InterviewSolution
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InterviewSolution
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A non-uniform bar of weight `W` is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are `36.9^(@) and 53.1^(@)` respectively. The bar is `2m` long. Calculate the distance `d` of the centre of gravity of the bar from its left end. |
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Answer» As it is clear from Fig., `theta_(1)=36.9^(@), theta_(2)=53.1^(@)`. It `T_(1),T_(2)` are the tensions in the two strings, then for equilibrium along the horizontal, `T_(1) sin theta_(1)=T_(2) sin theta_(2)` or ` (T_(1))/(T_(2))=(sin theta_(2))/(sin theta_(1))=(sin 53.1^(@))/(sin36.9^(@))` `=(0.7404)/(0.5477)=1.3523` Let d be the distance of centre of gravity C of the bar from the left end. For rotaional equilibrium about C, `T_(1) cos theta_(1) xxd=T_(2) cos theta_(2) (2-d)` `T_(1) cos 36.9^(@)xxd=T_(2) cos 53.1^(@) (2-d)` `T_(1)xx0.8366 d= T_(2)xx0.6718 (2-d)` Put `T_(1)=13523 T_(2)` and solve to get d=0.745 m `tau=I_(1)alpha_(1)=I_(2)alpha_(2)` or ` (alpha_(1))/(alpha_(2))=(I_(2))/(I_(1))=((2//5)MR^(2))/(MR^(2))=(2)/(5)` or ` alpha_(2)=(5)/(2) alpha_(1)`. |
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