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8. (a) 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 \( \frac{1}{3} ml^2 \) - Leaving Cert Applied Maths - Question 8 - 2016
Question 8
8. (a) 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} ml... show full transcript
Worked Solution & Example Answer:8. (a) 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 \( \frac{1}{3} ml^2 \) - Leaving Cert Applied Maths - Question 8 - 2016
Step 1
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 \( \frac{1}{3} ml^2 \).
Answer
To derive the moment of inertia ( I ) of a uniform rod:
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Let the mass per unit length be ( M = \frac{m}{2l} ).
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Consider an infinitesimally small element of the rod, with length ( dx ), at distance ( x ) from the center. The mass of this element is:
[ dm = M \cdot dx = \frac{m}{2l} , dx ]
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The moment of inertia of this element about the center is:
[ dI = dm \cdot x^2 = \frac{m}{2l} , dx , x^2 ]
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Therefore, the total moment of inertia of the rod is:
[ I = \int_{-l}^{l} \frac{m}{2l} , x^2 , dx = \frac{m}{2l} \left[ \frac{x^3}{3} \right]_{-l}^{l} = \frac{m}{2l} \cdot \frac{(l^3 - (-l)^3)}{3} = \frac{m l^2}{3} ]
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Simplifying gives:
[ I = \frac{1}{3} ml^2 ]
Step 2
Find the moment of inertia of the wheel about an axis through the centre of the axle, perpendicular to its plane.
Answer
To calculate the moment of inertia ( I ) of the wheel:
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Moment of inertia of the rim (hoop): [ I_{rim} = m_{rim} r^2 = 4 , \text{kg} \cdot (0.6 , \text{m})^2 = 1.44 , \text{kg m}^2 ]
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Moment of inertia of the spokes: Each spoke has: [ I_{spoke} = \frac{1}{3} m_l l^2 = \frac{1}{3} \cdot 0.05 , \text{kg} \cdot (0.5 , \text{m})^2 = 0.00208 , \text{kg m}^2 ] For six spokes: [ I_{spokes} = 6 \cdot 0.00208 = 0.01248 , \text{kg m}^2 ]
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Moment of inertia of the axle (disc): [ I_{axle} = \frac{1}{2} m_{axle} r^2 = \frac{1}{2} \cdot 1 , \text{kg} \cdot (0.1 , \text{m})^2 = 0.005 \text{kg m}^2 ]
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Total moment of inertia of the wheel: [ I_{total} = I_{rim} + I_{spokes} + I_{axle} = 1.44 + 0.01248 + 0.005 = 1.45748 , \text{kg m}^2 \approx 1.488 , \text{kg m}^2 ]
Step 3
Calculate the kinetic energy of the wheel.
Answer
To find the kinetic energy ( E_k ) of the wheel:
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First, find the angular velocity ( \omega ): The relationship between linear speed ( v ), radius ( r ), and angular velocity is: [ v = r \omega \Rightarrow \omega = \frac{v}{r} = \frac{5 , \text{m/s}}{0.6 , \text{m}} = 8.33 , \text{rad/s} ]
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Now, calculate the kinetic energy: [ E_k = \frac{1}{2} I \omega^2 = \frac{1}{2} \cdot 1.488 \cdot (8.33)^2 = 51.17 , \text{J} ]
Step 4
If the wheel comes to an incline of \( \sin^{-1} \frac{1}{5} \), how far will it travel up the incline before it stops?
Answer
To determine how far the wheel will travel up the incline:
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The potential energy gained when going up is equal to the initial kinetic energy: [ mgh = E_k \Rightarrow m \cdot g \cdot h = 117.917 , J ]
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From the angle, it can be expressed as: [ h = d \sin \phi \Rightarrow d = \frac{117.917}{m g \sin \phi} = \frac{117.917}{(1 , kg)(9.81 , m/s^2)(0.2)} = 11.35 , m ]
Thus, the wheel will travel approximately 11.35 m up the incline before it stops.