Get Answers to all your Questions

header-bg qa

Q5.39 A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Answers (1)

best_answer

Frequency of rotation is :

                                               =\ \frac{200}{60}\ =\ \frac{10}{3}\ rev/sec

The required condition so that man will not fall :

                                              mg\ <\ f

or                                         mg\ <\ \mu F_N

or                                         mg\ <\ \mu mr\omega^2

or                                           \omega\ > \sqrt{\frac{g}{\mu r}}

And thus :

                                    \omega\ _{min}\ =\ \sqrt{\frac{10}{0.5\times 3}}\ =\ 4.71\ rad\ s^{-1}                              

Thus the required minimum rotational speed is 4.71 rad/s.

Posted by

Devendra Khairwa

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads