A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless)

Q: Why don't the sphere and the ring roll down the incline?

Rolling requires a torque to get the object spinning. On an inclined plane, this torque is usually provided by static friction. Since the plane is perfectly frictionless, there is no force to create this torque, so the objects simply slide down without rotating.

Q: Doesn't the solid sphere have a smaller moment of inertia, making it faster?

You are absolutely right that a solid sphere has a smaller moment of inertia compared to a hollow sphere or a ring. However, moment of inertia only matters when the object is *rolling*. Since these objects are purely sliding, their rotational properties don't come into play at all.

Q: What if the objects had different masses? Would the heavier one slide faster?

Not at all! When you set up the equation $mg \sin\theta = ma$, you'll notice that the mass $m$ cancels out from both sides. This means the acceleration is simply $g \sin\theta$, which is completely independent of the object's mass.

Q: Which object would win the race if the plane *did* have friction?

If there was enough friction to cause pure rolling, the acceleration would be $a = \frac{g \sin\theta}{1 + I/mR^2}$. In this case, the object with the smallest $I/mR^2$ ratio would have the highest acceleration. The solid sphere ($I/mR^2 = 0.4$) would win, followed by the hollow sphere ($0.67$), and the ring ($1.0$) would come in last.

Q: Is the normal force the same for all three objects?

The normal force is given by $N = mg \cos\theta$. So, it depends entirely on the mass $m$ of the object. If the objects have different masses, they will experience different normal forces, even though their acceleration down the plane is exactly the same.

Question

A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless), so that they slide down the plane. Then, maximum acceleration down the plane is for (no rolling)

Options:

(A) solid sphere

(B) hollow sphere

(C) ring

(D) All same

Sigma Percentile · Accepted Answer

### The Great Frictionless Race

Imagine a thrilling race at the top of an inclined plane. The competitors are a solid sphere, a hollow sphere, and a ring. They are all lined up, ready to be released from rest. However, there is a massive twist in this particular race: the inclined plane is perfectly **frictionless**.

To understand who will win this race, we need to dive into the physics of the motion. Let's draw a free body diagram for any of these objects. The primary force acting on the object is its weight, $mg$, which points vertically downwards.

We can resolve this weight into two perpendicular components. The component perpendicular to the inclined plane is $mg \cos	heta$, and the component parallel to the incline, driving the object downwards, is $mg \sin	heta$.

### The Absence of Torque

The normal force $N$ exerted by the plane perfectly balances the perpendicular component, so $N = mg \cos	heta$. Now, here is where the magic happens. Because the surface is completely frictionless, there is absolutely no friction force ($f = 0$) acting on the contact point between the object and the plane.

For an object to start rolling, it requires a **torque** about its center of mass. On an inclined plane, this torque is typically provided by static friction. Without friction, there is zero torque. Consequently, the objects will not roll at all; they will purely **slide** down the incline.

### The Master Equation

Since the objects are only sliding, we can treat them as simple point masses. We apply Newton's Second Law of Motion along the direction of the incline:

$$F_{	ext{net}} = ma$$

The only force acting along the incline is the parallel component of gravity. Therefore:

$$mg \sin	heta = ma$$

Look closely at this elegant equation. The mass $m$ beautifully cancels out from both sides! This leaves us with a incredibly simple expression for the acceleration:

$$a = g \sin	heta$$

### The Final Verdict

Notice what is missing from this final equation. The acceleration $a$ depends *only* on the acceleration due to gravity $g$ and the angle of inclination $	heta$. It is completely independent of the object's mass, its shape, or its moment of inertia.

Therefore, whether it is a solid sphere, a hollow sphere, or a ring, they will all slide down with the exact same acceleration of $g \sin	heta$. It is a perfect, unbreakable tie!

**Food for thought:** What if the plane *did* have friction? In that case, the bodies would roll, and their acceleration would be governed by $a = \frac{g \sin	heta}{1 + I/mR^2}$. The object with the smallest moment of inertia (the solid sphere) would have the highest acceleration and win the race. But in our frictionless world, geometry takes a backseat to pure gravity.

Animated Solution for Physics - Laws of Motion: A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless), so that they slide down the plane. Then, maximum acceleration down the plane is for (no rolling)

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Free Body Diagram}

\text{Analyzing the Forces}

\text{Newton's Second Law}

\text{Calculating Acceleration}

\text{Final Conclusion}

\text{The Way Forward}

The Sigma Insight: Newton's Laws of Motion

The Great Frictionless Race

The Absence of Torque

The Master Equation

The Final Verdict

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