A thin hoop of mass m slides frictionlessly on a smooth horizontal surface with velocity $v_{0}$, without rotation. The hoop axis is horizontal, vector $v_{0}$ lies in the plane of the hoop. At time $t=0$, the smooth surface is replaced by a rough surface with friction coefficient $\mu$. Find the steady-state velocity of the center of the hoop.
Decision:
As soon as the hoop hits the rough surface, a sliding friction force $F_{T}=\mu mg$ with momentum $M=F_{T}R$ relative to the hoop axis begins to act on it from the surface. Under the action of a constant force $F_{T}$ the velocity $v$ of the hoop center of mass decreases and can be found by the formula.
$v=v_{0}-at=v_{0}-\frac{F_{T}t}{m}=v_{0}-\mu gt$. (1)
on the other hand, the momentum $M$ gives the hoop angular acceleration
$\beta = \frac{M}{J} =\mu gR$,
($I = mR^{2}$ is the hoop's moment of inertia), and the hoop begins to rotate about its axis with an angular velocity of
$\omega=\beta t = \frac{ \mu gt}{R}$,
i.e., all its points in a stationary frame of reference acquire linear velocity
$v^{\prime}=\omega R = \mu gt$. (2)
The frictional force acts until the velocities $v$ and $v^{\prime}$ are equal and the hoop begins to roll without slipping. From formulas (1) and (2) we get equation
$v_{0} - \mu gt= \mu gt$
for finding the moment of time $t$ when the force $F$ ceases. Solving it, we obtain
$t=\frac{v_{0}}{2} \mu g$. (3)
Substituting into formula (1) the value of $t$ from equality (3), we find the speed of steady motion of the hoop center
$v= \frac{v_{0}}{2}$
Decision:
As soon as the hoop hits the rough surface, a sliding friction force $F_{T}=\mu mg$ with momentum $M=F_{T}R$ relative to the hoop axis begins to act on it from the surface. Under the action of a constant force $F_{T}$ the velocity $v$ of the hoop center of mass decreases and can be found by the formula.
$v=v_{0}-at=v_{0}-\frac{F_{T}t}{m}=v_{0}-\mu gt$. (1)
on the other hand, the momentum $M$ gives the hoop angular acceleration
$\beta = \frac{M}{J} =\mu gR$,
($I = mR^{2}$ is the hoop's moment of inertia), and the hoop begins to rotate about its axis with an angular velocity of
$\omega=\beta t = \frac{ \mu gt}{R}$,
i.e., all its points in a stationary frame of reference acquire linear velocity
$v^{\prime}=\omega R = \mu gt$. (2)
The frictional force acts until the velocities $v$ and $v^{\prime}$ are equal and the hoop begins to roll without slipping. From formulas (1) and (2) we get equation
$v_{0} - \mu gt= \mu gt$
for finding the moment of time $t$ when the force $F$ ceases. Solving it, we obtain
$t=\frac{v_{0}}{2} \mu g$. (3)
Substituting into formula (1) the value of $t$ from equality (3), we find the speed of steady motion of the hoop center
$v= \frac{v_{0}}{2}$