A person with a mass of M kg stands in contact against the wall of the cylindrical drum…
Physics · JEE Main · NTA Exams — Rotational Motion
A person with a mass of M kg stands in contact against the wall of the cylindrical drum of radius r rotating with an angular velocity ω. The coefficient of friction between the wall and the clothing is µ. The minimum rotational speed of the cylinder which enables the person to remain stuck to the wall when the floor is suddenly removed is -
- ωmin = \(\sqrt{\frac{\mathrm{g}}{\mu \mathrm{r}}}\)
- ωmin = \(\sqrt{\frac{\mu \mathrm{r}}{\mathrm{~g}}}\)
- ωmin = \(\sqrt{\frac{2 \mathrm{~g}}{\mu \mathrm{r}}}\)
- ωmin = \(\sqrt{\frac{\mathrm{gr}}{\mu}}\)
Answer
(A) ω min = g r
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