Velocity of a ring tied to an inextensible rope

Question

In the diagram, the pulley is weightless and friction less and the thread connecting the pulley with the ring is inextensible. The thread is attached to the ring which in turn can move only horizontally. Suppose we start pulling the thread with a velocity $v$ $ ms^{-1}$. At this instant the thread makes an angle $\Theta$ with the horizontal. My question is : Why dont we take the velocity of the ring as $v$ $cos$ $\Theta$?

Instead my book says that the velocity of the rope is actually $$v_{rope}= v_{ring} cos \Theta $$ where angle $\Theta$ is the instantaneous angle between the horizontal and the thread, which is completely the opposite to I am thinking.

My logic: The rope is pulling the ring at an angle so only the component of the the velocity of the rope in the direction of movement of ring should act make it move, which ultimately makes the velocity of ring as $v_{ring} = v_{rope} cos \Theta$.

Can anyone provide me with a mathematical description of this fact?

Answer 1

Refer to the image above.

Applying Pythagoras' Theorem,

$$x^2 + y^2 = z^2$$

Now, differentiating this w.r.t time, we obtain:

$$2x\frac{\mathrm dx}{\mathrm dt} + 2y\frac{\mathrm dy}{\mathrm dt} = 2z\frac{\mathrm dz}{\mathrm dt}$$

Obviously, the length $y$ remains constant with time. Therefore,

$$\frac{\mathrm dy}{\mathrm dt} = 0$$

This condenses our equation to:

$$2x\frac{\mathrm dx}{\mathrm dt} = 2z\frac{\mathrm dz}{\mathrm dt}$$

Clearly, $\frac{\mathrm dx}{\mathrm dt}$ is equal to $v_{\rm ring}$, while $\frac{\mathrm dz}{\mathrm dt}$ is equal to $v_{\rm rope}$.

Therefore, the equation now becomes:

$$x v_{\rm ring} = z v_{\rm rope}$$

which can be written as,

$$v_{\rm ring} = \frac{v_{\rm rope}}{\cos\theta}$$

since $\cos\theta = \dfrac zx$

Rearranging,

$$v_{\rm rope} = v_{\rm ring} \cos\theta$$

Answer 2

The problem with your idea is that it does not take account of the inclination of the rope relative to the rod changing.

The relationship $v_{\rm rope}= v_{\rm ring} \cos \theta $ predicts the right values for the speed of the rope as $\theta \rightarrow 0^\circ$ and $\theta \rightarrow 90^\circ$, $v_{\rm ring}$ and zero respectively.

Consider what happens from a time $t$ to a time $t+\Delta t$ as shown in the diagram.

$\Delta x_{\rm rope}\approx \Delta x_{\rm ring} \cos \theta $ and the approximation gets better as the time interval gets smaller and smaller.
When divided by the time interval $\Delta t$ and taking the limit of the time interval $\Delta t$ tending to zero gives the equation which have quoted.