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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 \( \frac{1}{2} m r^2 \) - Leaving Cert Applied Maths - Question 8 - 2010
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} m ... 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 \( \frac{1}{2} m r^2 \) - Leaving Cert Applied Maths - Question 8 - 2010
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 \( \frac{1}{2} m r^2 \).
Answer
To find the moment of inertia of a circular disc, we consider a differential mass element of the disc, which can be expressed as:
[ M = \frac{m}{\pi r^2} \cdot 2\pi x , dx ]
where ( x ) is the distance from the center. Then, the moment of inertia of this differential element about the axis can be written as:
[ dI = x^2 , dm = x^2 \left( \frac{m}{\pi r^2} , 2\pi x , dx \right) = \frac{2m}{r^2} x^3 , dx ]
Thus, integrating from 0 to r, we have:
[ I = \int_{0}^{r} dI = \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{1}{2} m r^2 ]
Step 2
Show that the moment of inertia of the annulus about an axis through its centre and perpendicular to its plane is \( \frac{M (a^2 + b^2)}{2} \).
Answer
The moment of inertia of an annulus can be derived similarly to that of a disc. The mass of the annulus can be expressed as:
[ M = m_{disc} - m_{hole} ]
The moment of inertia of the whole disc is:
[ I_{disc} = \frac{1}{2} m_{disc} a^2 ]
The moment of inertia of the hole (the removed part) is:
[ I_{hole} = \frac{1}{2} m_{hole} b^2 ]
The moment of inertia of the annulus is then:
[ I_{annulus} = I_{disc} - I_{hole} = \frac{1}{2} (M + m_{hole}) a^2 - \frac{1}{2} m_{hole} b^2 ]
This simplifies to:
[ I_{annulus} = \frac{M (a^2 + b^2)}{2} ]
Step 3
Find its angular velocity, in terms of g, a and b, when it has rolled a distance \( \frac{a}{2} \).
Answer
To find the angular velocity when rolling down an incline, we need to equate the gain in kinetic energy to the loss in potential energy. The potential energy lost is:
[ PE = Mgh = M g \left( \frac{a}{2} \sin 30^{\circ} \right) ]
As the annulus rolls, the kinetic energy gained is the sum of translational and rotational:
[ KE = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 ]
Substituting ( I = \frac{M (a^2 + b^2)}{2} ) and ( v = a \omega ), we obtain:
[ KE = \frac{1}{2} M v^2 + \frac{1}{2} \cdot \frac{M (a^2 + b^2)}{2} \left( \frac{v}{a} \right)^2 ]
Hence, we set:
[ Mgh = \frac{1}{2} MU^2 + \frac{1}{4} M \cdot (a^2 + b^2) \cdot \frac{U^2}{a^2} ]
This simplifies to:
[ U^2 = \frac{g a}{3a^2 + 3b^2} ]
Thus:
[ \omega = \sqrt{\frac{ga}{3a^2 + 3b^2}} ]