Animated Solution for Physics - Kinematics: A boat is moving with uniform velocity $v_b=20\text{ m/s}$ pulling a water skier with the help of a tug-rope of length $l=10\text{ m}$. To increase his speed the water skier tilts the skies slightly away from the direction of motion of the boat. As he does so, the tug rope rotates. What is the speed $v_s$ of the skier with respect to the ground and angular velocity $\omega$ of the rope, when $\theta =30^{\circ }$ and $\varphi =60^{\circ }$?

Mastering Relative Velocity

Imagine you are standing on the shore, watching a boat pull a water skier across a pristine lake. It looks like a simple, graceful dance, but beneath the surface, it is a beautiful symphony of vectors, constraints, and relative motion. In this problem, we are tasked with finding the skier's speed and the angular velocity of the rope. Let's dive into the mechanics of this setup.

The Inextensible Rope

The absolute key to unlocking this problem lies in the rope itself. We are told it is a "tug-rope," which in the realm of physics means it is inextensible—its length $l$ cannot change.

Why is this so crucial? Because it dictates how the boat and the skier must move relative to one another. If you draw a straight line connecting the boat to the skier (the radial axis), the distance between them along this line is locked. Therefore, the rate at which the boat pulls the rope must be exactly matched by the rate at which the skier moves along that same line. If the skier moved slower, the rope would stretch and snap; if the skier moved faster, the rope would go slack.

The Radial Axis

To apply our constraint, we must resolve the velocities of both the boat ($v_b$) and the skier ($v_s$) into components that are parallel to the rope (radial) and perpendicular to the rope (tangential).

The boat is moving horizontally to the left, and the rope makes an angle $\theta$ with this horizontal line. Using basic trigonometry, the component of the boat's velocity pulling directly along the rope is $v_b\cos \theta$.

Similarly, the skier's velocity vector $v_s$ makes an angle $\varphi$ with the rope. The component of the skier's velocity directed along the rope is $v_s\cos \varphi$.

Equating these two radial components gives us our master equation for speed:

Rearranging to solve for the skier's speed:

Plugging in the given values ($v_b=20\text{ m/s}$, $\theta =30^{\circ }$, $\varphi =60^{\circ }$):

The Tangential Axis

Now, what about the rotation of the rope? A rigid object (or a taut rope) rotates when its two ends have different velocities perpendicular to the line connecting them.

Let's look at the tangential components. The boat has a velocity component perpendicular to the rope of $v_b\sin \theta$. The skier has a perpendicular component of $v_s\sin \varphi$.

Because these two ends are moving at different speeds in the perpendicular direction, the rope must sweep out an angle. The relative perpendicular velocity between the two ends is the difference between these components:

In circular motion, we know that velocity is related to angular velocity by $v=\omega r$. Here, the "radius" is the length of the rope $l$, and the velocity causing the rotation is our relative perpendicular velocity. Therefore:

Let's calculate the individual pieces:

The relative perpendicular velocity is $30-10=20\text{ m/s}$.

Finally, dividing by the rope's length ($l=10\text{ m}$):

By breaking the complex 2D motion into independent radial and tangential axes, a seemingly daunting problem transforms into a straightforward application of constraints and relative velocity. This is the true elegance of vector resolution in kinematics!