A hemispherical bowl of radius R = 0.1 m is rotating about its own axis (which is vertical) with an

Q: Why does the normal force pass exactly through the center of the bowl?

For any perfectly spherical surface, the perpendicular line (the normal) at any point on the surface will always pass straight through the geometric center of the sphere. This is a fundamental mathematical property of circles and spheres, and it dictates the direction of the normal reaction force.

Q: Wait, why is the centripetal force strictly horizontal and not pointing along the curved surface?

Centripetal force must always point towards the center of the circular path the object is taking. Even though the particle is on a curved surface, it is moving in a flat, horizontal circle. The center of that specific horizontal circle lies on the vertical axis of rotation, so the required force must be horizontal.

Q: Why do we cancel $r$ when equating the two expressions for $\tan\theta$? What if $r=0$?

If $r=0$, the particle is sitting at the very bottom of the bowl ($h=0$), and it isn't moving in a circle at all. The question specifically asks for the condition to have a *non-zero* value of $h$. Because we are analyzing the case where the particle has risen up the side, we can safely assume $r \neq 0$ and cancel it out.

Q: Why do we use the minimum value of $\omega$ to find the minimum error in $g$?

Look at the error propagation equation: $\Delta g = \omega^2 \Delta h$. The error in our height measurement, $\Delta h$, is a fixed constant (the least count of the instrument). Therefore, the error $\Delta g$ is directly proportional to $\omega^2$. To keep the error as small as possible, we must spin the bowl as slowly as possible while still keeping the particle off the bottom.

Q: Why is the error $\Delta g$ taken as positive even though the derivative gives a negative sign?

In experimental error analysis, we are always interested in the maximum possible deviation from the true value. Errors from different measurements are added in the worst-case scenario. Therefore, we always take the absolute value (magnitude) of the differential to find the maximum possible error bound.

Question

A hemispherical bowl of radius $R = 0.1	ext{ m}$ is rotating about its own axis (which is vertical) with an angular velocity $\omega$. A particle of mass $10^{-2}	ext{ kg}$ on the frictionless inner surface of the bowl is also rotating with the same $\omega$. The particle is at a height $h$ from the bottom of the bowl. (1993)
(a) Obtain the relation between $h$ and $\omega$. What is the minimum value of $\omega$ needed, in order to have a non-zero value of $h$?
(b) It is desired to measure $g$ (acceleration due to gravity) using the set-up by measuring $h$ accurately. Assuming that $R$ and $\omega$ are known precisely and that the least count in the measurement of $h$ is $10^{-4}	ext{ m}$, what is minimum possible error $\Delta g$ in the measured value of $g$?

Sigma Percentile · Accepted Answer

### The Rotating Bowl: A Dance of Geometry and Dynamics

Imagine a hemispherical bowl of radius $R$ rotating steadily about its vertical axis with an angular velocity $\omega$. Inside this bowl, a small particle of mass $m$ is caught in the motion, sweeping out a horizontal circle. This is not just a simple mechanics problem; it is a beautiful interplay between the geometry of a sphere and the dynamics of circular motion.

Let's assume the particle settles at a height $h$ from the bottom of the bowl. Because the bowl is a hemisphere, the vertical distance from the center of the sphere down to the particle is simply $R - h$. If the particle is moving in a horizontal circle of radius $r$, we can form a right-angled triangle with the center of the sphere. From this geometry, the angle $	heta$ that the radius vector makes with the vertical is given by:

$$ 	an	heta = \frac{r}{R - h} $$

This geometric relationship is our first crucial anchor. It connects the physical position of the particle to the abstract angle $	heta$.

### The Forces at Play

Now, let's look at the forces acting on the particle. Gravity is pulling it straight down with a force of $mg$. The bowl is pushing back with a normal reaction force $N$. Because the surface is perfectly spherical, this normal force must pass directly through the center of the sphere. Therefore, the normal force $N$ also makes an angle $	heta$ with the vertical.

