Physics: 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

A large turntable with radius 6.00 m rotates about

denym58

26.01.2022

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

0,0(0 оценок)