91影视

When a stone is tied to a cord and rotated in the horizontal circle, then the force is experienced by the stone. The force acts in the outward direction.

The room radius is \(R = 5.5\;{\rm{m}}\).

The rotation frequency is \(N = 0.5\;{\rm{rev/s}}\).

The free-body diagram for a person is shown below,

Applying force balance in horizontal direction,

\(N = \frac{{m{v^2}}}{R}\)

Here, N is the normal reaction force, and v is the velocity.

The velocity in terms of frequency of rotation can be written as,

\(\begin{aligned}v &= R\omega \\v &= R\left( {2\pi N} \right)\\v &= 2\pi NR\end{aligned}\)

Applying force balance in vertical direction,

\(\begin{aligned}f &= mg\\\mu N &= mg\\\mu &= \frac{{mg}}{N}\end{aligned}\)

Here, \(\mu \) is the coefficient of friction.

Substitute the values of N in above equation,

\(\begin{aligned}\mu &= \frac{{mg}}{{\frac{{m{v^2}}}{R}}}\\\mu &= \frac{{Rg}}{{{v^2}}}\\\mu &= \frac{{Rg}}{{{{\left( {2\pi NR} \right)}^2}}}\end{aligned}\)

Substitute the given values in above equation,

\(\begin{aligned}\mu &= \frac{{\left( {5.5\;{\rm{m}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{{{\left( {2\pi \left( {0.5\;{\rm{rev/s}}} \right)\left( {5.5\;{\rm{m}}} \right)} \right)}^2}}}\\\mu &= 0.18\end{aligned}\)

Thus, theminimum coefficient of static friction is0.18.

The force pressing against the wall is present on the person, this force is known as centripetal force. The source of the force is the rotation of the cylinder.