At time t = 0, a disk of radius 1 m starts to roll without slipping on a horizontal plane with an an

Q: Why does the stone move backwards relative to the center?

When a disk rolls forward, it rotates clockwise. A point starting at the bottom will be lifted up and rotated towards the back of the disk. This is why the stone is at the bottom-left when it detaches.

Q: Why is the vertical velocity $v_y = \omega R \sin 60^\circ$?

The total velocity is $\vec{v}_c + \vec{v}'$. The center's velocity $\vec{v}_c$ is purely horizontal. The tangential velocity $\vec{v}'$ has a magnitude of $\omega R$ and is directed perpendicular to the radius. At an angle of $60^\circ$ from the bottom, the vertical component of this tangential velocity is exactly $\omega R \sin 60^\circ$.

Q: Why do we ignore the horizontal velocity for the maximum height?

In projectile motion, the maximum height depends entirely on the initial vertical velocity and the initial height. The horizontal velocity only affects the range, not how high the stone goes.

Q: What does 'rolling without slipping' physically mean here?

It means the point of the disk in contact with the ground has zero instantaneous velocity. This enforces the strict relationship $v_c = \omega R$ and $a_c = \alpha R$, which is the crucial link connecting the rotational and translational motion.

Q: Why is the initial height $y_0 = R - R \cos 60^\circ$?

Imagine a right triangle formed by the radius to the stone, the vertical line from the center, and a horizontal line. The vertical distance from the center down to the stone is $R \cos 60^\circ$. Since the center is at height $R$ from the ground, the stone's height is $R - R \cos 60^\circ$.

Question

At time $t = 0$, a disk of radius $1	ext{ m}$ starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha = \frac{2}{3}	ext{ rad s}^{-2}$. A small stone is stuck to the disk. At $t = 0$, it is at the contact point of the disk and the plane. Later, at time $t = \sqrt{\pi}	ext{ s}$, the stone detaches itself and flies off tangentially from the disk. The maximum height (in $	ext{m}$) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$. The value of $x$ is _______. [Take $g = 10	ext{ m s}^{-2}$.]

Sigma Percentile · Accepted Answer

### The Anatomy of Rolling Motion

Imagine a disk resting peacefully on the ground. At the exact moment $t = 0$, a small stone is stuck to the very bottom of the disk, right where it touches the floor.

Suddenly, the disk begins to roll forward without slipping. It accelerates with an angular acceleration of $\alpha = \frac{2}{3} 	ext{ rad s}^{-2}$.

To understand where the stone goes, we first need to track its angular journey. We can use the fundamental equation of rotational kinematics:
$$ 	heta = \frac{1}{2}\alpha t^2 $$

### Tracing the Stone's Journey

We are told the stone detaches at $t = \sqrt{\pi} 	ext{ s}$. Let's plug this time into our equation.

Substituting the values, we get:
$$ 	heta = \frac{1}{2} \left(\frac{2}{3}ight) (\sqrt{\pi})^2 = \frac{\pi}{3} 	ext{ rad} $$

This means the disk has rotated exactly $60^\circ$. Because the disk rolls forward (clockwise), the stone, which started at the bottom, is lifted up and moves towards the back.

Its new height from the ground is a simple geometry problem. The vertical distance from the center down to the stone is $R \cos 60^\circ$. Since the center is at height $R$, the stone's height is:
$$ y_0 = R - R \cos 60^\circ = 1 - 0.5 = 0.5 	ext{ m} $$

### The Projectile Phase

Now, how fast is the stone moving when it breaks free? In pure rolling, the velocity of any point is the vector sum of the center's velocity ($\vec{v}_c$) and the tangential velocity relative to the center ($\vec{v}'$).

The center moves horizontally with $v_c = \omega R$. The tangential velocity also has a magnitude of $\omega R$, but it points perpendicular to the radius.

Since we only care about how high the stone will fly, we only need the **vertical component** of this total velocity. The center's velocity has no vertical component. The tangential velocity, angled at $60^\circ$ from the vertical, gives us:
$$ v_y = \omega R \sin 60^\circ $$

First, let's find $\omega$ at the moment of detachment:
$$ \omega = \alpha t = \left(\frac{2}{3}ight)\sqrt{\pi} 	ext{ rad/s} $$

Now, substitute this into our vertical velocity equation:
$$ v_y = \left(\frac{2}{3}\sqrt{\pi}ight)(1)\left(\frac{\sqrt{3}}{2}ight) = \frac{\sqrt{\pi}}{\sqrt{3}} 	ext{ m/s} $$

### The Final Calculation

The moment the stone detaches, it becomes a free projectile under gravity. The maximum height of a projectile is given by its initial height plus the height gained from its vertical velocity:
$$ H = y_0 + \frac{v_y^2}{2g} $$

Let's carefully substitute our known values into this master equation:
$$ H = 0.5 + \frac{\pi/3}{2 	imes 10} = 0.5 + \frac{\pi}{60} $$

The problem asks us to express this height in the format $\frac{1}{2} + \frac{x}{10}$. By comparing the two expressions, we can easily isolate $x$:
$$ \frac{x}{10} = \frac{\pi}{60} \implies x = \frac{\pi}{6} $$

Calculating the numerical value, we get $x \approx 0.52$.

This beautiful problem perfectly weaves together rotational kinematics, relative velocity, and projectile motion into one elegant sequence!

Enter Numerical Value:

Visualized Solution

$t = 0, ω_{0} = 0$

$θ = \frac{1}{2} α t^{2}$

$θ = \frac{1}{2} (\frac{2}{3}) (π)^{2}$

$θ = \frac{π}{3} = 6 0^{\circ}$

$y_{0} = R - R cos θ$

$v = v_{c} + v^{'}$

$v_{y} = ω R sin 6 0^{\circ}$

$H = y_{0} + \frac{v _{y}^{2}}{2 g}$

$H = \frac{1}{2} + \frac{π /3}{20}$

$x = \frac{π}{6} \approx 0.52$

$v_{c} \neq = ω R$

The Sigma Insight: Rolling Motion

The Anatomy of Rolling Motion

Tracing the Stone's Journey

The Projectile Phase

The Final Calculation

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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; } t=0,ω0​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=21​αt2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=21​(32​)(π​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=3π​=60∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y0​=R−Rcosθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=vc​+v′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vy​=ωRsin60∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H=y0​+2gvy2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H=21​+20π/3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=6π​≈0.52

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vc​=ωR