Question

Question 5.39 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?

Chapter Laws Of Motion Page 113

Answer 1

Hi

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction, μ = 0.15

Frequency of rotation, ν = 200 rev/min = 200 / 60 = 10 / 3 rev/s

The necessary centripetal force required for the rotation of the man is provided by the normal force (F).

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force.

(f = μF ) acting upward.
Hence, the man will not fall until:

mg < f
mg < μF= μmrω^2
g < μrω^2
ω > (g / μr) ^1/2

The minimum angular speed is given as:
ωmin = (g / μr)^ 1/2
= ( 10 / (0.15 X 3) )^1/2 = 4.71 rad s^-1
hope u got ur answer

Answer 2

Hey there !!!!!!

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Mass of man , m = 70kg

radius of hollow drum =3m 

Frequency 200rev/min = 200/60 = 20/3 rev/sec

coefficient of friction (u) = 0.15

Normal Force (N) provides the required centripetal force for rotation 

As man revolves , weight of man acts downwards which balances with frictional force 

If frictional force > mg then man will not fall

So mg<f

     mg<uN

    mg<umrw²

     w>(g/ur)¹/²

w is minimum when it equals the RHS

   w= (g/ur)¹/²= (10/3*0.15)¹/²= (4.74) rad/sec

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Hope this helped you.................