An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts

Q: Why does the insect eventually slip? Why can't it just keep climbing?

As the insect climbs higher, the angle $\theta$ increases. This causes the tangential component of gravity ($mg \sin\theta$) to increase, meaning it needs more friction to stay put. Simultaneously, the normal force ($mg \cos\theta$) decreases, which actually reduces the maximum available friction ($f_{\max} = \mu N$). Eventually, the required friction exceeds the available friction, and it slips.

Q: Why is the normal force $N = mg \cos\theta$ and not just $mg$?

The normal force is the surface's reaction to the force pressing perpendicularly against it. On a flat horizontal surface, that's the full weight $mg$. But on a slope, only a component of the weight ($mg \cos\theta$) presses perpendicularly into the surface. The other component ($mg \sin\theta$) acts parallel to the surface.

Q: How did we get the formula $h = R - R \cos\theta$?

Imagine the vertical axis passing through the center of the hemisphere. The total distance from the center down to the very bottom of the ditch is the radius $R$. The vertical position of the insect below the center forms a right triangle with the radius, making that vertical depth $R \cos\theta$. The height $h$ from the bottom is just the total depth $R$ minus the insect's depth $R \cos\theta$.

Q: Why didn't we need the mass of the insect to solve this?

Because both the force pulling the insect down the slope ($mg \sin\theta$) and the maximum friction holding it up ($\mu mg \cos\theta$) are directly proportional to its mass. When you equate them, the mass $m$ cancels out entirely. A heavy beetle and a light ant would slip at the exact same height if their feet had the same coefficient of friction!

Q: What would happen if the coefficient of friction $\mu$ was larger?

A larger $\mu$ means the surface is rougher. Mathematically, since $\tan\theta = \mu$, a larger $\mu$ results in a larger critical angle $\theta$. This means the insect could climb to a steeper part of the ditch (a greater height $h$) before the tangential pull of gravity overcomes the frictional grip.

Question

An insect is at the bottom of a hemispherical ditch of radius $1	ext{ m}$. It crawls up the ditch but starts slipping after it is at height $h$ from the bottom. If the coefficient of friction between the ground and the insect is $0.75$, then $h$ is (Take, $g = 10	ext{ ms}^{-2}$)

Options:

(A) $0.20	ext{ m}$

(B) $0.45	ext{ m}$

(C) $0.60	ext{ m}$

(D) $0.80	ext{ m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a tiny insect starting its journey from the very bottom of a hemispherical ditch. As it crawls upwards, the slope of the surface it stands on becomes increasingly steeper. Gravity is constantly trying to pull it back down, but the static friction between its tiny legs and the ground keeps it anchored.

However, friction has its limits. There will be a critical point—a maximum height $h$—where the slope is so steep that the maximum possible static friction is no longer enough to counteract the downward pull of gravity. At this exact moment, the insect is on the verge of slipping. Let's define the position of the insect at this critical point by an angle $	heta$ measured from the vertical axis passing through the center of the hemisphere.

### The Master Equation: Balancing the Forces
To find this critical angle, we must draw a free body diagram of the insect. There are three primary forces at play here:
1. **Weight ($mg$)**: Acting vertically downwards.
2. **Normal Reaction ($N$)**: Pushing outward from the surface, directed towards the center of the hemisphere.
3. **Static Friction ($f$)**: Acting tangentially upwards along the surface of the ditch to prevent slipping.

Because the insect is on a curved surface, it is highly beneficial to resolve the weight $mg$ into two components: one along the radial direction and one along the tangential direction.

Along the radial direction, the component of weight is $mg \cos	heta$. Since the insect is not flying off or sinking into the surface, this must be perfectly balanced by the normal reaction:
$$N = mg \cos	heta$$

Along the tangential direction, the component of weight trying to pull the insect down the slope is $mg \sin	heta$. This is balanced by the static friction:
$$f = mg \sin	heta$$

### The Limiting Condition
At the maximum height, the insect is just about to slip. This means the static friction has reached its absolute maximum limit, which is given by the coefficient of static friction $\mu$ multiplied by the normal reaction $N$:
$$f = \mu N$$

Substituting our balanced force equations into this limiting condition, we get:
$$mg \sin	heta = \mu (mg \cos	heta)$$

Notice how beautifully the mass $m$ and gravity $g$ cancel out from both sides! This tells us a fascinating physical truth: the maximum angle the insect can reach is completely independent of how heavy it is or what planet it is on. Rearranging the terms, we find:
$$\frac{\sin	heta}{\cos	heta} = \mu$$
$$	an	heta = \mu$$

### Final Calculation
We are given that the coefficient of friction $\mu = 0.75$, which can be written as the fraction $\frac{3}{4}$. Therefore:
$$	an	heta = \frac{3}{4}$$

If we imagine a right-angled triangle where the opposite side is $3$ and the adjacent side is $4$, the hypotenuse would be $\sqrt{3^2 + 4^2} = 5$. From this, we can easily determine the cosine of the angle:
$$\cos	heta = \frac{4}{5}$$

Now, we need to relate this angle $	heta$ back to the physical height $h$ from the bottom of the ditch. Looking at the geometry of the hemisphere, the vertical distance from the center to the bottom is simply the radius $R$. The vertical distance from the center to the insect is $R \cos	heta$. Therefore, the height $h$ from the bottom is the difference between these two:
$$h = R - R \cos	heta = R(1 - \cos	heta)$$

Substituting the given radius $R = 1	ext{ m}$ and our calculated $\cos	heta = \frac{4}{5}$:
$$h = 1 \left(1 - \frac{4}{5}ight) = \frac{1}{5} = 0.20	ext{ m}$$

The insect will start slipping exactly when it reaches a height of **$0.20	ext{ m}$**.

Select Answer:

Visualized Solution

\text{Visualizing the Ditch and Insect}

\text{Free Body Diagram}

\text{Resolving Forces}

\text{Equilibrium Equations}

\text{Limiting Friction}

\text{Finding } \theta

\text{Calculating } \cos\theta

\text{Relating } h \text{ and } \theta

\text{Final Calculation}

\text{Food for Thought}

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup Imagine a tiny insect starting its journey from the very bottom of a hemispherical ditch

The Master Equation

The Limiting Condition At the maximum height, the insect is just about to slip

Final Calculation We are given that the coefficient of friction $μ = 0.75$ , which can be written as the fraction $\frac{3}{4}$

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