Question Number 225820 by mr W last updated on 13/Nov/25
Commented by mr W last updated on 13/Nov/25
Answered by mr W last updated on 13/Nov/25
Commented by mr W last updated on 14/Nov/25
$${dT}=−\lambda{gR}\:\mathrm{cos}\:\theta\:{d}\theta \\ $$$$\Rightarrow{T}={T}_{\mathrm{0}} −\lambda{gR}\:\mathrm{sin}\:\theta \\ $$$$\left({p}+\lambda{g}\:\mathrm{sin}\:\theta\right){Rd}\theta={T}\:{d}\theta \\ $$$$\Rightarrow{p}=\frac{{T}_{\mathrm{0}} }{{R}}−\mathrm{2}\lambda{g}\:\mathrm{sin}\:\theta \\ $$$${at}\:\theta=\frac{\pi}{\mathrm{2}},\:{i}.{e}.\:{at}\:{point}\:{C}: \\ $$$${p}_{{min}} =\frac{{T}_{\mathrm{0}} }{{R}}−\mathrm{2}\lambda{g}\geqslant\mathrm{0} \\ $$$$\Rightarrow{T}_{\mathrm{0}} \geqslant\mathrm{2}\lambda{gR} \\ $$$${if}\:{T}_{\mathrm{0}} \:<\mathrm{2}\lambda{gR},\:{rope}\:{near}\:{point}\:{C} \\ $$$${will}\:{start}\:{to}\:{lose}\:{contact}\:{to}\:{the}\: \\ $$$${cylinder}. \\ $$
Commented by mr W last updated on 14/Nov/25
