I think you should hold a stretch for 10-30 seconds
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration.
The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N.
a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r.
The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec.
So a_N = 114 m/sec^2.
g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

Here



Mass inside the orbit in terms of Volume and Density is

Where,
V = Volume
Density
Now considering the volume of the star as a Sphere we have

Replacing at the previous equation we have,

Now replacing the mass at the gravitational acceleration formula we have that


For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

At the same time the general expression for the centripetal acceleration is

Where
is the orbital velocity
Using this expression in the left hand side of the equation we have that



Considering the constant values we have that


As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.
So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density