One end of a light inextensible thread of length l is held stationary over a frictionless horizontal

Q: Why is the initial radial velocity $\dot{r}$ equal to zero?

Before the thread is pulled, the hand is held stationary, so its vertical velocity $\dot{h} = 0$. According to the constraint equation $r\dot{r} + h\dot{h} = 0$, if $\dot{h}$ is zero, then $\dot{r}$ must also be zero to satisfy the equation.

Q: Why isn't the acceleration just the standard centripetal acceleration $v_0^2/r$?

Because the radius $r$ is no longer constant! The sudden pulling of the thread introduces a dynamic change in the radius, represented by the second derivative $\ddot{r}$, which adds an extra inward acceleration component on top of the standard centripetal requirement.

Q: Does the tension in the string change the moment it is pulled?

Yes, absolutely. The sudden upward acceleration of the hand requires a sharp increase in tension to both accelerate the mass of the hand (if it had any) and to provide the necessary inward force to accelerate the bead radially.

Q: Why is there no tangential acceleration?

The floor is frictionless, and the tension force from the string acts purely in the plane containing the vertical axis and the radial vector. Since there is no force component pointing in the tangential direction, the bead's tangential speed $v_0$ cannot change instantaneously.

Q: What does the negative sign in $\ddot{r} = -a_0 \cot\theta$ physically mean?

The negative sign indicates that the second derivative of the radius is negative. This means the radius $r$ is decreasing at an accelerating rate; the bead is being actively pulled inwards towards the center axis.

Question

One end of a light inextensible thread of length $l$ is held stationary over a frictionless horizontal floor while a small bead tied at the other end of the thread is describing a circular path with a uniform speed $v_0$ on the floor as shown in the figure. The upper end of the thread is suddenly pulled vertically upwards with a constant acceleration $a_0$. If the bead does not leave the floor, find magnitude of its acceleration immediately after the upper end of the thread is pulled.

Sigma Percentile · Accepted Answer

### The Deceptive Simplicity of the Pulled Bead

Imagine a bead spinning peacefully in a circle on a frictionless floor, tethered by a string. Suddenly, the center of the string is yanked violently upwards. What happens to the bead? Your intuition might scream that it just gets pulled inwards, but the exact mathematical reality of *how* it accelerates is a beautiful dance of geometry and polar kinematics.

This problem is a classic test of your ability to handle **constrained motion** in non-Cartesian coordinate systems. Let's break down the physics step-by-step.

### The Master Geometric Constraint

The absolute foundation of this problem is the string itself. It is inextensible, meaning its length $l$ is a strict constant. If we look at the setup, the string forms the hypotenuse of a right-angled triangle. The vertical height of the hand is $h$, and the horizontal distance from the center to the bead is the radius $r$.

By the Pythagorean theorem, we can write our master constraint equation:

$$l^2 = r^2 + h^2$$

This simple equation binds the horizontal motion of the bead to the vertical motion of the hand.

### Kinematics: Differentiating the Constraint

To find out how the accelerations are linked, we must differentiate our constraint equation with respect to time. Let's take the first derivative:

$$0 = 2r\frac{dr}{dt} + 2h\frac{dh}{dt}$$

Dividing by 2, we get:

$$r\dot{r} + h\dot{h} = 0$$

This tells us how the radial velocity $\dot{r}$ is related to the vertical velocity $\dot{h}$. But we need accelerations, so we must differentiate a second time. Applying the product rule carefully:

$$\dot{r}^2 + r\ddot{r} + \dot{h}^2 + h\ddot{h} = 0$$

Now, we apply the specific initial conditions given in the problem. We are asked for the acceleration *immediately* after the thread is pulled. Right before the pull, the hand is stationary, meaning $\dot{h} = 0$. Looking at our first derivative equation, if $\dot{h} = 0$, then $\dot{r}$ must also be $0$.

Substituting these zero velocities and the given upward acceleration of the hand ($\ddot{h} = a_0$) into our second derivative equation, the math simplifies beautifully:

$$0 + r\ddot{r} + 0 + h(a_0) = 0$$

Solving for the radial acceleration component $\ddot{r}$:

$$\ddot{r} = -a_0 \frac{h}{r}$$

Looking back at our right-angled triangle, the ratio $\frac{h}{r}$ is exactly $\cot	heta$. Therefore:

$$\ddot{r} = -a_0 \cot	heta$$

### The Polar Coordinate Trap

Here is where many students make a fatal error. They assume that $\ddot{r}$ is the total acceleration of the bead. **It is not.**

The bead is moving in a 2D plane, and we are tracking it using polar coordinates $(r, \phi)$. In polar coordinates, the total radial acceleration is not just the second derivative of the radius. It is given by the formula:

$$a_r = \ddot{r} - r\dot{\phi}^2$$

What is that second term? It is the **centripetal acceleration** required to keep the bead moving in a circle. We know the bead has a tangential speed $v_0$, so $r\dot{\phi} = v_0$. This means the centripetal term is:

$$r\dot{\phi}^2 = \frac{v_0^2}{r} = \frac{v_0^2}{l\sin	heta}$$

### Synthesizing the Final Result

Now we combine both effects. The pulling of the string forces the radius to shrink (contributing $\ddot{r}$), and the circular motion demands an inward pull (contributing $-r\dot{\phi}^2$).

$$a_r = -a_0 \cot	heta - \frac{v_0^2}{l\sin	heta}$$

The negative sign simply indicates that the net acceleration is directed inwards, towards the center of the circular path. The problem asks for the **magnitude** of this acceleration, which is:

$$|a_r| = a_0 \cot	heta + \frac{v_0^2}{l\sin	heta}$$

This elegant expression captures both the dynamic pulling effect and the inherent kinematic requirement of circular motion. It is a perfect demonstration of why rigorous coordinate geometry is essential in advanced mechanics.

Visualized Solution

$l^{2} = r^{2} + h^{2}$

$r \overset{r}{˙} + h \dot{h} = 0$

$\overset{r}{˙}^{2} + r \overset{r}{¨} + \dot{h}^{2} + h \ddot{h} = 0$

$\overset{r}{¨} = - a_{0} cot θ$

$a_{n e t} = a_{0} cot θ + \frac{v _{0}^{2}}{l s i n θ}$

$N \geq 0$

The Sigma Insight: Kinematics of Circular Motion

The Deceptive Simplicity of the Pulled Bead

The Master Geometric Constraint

Kinematics

The Polar Coordinate Trap

Synthesizing the Final Result

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Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rr˙+hh˙=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r˙2+rr¨+h˙2+hh¨=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r¨=−a0​cotθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } anet​=a0​cotθ+lsinθv02​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N≥0