Question

A large turntable with radius 6.00 m rotates about a fixed vertical axis, making one revolution in 8.00 s. The moment of inertia of the turntable about this axis is 1200 kg⋅m². You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is 73.0 kg. Since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point 3.00 m from the center of the turntable before your feet begin to slip.
What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?

Answer 1

Answer:

0.8024

Explanation:

From the given question; we can say that the angular momentum of the system is conserved if the net torque is  is zero.

So; $I_o \omega _o = I_2 \omega_2$

At the closest distance ; the friction is :

$f_s = \mu_s (mg)$

According to Newton's Law:

F = ma

F = mrω²

From conservation of momentum:

$I_o \omega _o = I_2 \omega_2$

$\omega_2= \frac { I_o \omega _o}{ I_2 }$

$\omega_2=( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })* \omega_o$

However ; since the static friction is producing the centripetal force :

$\mu_s (mg) = mr \omega_2^2$

$\mu _s = \frac{ \omega^2_2 *r }{g}$

The coefficient of static friction between the bottoms of your feet and the surface of the turntable can now be calculated by using the formula :

$\mu _s = \frac{ \omega^2_2 *r }{g}$

=  $( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })^2 * \frac{ \omega^2_o *r }{g}$

= $[\frac{1200+(73(6)^2)}{((1200)+(73)(3)^2)} ]^2*[\frac{(\frac{\pi}{4})^2(3)} {9.8}]$

= 0.8024