Question

In a rotor,a hollow vertical cylindrical structure rotates about its axis and a person rests 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 2m and the coefficient of friction between the wall and the person is 0.2,find the minimum speed at which the floor may be removed.(take g=10m/s²)
PLEASE HELP ME......​

Answer 1

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

Explanation :

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

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

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 = √(2 x 10/.2)

=> v = √100

=> v = 10 m/s

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

Answer 2

Explanation:

I hope it help you thank you