A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure. The h

Q: Why does the block lose contact at all? Shouldn't gravity keep it on the surface?

As the block slides down, its speed increases, requiring a larger centripetal force to keep it moving in a circle. However, the component of gravity providing this force ($mg \cos\theta$) actually *decreases* as $\theta$ increases. Eventually, gravity isn't strong enough to provide the required centripetal force, and the block flies off!

Q: Why do we set the normal force $N$ to zero?

The normal force is a contact force; it only exists when two surfaces are pressing against each other. When the block is just about to fly off, it is no longer pressing against the hemisphere, meaning the normal force drops exactly to zero.

Q: Can we use kinematic equations ($v^2 = u^2 + 2as$) instead of energy conservation?

No! The acceleration of the block is not constant. The tangential component of gravity ($mg \sin\theta$) changes continuously as the angle $\theta$ changes. Therefore, we must use the Work-Energy theorem, which handles varying forces perfectly.

Q: Does the mass of the block matter?

Not at all! If you look at both the energy equation and the centripetal force equation, the mass $m$ appears on both sides and cancels out. A tiny pebble and a massive boulder would lose contact at the exact same height (assuming no friction).

Q: What if there was friction between the block and the hemisphere?

Friction would do negative work, meaning the block would have less kinetic energy (and lower speed) at any given angle compared to the frictionless case. Because it's moving slower, it requires less centripetal force, so it would stay on the surface longer and lose contact at a lower height.

Question

A small block slides down from the top of hemisphere of radius $R = 3$ m as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is ………… m.
(Assume there is no friction between the block and the hemisphere)

Sigma Percentile · Accepted Answer

### The Physics of Losing Contact: Sliding Down a Hemisphere

Imagine a water slide shaped perfectly like a smooth dome. If you start sliding from the very top, you won't stay glued to the surface all the way down. At some specific point, you will feel a sudden moment of weightlessness as you fly off the slide! This classic physics problem asks us to find exactly where that happens.

### The Geometry of the Drop

Let's set up our coordinate system. The block starts at the top of a hemisphere of radius $R$. As it slides down to an angle $	heta$ (measured from the vertical), it drops by a certain vertical distance.

If we look at the right-angled triangle formed by the center of the hemisphere, the block, and the vertical axis, the vertical height of the block from the ground is $h = R \cos	heta$.

This means the vertical distance the block has dropped from the top is simply $R - h$.

### The Forces at Play

While the block is sliding, two main forces are acting on it:
1. **Gravity ($mg$)**: Pulling straight down.
2. **Normal Force ($N$)**: Pushing outward, perpendicular to the surface.

Because the block is moving in a circular path, it requires a **centripetal force** directed towards the center. The component of gravity pointing towards the center is $mg \cos	heta$. The normal force $N$ points away from the center. Therefore, the net inward force is:

$$F_{	ext{net}} = mg \cos	heta - N$$

According to Newton's Second Law for circular motion, this net force must equal $\frac{mv^2}{R}$:

$$mg \cos	heta - N = \frac{mv^2}{R}$$

### The Critical Moment

What does it physically mean to "lose contact"? It means the block is no longer pressing against the surface of the hemisphere. At this exact critical moment, the normal force drops to zero ($N = 0$).

Substituting $N = 0$ into our dynamics equation gives us the condition for losing contact:

$$mg \cos	heta = \frac{mv^2}{R}$$

Notice how the mass $m$ cancels out! This tells us that the speed required to fly off is independent of how heavy the block is.

$$v^2 = gR \cos	heta$$

### The Energy Connection

We have an equation for $v^2$, but we don't know the velocity $v$ yet. To find it, we use the **Work-Energy Theorem** (or Conservation of Mechanical Energy).

The block started from rest. As it dropped by a height of $(R - h)$, it lost potential energy and gained an equal amount of kinetic energy.

$$	ext{Loss in PE} = 	ext{Gain in KE}$$

$$mg(R - h) = \frac{1}{2}mv^2$$

Again, the mass $m$ cancels out. We can rearrange this to solve for $v^2$:

$$v^2 = 2g(R - h)$$

### Bringing It All Together

Now we have two different expressions for $v^2$. Let's equate them!

$$gR \cos	heta = 2g(R - h)$$

The acceleration due to gravity $g$ cancels out, leaving us with a purely geometric relationship:

$$R \cos	heta = 2(R - h)$$

Earlier, we established from our right-angled triangle that $\cos	heta = \frac{h}{R}$. Let's substitute this into our equation:

$$R \left(\frac{h}{R}ight) = 2(R - h)$$

$$h = 2R - 2h$$

Bringing the $h$ terms to one side:

$$3h = 2R$$

$$h = \frac{2R}{3}$$

This is a beautiful, universal result! Any object sliding from rest off a frictionless hemisphere will lose contact at exactly two-thirds of the radius.

Finally, we substitute the given radius $R = 3$ m:

$$h = \frac{2(3)}{3} = 2 	ext{ m}$$

The block will lose contact exactly 2 meters above the ground.

Enter Numerical Value:

Visualized Solution

\text{Geometry of the Setup}

\text{Forces at Angle } \theta

\text{Condition for Losing Contact}

\text{Centripetal Dynamics}

\text{Work-Energy Theorem}

\text{Energy Equation}

\text{Equating Velocities}

\text{Geometric Substitution}

\text{Solving for } h

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Vertical Circular Motion

The Physics of Losing Contact

The Geometry of the Drop

The Forces at Play

The Critical Moment

The Energy Connection

Bringing It All Together

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