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A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is |
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Answer» `(pL^(3))/(8pi^(2))` `therefore` Mass of wire = pL Since the wire of length L is TURNED into a circle, `2piR=LrArrR=L/(2PI)` M.I. of loop about given axis = `MR^(2)+(MR^(2))/2=3/2MR^(2)` M.I. of loop =`3/2xx(pL)(L/(2pi))^(2)=3/8.(pL^(3))/(pi^(2))` |
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