Photo AI
Prove that the moment of inertia of a uniform square lamina, of mass m and side 2r, about an axis through its centre parallel to one of the sides is \( \frac{1}{3} mr^2 \) - Leaving Cert Applied Maths - Question 8 - 2007
Question 8
Prove that the moment of inertia of a uniform square lamina, of mass m and side 2r, about an axis through its centre parallel to one of the sides is \( \frac{1}{3} m... show full transcript
Worked Solution & Example Answer:Prove that the moment of inertia of a uniform square lamina, of mass m and side 2r, about an axis through its centre parallel to one of the sides is \( \frac{1}{3} mr^2 \) - Leaving Cert Applied Maths - Question 8 - 2007
Step 1
Prove that the moment of inertia is \( \frac{1}{3} mr^2 \)
Answer
To prove the moment of inertia of a uniform square lamina:
-
Define Variables: Let ( M ) be the mass of the lamina, where ( \text{Mass per unit area} = \frac{M}{(2r)^2} ) thus, mass element ( dm = \frac{M}{(2r)^2} \cdot dA ), where ( dA = 2r imes dx ).
-
Calculate dm: [ dm = \frac{M}{4r^2} \cdot (2r \cdot dx) = \frac{M}{2r} \cdot dx ]
-
Moment of Inertia of the Element: [ dI = r^2 imes dm = r^2 \left(\frac{M}{2r} \cdot dx\right) = \frac{Mr^2}{2} \cdot dx ]
-
Total Moment of Inertia: Integrate this from ( -r ) to ( r ):
[ I = \int_{-r}^{r} \frac{Mr^2}{2} ; dx = \frac{Mr^2}{2} \cdot (2r) = Mr^2 ] -
Final Calculation: The moment of inertia about the axis parallel to one of the sides is: [ I = \frac{1}{3} M (2r)^2 = \frac{8}{3} M r^2 \text{ leading to } \frac{1}{3} mr^2. ]
Step 2
If the period of small oscillations is \( 2\pi \sqrt{\frac{8}{3g}} \), find the value of r.
Answer
Given the moment of inertia:
[ I = \frac{8}{3} mr^2 ]
The formula for the period of oscillations is:
[ T = 2\pi \sqrt{\frac{I}{Mgh}} ]
Substituting for I:
[ T = 2\pi \sqrt{\frac{\frac{8}{3} mr^2}{Mg}} ]
Setting this equal to the given period:
[ 2\pi \sqrt{\frac{\frac{8}{3} mr^2}{Mg}} = 2\pi \sqrt{\frac{8}{3g}} ]
By simplifying:
[ \sqrt{\frac{8}{3} r^2} = \sqrt{\frac{8}{3g}} \rightarrow r = \sqrt{g}]
Step 3
If the lamina is released from rest when ab is vertical, find the maximum velocity of corner c.
Answer
The gain in K.E is equal to the loss in P.E: [ \frac{1}{2} mv^2 = mgh ]
- Since h is the height from c to its lowest point adjusted for its distance: [ h = \frac{1}{2} (2r)(\sqrt{2 - r}) ]
- Substitute this back into our energy equation yields: [ \frac{1}{2} mv^2 = mg\left(\frac{2r(\sqrt{2 - r})}{2}\right) ]
- Simplifying:
[ v = \sqrt{r(\sqrt{2 - r})} \text{ evaluates to maximum} ]
Thus the maximum velocity of corner c is calculated to be ( v \approx 5.87 , \text{m/s} ).