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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 - 2008
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 - 2008
Step 1
Prove that the moment of inertia is \( \frac{1}{2} m r^{2} \)
Answer
To prove that the moment of inertia ( I ) of a uniform circular disc is ( \frac{1}{2} m r^{2} ), we start by considering a mass element ( dm ) at a distance ( x ) from the axis, with the total mass ( m ) distributed uniformly over the area.
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Define Mass per Unit Area:
Let ( M ) be the mass per unit area. Thus, ( dm = M(2\pi x) dx ). -
Calculate Moment of Inertia of Element:
The moment of inertia of this element about the axis is given by:
[ dI = x^2 dm = x^2 (M(2\pi x) dx) = 2\pi M x^3 dx ] -
Integrate to find Moment of Inertia of the Disc:
The total moment of inertia of the disc is:
[ I = \int_0^r 2\pi M x^3 dx = 2\pi M \left[ \frac{x^4}{4} \right]_0^r = \frac{1}{2} M \pi r^4 ] -
Express Moment of Inertia in terms of Total Mass:
Since the area of the disc is ( A = \pi r^2 ), and total mass ( m = M \cdot A ), we have ( M = \frac{m}{\pi r^2} ). Substituting this in, we get:
[ I = \frac{1}{2} \left( \frac{m}{\pi r^2} \right) \pi r^4 = \frac{1}{2} m r^{2} ]
Step 2
Find (i) the common acceleration of the masses
Answer
To find the common acceleration of the masses, we can use the principles of energy conservation.
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Calculate Gain in Kinetic Energy (KE):
The system gains kinetic energy as both masses accelerate. The total kinetic energy gained is:
[ \text{Gain in KE} = \frac{1}{2} I \omega^2 + \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 ]
Substituting ( I = 0.08 \text{ kg m}^2 ), ( m_1 = 6 ext{ kg} ), ( m_2 = 4 ext{ kg} ), and ( v = 1 ext{ m/s} ):
[ = \frac{1}{2} (0.08)(1^2) + \frac{1}{2}(6)(1)^2 + \frac{1}{2}(4)(1^2) = 6 ext{ J} ] -
Calculate Loss in Potential Energy (PE):
When released, the potential energy lost is equal to:
[ \text{Loss in PE} = mgh = (6 + 4)g h = 20g h ]
Setting Gain in KE equal to Loss in PE:
[ 6 = 20g h = 2gh \Rightarrow h = \frac{6}{2g} = 3 ext{ m} ] -
Apply Kinematic Equation:
Using ( v^2 = u^2 + 2as ) where ( u = 0, v = 1 ):
[ 1^2 = 0 + 2a(3) \Rightarrow 1 = 6a \Rightarrow a = \frac{1}{6} \text{ m/s}^2 ]
Step 3
Find (ii) the tensions in the vertical portions of the string
Answer
To find the tensions in the strings, we can use the equations of motion for each mass.
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For the 6 kg Mass:
Applying Newton's second law:
[ 6g - T_2 = 6a \Rightarrow 6g - T_2 = 6(\frac{1}{6}) \Rightarrow T_2 = 6g - 1 ]
Substituting ( g \approx 9.81 ):
[ T_2 = 6(9.81) - 1 = 58.86 - 1 \Rightarrow T_2 = 57.86 \text{ N} ] -
For the 4 kg Mass:
Similarly, the equation for the 4 kg mass is:
[ T_1 - 4g = 4a \Rightarrow T_1 - 4(\frac{9.81}{6}) = 4(\frac{1}{6}) \Rightarrow T_1 = 4g + 4(\frac{1}{6}) ]
Thus, substituting values:
[ T_1 = 4(9.81) + \frac{14}{6} \approx 39.24 + 2.33 = 41.57 \text{ N} ] -
Final Tension Results:
Therefore, the tensions in the vertical portions of the strings are:- ( T_1 \approx 41.57 \text{ N} )
- ( T_2 \approx 57.86 \text{ N} )