A small bob tied at one end of a thin string of length 1m is describing a vertical circle, so that t

Q: Why is the tension maximum at the bottom of the circle?

At the bottom, the string must pull upwards to counteract the downward pull of gravity AND provide the extra upward force needed for centripetal acceleration. Hence, $T_{\text{max}} = mg + mv^2/r$.

Q: Why is the tension minimum at the top?

At the top, gravity is already pulling the bob downwards towards the center of the circle. This means gravity is helping to provide the required centripetal force, so the string doesn't need to pull as hard. Hence, $T_{\text{min}} = mv^2/r - mg$.

Q: Can the tension at the top ever be negative?

No, a string can only pull, it cannot push. If the required centripetal force ($mv^2/r$) becomes less than the weight ($mg$), the string will go slack, and the tension becomes zero. The bob will then fall out of its circular path.

Q: Why do we use the conservation of mechanical energy here?

In this system, the only forces doing work are conservative (gravity). The tension force is always perpendicular to the instantaneous velocity of the bob, so it does zero work. Therefore, the total mechanical energy (Kinetic + Potential) remains constant.

Q: What does 'critical velocity' mean in this context?

Critical velocity is the absolute minimum speed the bob must have at the highest point to just barely keep the string taut ($T=0$). Setting $T_{\text{min}} = 0$ gives $v_{\text{critical}} = \sqrt{gl}$.

Question

A small bob tied at one end of a thin string of length 1m is describing a vertical circle, so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ……… m/s.
(Take, $g = 10 	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

### The Thrill of the Vertical Circle

Imagine you are swinging a small stone tied to a string in a vertical circle. You can feel the string pulling hardest when the stone is at the very bottom, and it feels the slackest when the stone reaches the very top. This intuitive feeling is the heart of vertical circular motion, and it's exactly what we need to decode to solve this problem.

### Analyzing the Setup

Let's break down the physics at the two extreme points of the circle. At the lowest point, the tension $T_{	ext{max}}$ must not only support the weight of the bob $mg$ but also provide the necessary centripetal force to keep it moving in a circle. Therefore, the equation is:

$$T_{	ext{max}} = \frac{mv_1^2}{l} + mg$$

Conversely, at the highest point, gravity is already pulling the bob towards the center of the circle. This means the string doesn't have to work as hard. The tension $T_{	ext{min}}$ is simply the required centripetal force minus the contribution from gravity:

$$T_{	ext{min}} = \frac{mv_2^2}{l} - mg$$

### The Master Equation

We are given a crucial piece of information: the ratio of the maximum tension to the minimum tension is $5:1$. Let's set up our master equation by substituting our tension expressions into this ratio:

$$\frac{\frac{mv_1^2}{l} + mg}{\frac{mv_2^2}{l} - mg} = 5$$

This equation looks a bit intimidating because it has two unknown velocities, $v_1$ and $v_2$. But physics always provides a way out! We can relate these two velocities using the **Conservation of Mechanical Energy**. As the bob travels from the bottom to the top, it gains potential energy ($mg 	imes 2l$) and loses an equal amount of kinetic energy.

$$\frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg(2l)$$

Simplifying this, we get a beautiful relation:

$$v_1^2 = v_2^2 + 4gl$$

### Final Calculation

Now, we substitute this energy relation back into our master tension equation. Notice how elegantly the mass $m$ cancels out from every term:

$$\frac{v_2^2 + 4gl + gl}{v_2^2 - gl} = 5$$

$$\frac{v_2^2 + 5gl}{v_2^2 - gl} = 5$$

Cross-multiplying and solving for $v_2^2$:

$$v_2^2 + 5gl = 5v_2^2 - 5gl$$

$$10gl = 4v_2^2$$

$$v_2^2 = \frac{5}{2}gl$$

Finally, we plug in the given values: $l = 1 	ext{ m}$ and $g = 10 	ext{ m/s}^2$:

$$v_2 = \sqrt{\frac{5}{2} 	imes 10 	imes 1} = \sqrt{25} = 5 	ext{ m/s}$$

And there we have it! The velocity of the bob at the highest position is exactly $5 	ext{ m/s}$.

Enter Numerical Value:

Visualized Solution

\text{Vertical Circular Motion Setup}

\text{Tension at Extreme Points}

\text{Conservation of Mechanical Energy}

\text{Using the Given Ratio}

\text{Substituting } v_1^2

\text{Solving for } v_2

\text{Calculating the Final Velocity}

\text{Conclusion & Extension}

The Sigma Insight: Vertical Circular Motion

The Thrill of the Vertical Circle

Analyzing the Setup

The Master Equation

Final Calculation

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