NLM BYU 54

In the first case, the ring is given only an angular velocity $\omega$. It stops in time $t_0$.
The torque due to kinetic friction $f$ provides the angular deceleration $\alpha$. $$ \tau = f \cdot r = I \alpha $$ $$ f \cdot r = (m r^2) \frac{\omega}{t_0} $$ $$ f = \frac{m r \omega}{t_0} $$ The magnitude of the kinetic friction force $f$ is constant and equal to $\frac{m r \omega}{t_0}$.

In the second case, the ring has an initial axial velocity $u$ and tangential velocity $v_t = \omega r$.
The resultant initial velocity is $V = \sqrt{u^2 + (\omega r)^2}$.
Kinetic friction always acts opposite to the direction of the resultant instantaneous velocity vector. Since the magnitude of friction $f$ is constant, the ring undergoes a constant deceleration $a_{res}$ along its helical path.

$$ a_{res} = \frac{f}{m} = \frac{r \omega}{t_0} $$

Since the frictional force opposes the total velocity vector, both the axial and tangential components of velocity decay proportionally. This means the path of the ring, if the cylinder surface were unrolled, would be a straight line.
The total stopping time $T$ depends on the initial resultant speed $V$:

$$ T = \frac{V}{a_{res}} = \frac{\sqrt{u^2 + (\omega r)^2}}{r \omega / t_0} $$

The distance traveled along the axial direction, $x$, is related to the total distance $S$ by the geometry of the velocity triangle (ratio of $u$ to $V$). Using the equation $S = \frac{V^2}{2a_{res}}$:

$$ x = S \cdot \frac{u}{V} = \frac{V^2}{2 a_{res}} \cdot \frac{u}{V} = \frac{u V}{2 a_{res}} $$

Substitute $V$ and $a_{res}$ back into the equation:

$$ x = \frac{u \sqrt{u^2 + (\omega r)^2}}{2 (r \omega / t_0)} = \frac{u t_0 \sqrt{u^2 + (\omega r)^2}}{2 \omega r} $$

Given values:
$r = 0.3 \text{ m}, \quad \omega = 10 \text{ rad/s}, \quad t_0 = 3.0 \text{ s}, \quad u = 4.0 \text{ m/s}$

$$ x = \frac{4.0 \times 3.0 \sqrt{4.0^2 + (10 \times 0.3)^2}}{2 \times 10 \times 0.3} $$ $$ x = \frac{12 \sqrt{16 + 9}}{6} = \frac{12 \times 5}{6} = 10 \text{ m} $$