Let there is a ball with radius equal to the Earth radius and mass equal to the Earth mass. Some small body is located on the ball. The ball begins to rotate around an axis perpendicular to the radius connecting the body to the center of the ball. Describe what will happen to the body as the angular velocity of the ball slowly increases. Between
surfaces of the body and the ball there is friction.
Decision:
When the ball is at rest, there are counterbalancing forces acting on the body. These are the gravitational force $F$ and the reaction force $N$. From the moment the ball begins to rotate, a third force, the friction force $F_{T}$ directed tangentially to the surface of the ball in the direction of rotation, begins to act on the body. This force initially gives the body an angular acceleration equal to the angular acceleration of the ball. Thanks to this, the body first moves with the ball without slipping, maintaining a constant position on its surface. The frictional force, or more precisely, the resting frictional force $F_{T}$ (the body is stationary relative to the surface of the ball the surface of the ball is stationary) cannot take any large value. It's limited to $F_{max} = \mu Q$, where $\mu$ is the coefficient of friction, and $Q$ is the normal pressure force equal in magnitude to the reaction force $N$ ($Q = N$). The force $N$ is not difficult to find. Under the action of the difference of forces $F$ and $N$, the body receives centripetal acceleration $a=\omega^{2}r$ (here $\omega$ is the angular velocity, $r$ is the radius of the ball). By Newton's II law, $F - N = m \omega^{2}r$ ($m$ is the mass of the body). Given that $F = \frac{GmM}{r^{2}}$ ($G$ is the gravitational constant, $M$ is the mass of the ball), we find:
$N=m \left ( G \frac{M}{r} - \omega^{2}r \right)$. (1)
Formula (1) shows that as the angular velocity of the ball $\omega$ increases, the reaction force $N$, and consequently the force of the normal pressure $Q$ of the body on the ball, and the value of the maximum rest friction force $F_{max}$ decrease. As the angular velocity of the ball increases further, the body will begin to slip, i.e., the angular velocity of the ball will be greater than the angular velocity of the body. Finally, at some value of the angular velocity of the body, the support reaction force and the friction force will turn to zero. It follows from (1) that
$\omega^{2}_{max}=G \frac{M}{r^{2}}$
At this angular velocity, the body stops pressing on the ball, acquires first space velocity and becomes a satellite orbiting in a circular orbit. Thereafter, only the angular velocity of the ball grows, and the angular velocity of the body remains constant.
surfaces of the body and the ball there is friction.
Decision:
When the ball is at rest, there are counterbalancing forces acting on the body. These are the gravitational force $F$ and the reaction force $N$. From the moment the ball begins to rotate, a third force, the friction force $F_{T}$ directed tangentially to the surface of the ball in the direction of rotation, begins to act on the body. This force initially gives the body an angular acceleration equal to the angular acceleration of the ball. Thanks to this, the body first moves with the ball without slipping, maintaining a constant position on its surface. The frictional force, or more precisely, the resting frictional force $F_{T}$ (the body is stationary relative to the surface of the ball the surface of the ball is stationary) cannot take any large value. It's limited to $F_{max} = \mu Q$, where $\mu$ is the coefficient of friction, and $Q$ is the normal pressure force equal in magnitude to the reaction force $N$ ($Q = N$). The force $N$ is not difficult to find. Under the action of the difference of forces $F$ and $N$, the body receives centripetal acceleration $a=\omega^{2}r$ (here $\omega$ is the angular velocity, $r$ is the radius of the ball). By Newton's II law, $F - N = m \omega^{2}r$ ($m$ is the mass of the body). Given that $F = \frac{GmM}{r^{2}}$ ($G$ is the gravitational constant, $M$ is the mass of the ball), we find:
$N=m \left ( G \frac{M}{r} - \omega^{2}r \right)$. (1)
Formula (1) shows that as the angular velocity of the ball $\omega$ increases, the reaction force $N$, and consequently the force of the normal pressure $Q$ of the body on the ball, and the value of the maximum rest friction force $F_{max}$ decrease. As the angular velocity of the ball increases further, the body will begin to slip, i.e., the angular velocity of the ball will be greater than the angular velocity of the body. Finally, at some value of the angular velocity of the body, the support reaction force and the friction force will turn to zero. It follows from (1) that
$\omega^{2}_{max}=G \frac{M}{r^{2}}$
At this angular velocity, the body stops pressing on the ball, acquires first space velocity and becomes a satellite orbiting in a circular orbit. Thereafter, only the angular velocity of the ball grows, and the angular velocity of the body remains constant.