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A small object loops a vertical loop in which a symmetrical section of angle 2alphahas been removed (Fig.) Find the maximum and the minimum heights from which the object, after loosing contact with the loop at point A and flying through the air, will reach point B. Find the corresponding angles of the section removed for which this is possible. |
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Answer» `L=(2v_0^2sinalphacosalpha)/g` So `v_0^2=(gR)/(COSALPHA)` To find the speed `v_0` apply the law of conservation of energy `mgH = mgH (1+ cos alpha)+1/2 mv_0^2` for which `k=H/R=1+cos alpha+1/(2 cosalpha)`. For the computation of `cos alpha ` we obtain an equation `2 cos^2alpha-2(k-1) cos alpha+1=0` `cosalpha=1/2(k-1pmsqrt((k-1)^2-2))` Since the number under the root sign must be non - negative , we obtain `k-1gesqrt2` i.e. `kge1+sqrt2`. On the other hand `0 lt cos alpha le 1`, i.e. `k-1 +sqrt((k-1)^(2)-2) le le 2 and k le 2.5`. Thus `1+sqrt2 le k le 2.5 `, i.e. `(1+sqrt2)R le H le 2.5R` For the limiting VALUES of the cosines we have `cosalpha=sqrt2/2,e. alpha_1=45^@` `cos alpha=(1.5pm0.5)//2,cosalpha_(2)=0.5, cosalpha_3=1` Obviously the solution `cos alpha_(3) =1` does not satisfy the condition of the problem since for `alpha_3=0` there will be no cut. So the remaining solution is `cos alpha_(2)=0.5 , alpha_(2) = 60^@` Accordingly for the heights in the range `(1+sqrt2)R le H le le 2.5R` corresponding suitable cuts are those with angles of`45^@ le le alpha le 60^@` chosen so as to satisfy the condition `cosalpha=1/2(k-1-sqrt(k^2-2k-1))` |
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