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  • A man of 50 pounds was made to stand against the inner wall of a hollow cylinder and the rotor of radius 5m is rotating at its vertical axis, and the coefficient of friction between the wall and hi

A man of 50 pounds was made to stand against the inner wall of a hollow cylinder and the rotor of radius 5m is rotating at its vertical axis, and the coefficient of friction between the wall and his clothing is 0.35. What is the minimum angular speed of the rotor so that the person does not slip downward?

Option: 1

2rads^{-1}


Option: 2

6rads^{-1}


Option: 3

5rads^{-1}


Option: 4

0.3rads^{-1}


Answers (1)

best_answer

The cylinder is in a vertical position. The normal reaction of the wall on man acts horizontally and it is equal to the centripetal force acting on it.

R=mr\omega^2

Then the frictional force acts upward

F\ =\ mg

The man will still remain stuck to the wall after the floor is removed he will continue to rotate with the cylinder

\mu\geq\frac{F}{R}

F\le\mu R

mg\le\mu mr\omega^2

\omega^2\le\frac{g}{\mu R}

\omega\le\sqrt{\frac{g}{\mu R}}

The minimum angular speed of rotation

 \omega=\sqrt{\frac{g}{\mu R}}

\mu=0.35,\ r=5m,\ g=9.8m/s^2

\omega=\frac{\sqrt{9.8}}{0.35}\times5

\omega=2.31rads^{-1}=2rads^{-1}

 

Posted by

Anam Khan

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