circular motion - 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 ?

Grigori (last edited 1 year ago) (Student, UG2 Grade)

Mass of man m = 70 kg

Radius of the cylindrical drum r = 3 m

Frequency of the vertical axis f = 200 rev/min

Coefficient of the friction \mu = 0.15

Step 1: Note down the forces acting on the man

Frictional force acting upward is balanced by the weight of the man acting down ward

f_r = mg .....................(1)

The normal force acting towards the central axis gives centripetal force.

N = F_c

N = \frac{mv^2}{r} = mr \omega^2 ........(2)

Step 2: Set up an equation for the minimum angular velocity

Frictional force F_r = \mu * N

f_r = \mu * (mr \omega^2) \because \text{ equation (2) }

mg = \mu * (mr \omega^2) \because \text{ equation (1) }

g = \mu * (r \omega^2)

\omega = \sqrt{\frac{g}{\mu * r}}

Step 3: Find the minimum angular speed at which the drum rotates

\omega = \sqrt{\frac{10}{0.15 * 3}}

\omega = 4.7 rad/sec

Hence, minimum angular speed at which the drum rotates \omega = 4.7 rad/sec