Solution to Question 27
The force at one end is removed, and the rope is pulled by a force $F$ at the other end. The entire rope of mass $m$ accelerates together. $$ a = \frac{F}{m} $$ Let $t$ be the time elapsed. The speed $v$ and displacement $S$ of the rope are: $$ v = at, \quad S = \frac{1}{2}at^2 $$
Consider an element of the rope on the curved surface. The normal force $N$ vanishes when the inward radial component of tension is insufficient to provide the necessary centripetal force.
Radial Equation: $\frac{T}{r} – N = \lambda v^2$
For contact to be lost ($N=0$), the condition is:
$$ T \le \lambda v^2 $$
Since tension $T$ varies along the rope, contact is lost first at the point on the curve where tension is minimum. This minimum tension occurs at the “rear” junction where the straight tail segment meets the cylinder (marked red in the diagram).
Let $L_0$ be the initial length of the straight segments. Since the rope makes a half turn ($\pi r$) and the total length is $l$: $$ L_0 = \frac{l – \pi r}{2} $$ At time $t$, the rope has moved a distance $S$. The remaining length of the straight tail $y$ is: $$ y = L_0 – S $$
The tension at distance $y$ from the free end is simply the force required to accelerate the mass of that segment ($m_{segment} = \lambda y$). $$ T(y) = (\lambda y) a $$ Substituting this into the losing contact condition ($T = \lambda v^2$): $$ \lambda y a = \lambda v^2 $$ $$ y a = v^2 $$ Now, substitute kinematic variables $y = L_0 – S$ and $v^2 = 2aS$: $$ (L_0 – S) a = 2aS $$ $$ L_0 – S = 2S \implies L_0 = 3S $$ $$ S = \frac{L_0}{3} $$
Substitute $S = \frac{1}{2}at^2$ into the equation: $$ \frac{1}{2}at^2 = \frac{L_0}{3} $$ $$ t^2 = \frac{2L_0}{3a} $$ Finally, substitute $L_0 = \frac{l – \pi r}{2}$ and $a = \frac{F}{m}$: $$ t^2 = \frac{2}{3(F/m)} \left( \frac{l – \pi r}{2} \right) = \frac{m(l – \pi r)}{3F} $$ $$ t = \sqrt{ \frac{ml}{3F} \left( 1 – \frac{\pi r}{l} \right) } $$