First slide

Static friction

Question

In a certain machine at a fun fair, a man stands against the wall of a circular room of radius 3 m. The room rotates about a vertical axis through the centre of the floor at 2 rad/sec., then the floor is lowered- If the man does not slip vertically downwards, the least value of the coefficient of friction between the man and wall is

Moderate

A

$g / {Ï}^{2} r$
B

$g / Ïr$
C

$g^{2} / Ïr$
D

$g / {Ïr}^{2}$

Solution

When the machine rotates. the man also

rotates with the wall. The contact force R between the wall and man is centripetal force. Due to this contact force R, there is frictional force Âµ R. The upward frictional force is balanced by weight of the man. Hence
$\begin{array}{l} R = \frac{{mv}^{2}}{r} = {mÏ}^{2} r \\ and Î¼R = mg \\ Î¼ ({mÏ}^{2} r) = mg or Î¼ = \frac{g}{{Ï}^{2} r} . \end{array}$

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