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Question
A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have identical areas and masses (Figure).
Show that
- IxR/IxS < 1
- IyR/IyS > 1
- IzR/IzS > 1
Solution
According to the problem,
Area of square = area of rectangular plate
⇒ `c^2 = a xx b`
⇒ `c^2 = ab`
a. `I_(xR)/I_(xS) = b^2/c^2` ......[∵ I ∝ (area)2]
It is clear from the diagram that b < c
⇒ `I_(xR)/I_(xS) = (b/c)^2 < 1`
⇒ `I_(xR) < I_(xS)`
b. `I_(yR)/I_(yS) = a^2/c^2` ......(It is clear that a < c)
`I_(yR)/I_(yS) - (a/c)^2 > 1`
c. `I_(zR) = 1/12 M(a^2 + b^2)`
`I_(zS) = 1/12 M(c^2 + c^2)`
Now, `I_(zR) - I_(zS) = 1/12 M[a^2 + b^2 - 2c^2]`
= `1/12 M(a^2 + b^2 - 2ab)`
`I_(zR) - I_(zS) = 1/12 M(a - b)^2 > 0`
⇒ `I_(zR)/I_(zS) > 1`
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