A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes a

Q: Why do the ring and disc have different accelerations if gravity is the same for both?

While gravity pulls them down equally, rolling requires some of that gravitational potential energy to be converted into rotational kinetic energy. Since a ring has its mass distributed further from the center, it has a higher moment of inertia and requires more energy to rotate, leaving less energy for linear acceleration.

Q: How did we get the distance $s = \frac{h}{\sin\theta}$?

If you draw a right-angled triangle representing the inclined plane, the height $h$ is the opposite side to the angle $\theta$, and the length of the incline $s$ is the hypotenuse. Using trigonometry, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{s}$, which rearranges to $s = \frac{h}{\sin\theta}$.

Q: Why does the disc reach the bottom first?

The disc has a smaller moment of inertia ($I = \frac{1}{2}MR^2$) compared to the ring ($I = MR^2$). This means it resists rotational changes less, allowing it to accelerate faster linearly down the incline.

Q: What does 'rolling without slipping' mean in this context?

It means the point of the object in contact with the incline is momentarily at rest. This condition gives us the crucial relationship between linear and angular acceleration: $a = \alpha R$, which allows us to derive the acceleration formula.

Q: Can we use energy conservation instead of kinematics to solve this?

Energy conservation is great for finding the final velocity at the bottom of the incline, but it doesn't directly give you the time taken. Since the problem specifically gives a time difference, using kinematics ($s = \frac{1}{2}at^2$) is the most direct and efficient approach.

Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^\circ$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2 - \sqrt{3}) / \sqrt{10}	ext{ s}$, then the height of the top of the inclined plane, in meters, is _______. (Take $g = 10	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### The Setup: A Race Against Gravity

Imagine standing at the top of a steep, $60^\circ$ inclined plane. At the starting line, we have two competitors: a hollow ring and a solid disc. They are released at the exact same moment, beginning a race to the bottom.

Our mission is to determine the height of this inclined plane, $h$, given a very specific clue: the time difference between their arrivals at the bottom is $\frac{2 - \sqrt{3}}{\sqrt{10}} 	ext{ s}$.

To solve this, we must dive into the fascinating world of rotational dynamics. Why don't they reach the bottom at the same time? After all, Galileo famously demonstrated that objects fall at the same rate regardless of mass. The secret lies in the fact that these objects aren't just falling; they are **rolling**.

### The Physics of Rolling: Why Shape Matters

When an object rolls down an incline without slipping, gravity pulls it down, but static friction at the contact point forces it to rotate. This means the gravitational potential energy is converted into two forms of kinetic energy: translational (moving forward) and rotational (spinning).

The acceleration $a$ of a rolling body is given by the master equation:
$$a = \frac{g \sin	heta}{1 + \frac{I}{MR^2}}$$

Here, $I$ is the moment of inertia, which measures how mass is distributed relative to the center. The larger the moment of inertia, the more energy is "stolen" by rotation, leaving less for linear acceleration.

### Calculating the Accelerations

Let's evaluate our two racers.

**The Ring:**
A ring has all its mass concentrated at its outer edge. This gives it the maximum possible moment of inertia for a circular object: $I_{	ext{ring}} = MR^2$.
Substituting this into our formula:
$$a_{	ext{ring}} = \frac{g \sin 60^\circ}{1 + \frac{MR^2}{MR^2}} = \frac{g \sin 60^\circ}{1 + 1} = \frac{g \sin 60^\circ}{2}$$

**The Disc:**
A solid disc has its mass spread evenly from the center to the edge. Its moment of inertia is smaller: $I_{	ext{disc}} = \frac{1}{2}MR^2$.
Substituting this:
$$a_{	ext{disc}} = \frac{g \sin 60^\circ}{1 + \frac{1}{2}} = \frac{g \sin 60^\circ}{\frac{3}{2}} = \frac{2g \sin 60^\circ}{3}$$

Comparing the two, $\frac{2}{3}$ is greater than $\frac{1}{2}$. The disc accelerates faster because its mass is closer to the center, making it easier to spin!

### Kinematics: The Race to the Bottom

Now, let's connect their accelerations to the time it takes to reach the bottom. The length of the incline, $s$, is related to the height $h$ by trigonometry:
$$s = \frac{h}{\sin 60^\circ}$$

Using the second equation of motion, $s = \frac{1}{2}at^2$, we can solve for time $t$:
$$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2h}{a \sin 60^\circ}}$$

Let's find the time for each competitor.

**Time for the Ring ($t_1$):**
$$t_1 = \sqrt{\frac{2h}{\left(\frac{g \sin 60^\circ}{2}ight) \sin 60^\circ}} = \sqrt{\frac{4h}{g \sin^2 60^\circ}}$$

**Time for the Disc ($t_2$):**
$$t_2 = \sqrt{\frac{2h}{\left(\frac{2g \sin 60^\circ}{3}ight) \sin 60^\circ}} = \sqrt{\frac{3h}{g \sin^2 60^\circ}}$$

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we know $\sin^2 60^\circ = \frac{3}{4}$. Substituting this in:
$$t_1 = \sqrt{\frac{4h}{g \left(\frac{3}{4}ight)}} = \sqrt{\frac{16h}{3g}}$$
$$t_2 = \sqrt{\frac{3h}{g \left(\frac{3}{4}ight)}} = \sqrt{\frac{12h}{3g}} = \sqrt{\frac{4h}{g}}$$

### The Grand Finale: Solving for Height

We are given the time difference, $\Delta t = t_1 - t_2$. Let's subtract our expressions:
$$\Delta t = \sqrt{\frac{16h}{3g}} - \sqrt{\frac{4h}{g}}$$

We can factor out $\sqrt{\frac{4h}{g}}$:
$$\Delta t = \sqrt{\frac{4h}{g}} \left( \frac{2}{\sqrt{3}} - 1 ight) = \sqrt{\frac{4h}{g}} \left( \frac{2 - \sqrt{3}}{\sqrt{3}} ight)$$

Now, we equate this to the given time difference:
$$\sqrt{\frac{4h}{10}} \left( \frac{2 - \sqrt{3}}{\sqrt{3}} ight) = \frac{2 - \sqrt{3}}{\sqrt{10}}$$

Notice the beautiful symmetry! The $(2 - \sqrt{3})$ terms cancel out perfectly on both sides.
$$\frac{\sqrt{4h}}{\sqrt{10} \cdot \sqrt{3}} = \frac{1}{\sqrt{10}}$$

Multiply both sides by $\sqrt{10}$:
$$\frac{2\sqrt{h}}{\sqrt{3}} = 1$$

Rearranging and squaring both sides:
$$2\sqrt{h} = \sqrt{3} \implies 4h = 3 \implies h = 0.75 	ext{ m}$$

And there we have it! The height of the inclined plane is exactly **0.75 meters**. This problem is a beautiful symphony of rotational dynamics, kinematics, and elegant algebra.

Enter Numerical Value:

Visualized Solution

$The Setup$

$Acceleration of a Rolling Body$

$Acceleration of the Ring$

$Acceleration of the Disc$

$Kinematics on the Incline$

$Time for the Ring$

$Time for the Disc$

$Substituting Values$

$Time Difference$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Rolling Motion

The Setup

The Physics of Rolling

Calculating the Accelerations

Kinematics

The Grand Finale

Similar Questions

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Acceleration of a Rolling Body

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

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward