Question

Q. In a rotor, a hollow vertical cylinder rotates about its axis and a person rest against the inner wall. At a particular speed of the rotor, the floor below the person is removed and the person hangs resting against the wall without
any floor. If the radius of the rotor is 2 m and the coefficient of static firction between the wall and the person is 0.2. Find the minimum speed at which the floor may be removed.
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Answer 1

Answer:

10 m/s

Explanation:

$v = \sqrt{ \frac{rg}{ \alpha } } = \sqrt{ \frac{20}{0.2} } = 10ms$

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Answer 2

$\huge \color{blue} { \underline{ \underline \red{answer :-}}}$

The minimum speed at which the person remains at rest even when the floor is removed will be 10 m/s

$\bold \color{green} { \underline{ \underline \red{explanation :-}}}$

The minimum speed at which the person remains at rest even when the floor is removed, is given by,

$v = \sqrt{\frac{Rg}{u} } uRg$

where,

R = radius of the rotor = 2 m

g = acceleration due to gravity = 10 m/s²

u = co-efficient of friction = 0.2

Putting these value in the above formula we get,

$v = \sqrt{ \frac{2 \times 10}{0.2} } </p><p> \\ \implies \: v \: = \: \sqrt{100} \\ \implies \: v = 10 \frac{m}{s}$

Hence, The minimum speed at which the person remains at rest even when the floor is removed will be 10 m/s