Question

A child is playing in a park on a rotating cylinder of radius, r , is set in rotation at an angular speed of w. The Base of the cylinder is slowly moved away, leasing the child suspended against the wall in a vertical position.

What Is the minimum coefficient of friction between the child's clothing and wall is needed to prevent it from falling .

Answer 1

Answer:

$\mathbf{\mu_s = \dfrac{g}{\omega^2r}}$

Explanation:

From the given information:

The force applied to the child should be at equilibrium in order to maintain him vertically hung on the wall.

Also, the frictional force acting on the child against gravitational pull is:

$F_f = \mu _sN$

where,

the centripetal force $F_c$ acting outward on the child is equal to the normal force.

$F_c= N$

SO,

$F_f = \mu_s F_c$

Since the centripetal force $F_c = \dfrac{mv^2}{r}$

Then:

$F_f = \dfrac{ \mu_s \times mv^2}{r}$

Using Newton's law, the frictional force must be equal to the weight

$F_f = W$

$\dfrac{ \mu_s \times mv^2}{r} = mg$

$\dfrac{ \mu_s v^2}{r} = g$

Recall that:

The angular speed $\omega = \dfrac{v}{r}$

Therefore;

$g = \mu_s \omega^2 r$

Making the coefficient of friction $\mu_s$ the subject of the formula:

$\mathbf{\mu_s = \dfrac{g}{\omega^2r}}$