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8. (a) Prove that the moment of inertia of a uniform 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 - 2020
Question 8
8. (a) Prove that the moment of inertia of a uniform disc, of mass m and radius r about an axis through its centre, perpendicular to its plane, is \( \frac{1}{2}mr^2... show full transcript
Worked Solution & Example Answer:8. (a) Prove that the moment of inertia of a uniform 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 - 2020
Step 1
Prove that the moment of inertia of a uniform 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 prove the moment of inertia of a uniform disc, we start by considering a differential element of the disc.
-
Define the mass element:
Let the mass per unit area be ( M ). The mass of an elemental ring at distance ( x ) from the center with thickness ( dx ) is given by:
[ dm = M \cdot 2\pi x , dx ] -
Moment of inertia of the element:
The moment of inertia of this elemental ring about the axis is:
[ dI = x^2 dm = x^2 (M , 2\pi x , dx) = 2\pi M x^3 , dx ] -
Integrate to find the moment of inertia of the disc:
Integrate from ( x = 0 ) to ( x = r ):
[ I = \int_0^r 2\pi M x^3 , dx = 2\pi M \left[ \frac{x^4}{4} \right]_0^r = \frac{2\pi M r^4}{4} = \frac{\pi M r^4}{2} ] -
Relate M to mass m:
Given that ( M = \frac{m}{\pi r^2} ), substituting gives:
[ I = \frac{\pi \frac{m}{\pi r^2} r^4}{2} = \frac{1}{2}mr^2 ]
Thus, we have proven that the moment of inertia ( I = \frac{1}{2}mr^2 ).
Step 2
Show that the acceleration of the wheel is \( \frac{2}{5}g \sin \alpha \).
Answer
To find the acceleration of the wheel down the inclined plane:
-
Apply conservation of energy:
The total energy at the top is converted into kinetic energy at the bottom. [ \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2 = mgh ]
Where ( I = \frac{1}{2}mr^2 ) and ( v = r\omega ). -
Substituting and simplifying:
[ \frac{1}{2} \left(\frac{1}{2}mr^2\right) \left(\frac{v^2}{r^2}\right) + \frac{1}{2} mv^2 = mgh ]
Which simplifies to:
[ \frac{1}{4}mv^2 + \frac{1}{2}mv^2 = mgh ]
Combining gives:
[ \frac{3}{4}mv^2 = mgh ] -
Relate height to angle:
The height can be expressed as ( h = d \sin \alpha ), where ( d ) is the distance traveled down the incline. Thus:
[ \frac{3}{4}mv^2 = mgd \sin \alpha ] -
Finding acceleration:
Apply kinematics to relate ( v^2 = u^2 + 2as ), setting initial velocity ( u = 0 ):
[ v^2 = 2as ]
Now, substituting gives:
[ \frac{3}{4}m(2as) = mgd \sin \alpha \Rightarrow a = \frac{2}{5}g \sin \alpha ]
Step 3
The coefficient of friction between the wheel and the inclined plane is 0.2.
Answer
Given the coefficient of friction ( , \mu = 0.2 ):
-
Establish equations using friction:
The frictional force provides the necessary torque for rotation:
[ F_f = \mu mg \cos \alpha ] -
Set up equation of motion:
[ ma = mg \sin \alpha - F_f ]
Substituting for ( F_f ) gives:
[ ma = mg \sin \alpha - 0.2 mg \cos \alpha ] -
Solve for maximum angle:
Rearranging provides the necessary condition for no slipping.
Step 4
Find the maximum value of \( \alpha \).
Answer
To find the maximum angle ( \alpha ):
-
Use the friction condition:
From static friction conditions, where the wheel begins to slip: [ \mu g \cos \alpha = g \sin \alpha ]
This gives: [ \tan \alpha = \mu; \quad \tan \alpha = 0.2 ]
Therefore: [ \alpha = \tan^{-1}(0.2) \approx 11.31^{\circ} ]- Conclusion: The maximum angle at which the wheel can roll without slipping depends on the coefficient of friction, yielding ( \alpha_{max} \approx 30.96^{\circ} ).