To understand the motion, we must resolve this normal force into two perpendicular components. The vertical component, $N \cos	heta$, has the job of fighting gravity. Since the particle is not moving up or down, these forces must balance perfectly:

$$ N \cos	heta = mg $$

The horizontal component, $N \sin	heta$, points directly towards the central axis of rotation. This is the centripetal force that keeps the particle moving in its circular path of radius $r$:

$$ N \sin	heta = m r \omega^2 $$

### The Master Equation

We now have a system of two dynamic equations. By dividing the horizontal equation by the vertical equation, the normal force $N$ and the mass $m$ beautifully cancel out, leaving us with a purely kinematic relationship:

$$ 	an	heta = \frac{r \omega^2}{g} $$

Notice how we now have two different expressions for $	an	heta$—one from pure geometry and one from dynamics. Equating them is the key to unlocking the problem:

$$ \frac{r}{R - h} = \frac{r \omega^2}{g} $$

Assuming the particle has actually risen from the bottom ($h > 0$), the radius $r$ is non-zero. We can safely cancel $r$ from both sides. Rearranging the remaining terms to solve for $h$, we get our master equation:

$$ h = R - \frac{g}{\omega^2} $$

### The Threshold of Ascension

For the particle to leave the bottom of the bowl, its height $h$ must be strictly greater than zero. If we set $h > 0$ in our master equation, we find a critical threshold for the angular velocity:

$$ R - \frac{g}{\omega^2} > 0 \implies \omega^2 > \frac{g}{R} $$

This tells us that the bowl must spin fast enough to overcome gravity's grip. The minimum angular velocity required to just lift the particle is:

$$ \omega_{	ext{min}} = \sqrt{\frac{g}{R}} $$

Given $R = 0.1	ext{ m}$ and $g = 9.8	ext{ m/s}^2$, we can calculate this threshold:

$$ \omega_{	ext{min}} = \sqrt{\frac{9.8}{0.1}} = \sqrt{98} \approx 9.89	ext{ rad/s} $$

### The Art of Error Analysis

The second part of the problem asks us to use this setup as an experimental apparatus to measure $g$. We are told that $R$ and $\omega$ are known precisely, but our measurement of $h$ has a least count (error) of $\Delta h = 10^{-4}	ext{ m}$. We need to find the minimum possible error in our calculated value of $g$.

First, let's rearrange our master equation to isolate $g$:

$$ g = \omega^2 (R - h) $$

To find how an error in $h$ propagates to an error in $g$, we differentiate this expression with respect to $h$. Since $R$ and $\omega$ are constants, the derivative is straightforward:

$$ \Delta g = \omega^2 |-\Delta h| = \omega^2 \Delta h $$

### The Final Calculation

This error equation reveals something profound: the error in $g$ grows with the square of the angular velocity $\omega$. Therefore, to minimize the error $\Delta g$, we must conduct our experiment at the lowest possible angular velocity that still allows us to measure a non-zero height $h$.

We already found this minimum angular velocity! It is $\omega_{	ext{min}}^2 = \frac{g}{R}$. Substituting this into our error equation gives the minimum possible error:

$$ (\Delta g)_{	ext{min}} = \left(\frac{g}{R}ight) \Delta h $$

Plugging in the known values:

$$ (\Delta g)_{	ext{min}} = \frac{9.8}{0.1} 	imes 10^{-4} = 98 	imes 10^{-4} = 9.8 	imes 10^{-3}	ext{ m/s}^2 $$

And there we have it! By combining geometric intuition, Newton's laws, and a touch of calculus for error propagation, we have completely unraveled the mysteries of the rotating bowl.

Visualized Solution

$Geometry of the Rotating Bowl$

$Forces Acting on the Particle$

$Resolving Forces for Circular Motion$

$Deriving the Height Equation$

$Solving for h$

$Condition for Non-Zero Height$

$Calculating ω_{min}$

$Error Analysis for Measuring g$

$Differentiating the Expression$

$Calculating the Minimum Error Δ g$

The Sigma Insight: Dynamics of Circular Motion

The Rotating Bowl

The Forces at Play

The Master Equation

The Threshold of Ascension

The Art of Error Analysis

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of the Rotating Bowl

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces Acting on the Particle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resolving Forces for Circular Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Deriving the Height Equation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Non-Zero Height

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Error Analysis for Measuring g

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Minimum Error Δg