Question

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.​

Answer 1

Answer:

• Radius of Rotor = 2 m
• Acceleration = 10 m/s²
• μ = 0.2

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

$\displaystyle\sf \dashrightarrow v = \sqrt{\dfrac{Rg}{\mu}}$

• R = Radius of rotor
• g = Acceleration due to gravity
• μ = Coefficient of friction

$\\\displaystyle\sf \dashrightarrow v = \sqrt{\dfrac{2\times 10}{0.2}}$

$\displaystyle\sf \dashrightarrow v = \sqrt{\dfrac{20}{0.2}}$

$\displaystyle\sf \dashrightarrow v = \sqrt{\dfrac{20}{0.2}\times\dfrac{10}{10}}$

$\displaystyle\sf \dashrightarrow v = \sqrt{\dfrac{200}{2}}$

$\displaystyle\sf \dashrightarrow v = \sqrt{100}$

$\displaystyle \dashrightarrow \underline{\boxed{\sf v = 10 \ m/s}}$

$\displaystyle\therefore\:\underline{\textsf{The minimum speed will be \textbf{ 10 m}}/ \textsf{\textbf{s}}}$

Another Way

• Radius of rotor = 2 m
• Acceleration = 9.8 m/s²
• μ = 0.2

[You may note that we have used the acceleration as 9.8 m/s² here]

$\displaystyle\sf :\implies v = \sqrt{\dfrac{Rg}{\mu}}$

$\displaystyle\sf :\implies v = \sqrt{\dfrac{2\times 9.8}{0.2}}$

$\displaystyle\sf :\implies v = \sqrt{\dfrac{19.6}{0.2}}$

$\displaystyle\sf :\implies v = \sqrt{98}$

$\displaystyle\sf :\implies \underline{\boxed{\sf v = 9.9 \ m/s}}$

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,

$v = \sqrt{ \frac{2 \times 10}{0.2} }$

$\implies \: v \: = \: \sqrt{100} \\ \implies \: v = 10 \frac{m}{s}$

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