NLM CYU 26

Solution to Question 26

Let us analyze the motion of the bead in the horizontal plane (top view). The bead undergoes steady circular motion, meaning its speed $v$ and the radius of its path $r$ are constant.

Let $r$ be the radius of the circular path of the bead and $R$ be the radius of the cylinder. The string connects the bead to a tangent point on the cylinder.

From the geometry of the tangent, the angle $\theta$ between the string and the velocity vector $\vec{v}$ (which is tangential to the path of radius $r$) satisfies:

$$ \cos \theta = \frac{R}{r} \quad \dots(1) $$

The free end of the cord is pulled with a constant speed $u$. This implies that the component of the bead’s velocity along the direction of the string must equal $u$. Therefore, we have the kinematic constraint:

$$ v \cos \theta = u \quad \dots(2) $$

Combining (1) and (2), we can relate the speeds and radii:

$$ v \left(\frac{R}{r}\right) = u \implies \frac{v}{r} = \frac{u}{R} $$

The bead is acted upon by two horizontal forces: the Tension $T$ from the cord and the Viscous Drag force $F_d = kv$. The drag force opposes the velocity.

Resolving forces in the tangential (along $\vec{v}$) and normal (radial, towards center) directions:

Tangential Direction: The component of tension along the motion must balance the drag force to maintain constant speed.

$$ T \cos \theta = kv \quad \dots(3) $$

Radial Direction: The component of tension perpendicular to the motion provides the necessary centripetal force.

$$ T \sin \theta = \frac{mv^2}{r} \quad \dots(4) $$

To eliminate Tension $T$, we divide equation (4) by equation (3):

$$ \frac{T \sin \theta}{T \cos \theta} = \frac{mv^2 / r}{kv} $$ $$ \tan \theta = \frac{mv}{kr} $$

Substitute $r = \frac{vR}{u}$ (derived from the kinematic relation $\frac{v}{r} = \frac{u}{R}$) into the expression above:

$$ \tan \theta = \frac{mv}{k \left( \frac{vR}{u} \right)} = \frac{mu}{kR} $$

Now we express $v$ in terms of $u$ and $\theta$. From equation (2), $v = \frac{u}{\cos \theta}$. Using the identity $\frac{1}{\cos \theta} = \sqrt{1 + \tan^2 \theta}$:

$$ v = u \sqrt{1 + \tan^2 \theta} $$

Substituting $\tan \theta = \frac{mu}{kR}$:

Finally, to find the radius $r$, we use the relation $r = R \frac{v}{u}$. Substituting the expression for $v$:

$$ r = R \frac{u \sqrt{1 + \left(\frac{mu}{kR}\right)^2}}{u} $$