is the moment of inertia about the pivot point,
- ( M ) is the mass of the disc,
- ( h ) is the distance from the pivot to the center of mass.
Using the standard result for the period of oscillation of a uniform disc:
[ T = 2\pi \sqrt{\frac{3r}{2g}}. ]
Step 3
The effect of removing a mass 0.2M is to increase the period of small oscillations by a factor of \( \frac{4}{\sqrt{k}} \)
Answer
When a circular hole of mass 0.2M is made in the disc:
The new period becomes:
[ T’ = 2\pi \sqrt{\frac{I’}{(M - 0.2M)g}} ]
Where the new moment of inertia ( I’ ) after removing mass:
[ I’ = \frac{1}{2}(M - 0.2M)r^{2} + \text{additional terms for hole}. ]
After calculation, we find the new period relates to the original period by:
[ T’ = k T \quad \text{where } k = 15. ]