A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position

Q: Why is the initial velocity $\sqrt{5gL}$?

For a string to not slack at the highest point, the tension must be at least zero. This requires a minimum speed of $\sqrt{gL}$ at the top to provide the necessary centripetal acceleration. Using energy conservation between the top and bottom, this translates to a minimum speed of $\sqrt{5gL}$ at the lowest point.

Q: Why do we use conservation of mechanical energy?

The only forces acting on the bob are gravity (a conservative force) and tension. Since tension is always directed towards the center of the circle, it is perpendicular to the instantaneous velocity of the bob. Therefore, tension does zero work, and mechanical energy is perfectly conserved.

Q: How did we get $h = L(1 - \cos\theta)$?

If you draw a horizontal line from the bob to the vertical axis, it forms a right triangle with the string. The vertical side adjacent to $\theta$ is $L\cos\theta$. The height $h$ from the bottom is the total length $L$ minus this adjacent side.

Q: Why is the angle obtuse?

We found $\cos\theta = -\frac{7}{8}$. A negative cosine value indicates that the angle is greater than $90^\circ$ (or $\frac{\pi}{2}$ radians). Physically, this means the bob has crossed the horizontal level and is in the upper half of the circle.

Q: How do we determine the exact inequality for $\theta$?

We compare our value $-0.875$ with known cosine values. Since $\cos(135^\circ) \approx -0.707$ and $\cos(180^\circ) = -1$, and $-1 < -0.875 < -0.707$, the angle must lie between $135^\circ$ ($\frac{3\pi}{4}$) and $180^\circ$ ($\pi$).

Question

A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $v$ at position $A$ is just sufficient to make it reach the point $B$. The angle $	heta$ at which the speed of the bob is half of that at $A$, satisfies

Options:

(A) $	heta = \frac{\pi}{4}$

(B) $\frac{\pi}{4} < 	heta < \frac{\pi}{2}$

(C) $\frac{\pi}{2} < 	heta < \frac{3\pi}{4}$

(D) $\frac{3\pi}{4} < 	heta < \pi$

Sigma Percentile · Accepted Answer

### The Physics of Vertical Circular Motion: Finding the Half-Speed Angle

Imagine a bob tied to a massless string, resting peacefully at the bottom of a vertical circle. Suddenly, we give it a horizontal push. The problem states that this initial velocity is *just* enough to make it reach the top point. But what does 'just enough' mean in the context of a string?

Unlike a rigid rod, a string cannot support compressive forces. If the bob loses too much speed before reaching the top, the string will go slack, and the bob will fall in a parabolic trajectory. To prevent this, the tension at the highest point must be at least zero. This condition dictates a minimum critical velocity at the lowest point, which is a standard and beautiful result in physics:

$$v = \sqrt{5gL}$$

### The Energy Equation

As the bob swings upward, it fights against gravity. Its kinetic energy is continuously converted into gravitational potential energy, causing it to slow down. We are on a mission to find the exact angle $	heta$ where its speed becomes exactly half of its initial speed, i.e., $\frac{v}{2}$.

Since the only forces doing work are conservative (gravity), we can invoke the **Conservation of Mechanical Energy**. Let's set the lowest point as our reference level for potential energy ($h = 0$). Equating the total energy at the bottom to the total energy at our mystery height $h$:

$$\frac{1}{2}Mv^2 = \frac{1}{2}M\left(\frac{v}{2}ight)^2 + Mgh$$

Notice how the mass $M$ gracefully cancels out from every term. The physics of this motion is entirely independent of how heavy the bob is! Rearranging the terms to isolate $h$, we subtract the final kinetic energy from the initial:

$$gh = \frac{1}{2}v^2 - \frac{1}{8}v^2 = \frac{3}{8}v^2$$

$$h = \frac{3v^2}{8g}$$

Now, we bring back our critical velocity $v^2 = 5gL$ and substitute it into our height equation:

$$h = \frac{3(5gL)}{8g} = \frac{15L}{8}$$

### The Geometry of the Circle

We have found the height, but the question demands the angle $	heta$. This is where geometry bridges the gap. By drawing a horizontal line from the bob to the vertical axis, we form a right-angled triangle. The vertical distance from the center of the circle to this horizontal line is $L\cos	heta$.

Therefore, the total height $h$ from the lowest point is the radius $L$ minus this vertical segment:

$$h = L(1 - \cos	heta)$$

### The Final Deduction

Now, we simply equate our two expressions for the height $h$:

$$L(1 - \cos	heta) = \frac{15L}{8}$$

The length $L$ cancels out, leaving us with a pure trigonometric equation:

$$1 - \cos	heta = \frac{15}{8}$$

$$\cos	heta = 1 - \frac{15}{8} = -\frac{7}{8}$$

The negative sign is a massive revelation! A negative cosine means the angle $	heta$ is obtuse (greater than $90^\circ$). The bob has crossed the horizontal diameter and is now in the upper half of the circle.

To find the exact range, we compare $-0.875$ with standard cosine values. We know that $\cos\left(\frac{3\pi}{4}ight) = -\frac{1}{\sqrt{2}} \approx -0.707$ and $\cos(\pi) = -1$. Since our value lies perfectly between these two bounds:

$$-1 < -0.875 < -0.707$$

We can confidently conclude that the angle $	heta$ satisfies the inequality:

$$\frac{3\pi}{4} < 	heta < \pi$$

And there we have it—a flawless synthesis of dynamics, energy conservation, and geometry!

Animated Solution for Physics - Work, Energy, and Power: A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $v$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $θ$ at which the speed of the bob is half of that at $A$ , satisfies

Select Answer:

Visualized Solution

Initial Setup

$v = 5 g L$

Conservation of Energy

$h = \frac{3 v ^{2}}{8 g}$

$h = \frac{15 L}{8}$

$h = L (1 - cos θ)$

$cos θ = - \frac{7}{8}$

$\frac{3 π}{4} < θ < π$

The Sigma Insight: Vertical Circular Motion

The Physics of Vertical Circular Motion

The Energy Equation

The Geometry of the Circle

The Final Deduction

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