Consider a large disk of radius R and two smaller disks, each of radius r = R/50, lying on its circu

Q: Why is the angular separation $\Delta\theta = 2r/(R+r)$ and not $2r/R$?

The centers of the small disks move along a circle of radius $R+r$, not $R$. Therefore, the arc length connecting their centers is $(R+r)\Delta\theta$.

Q: Why do we use $v = \omega r$ for the velocity of the center?

Since the small disks roll without slipping on the stationary large disk, the point of contact has zero velocity. The center of the small disk is at a distance $r$ from the point of contact, so its velocity is $\omega r$.

Q: Why is the relative angle covered $2\pi - 2\Delta\theta$ and not $2\pi - \Delta\theta$?

They start with their centers separated by $\Delta\theta$. When they meet again, they must be touching, so their centers are again separated by $\Delta\theta$. The total relative angle they close is the full circle minus the initial gap and the final gap, which is $2\pi - 2\Delta\theta$.

Q: How do we find the angular velocity of the small disk's center around the large disk?

The center of the small disk moves with a linear speed $v$ along a circular path of radius $R+r$. Its angular velocity around the origin is simply $\Omega = v / (R+r)$.

Q: Does the direction of rolling matter?

Yes! The problem states they roll in opposite directions. This means their centers move in opposite directions around the large disk, so we add their angular velocities to find the relative angular velocity.

Question

Consider a large disk of radius R and two smaller disks, each of radius r = R/50, lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with an angular separation $\Delta	heta$ between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities $\omega$ and $2\omega$ while the large disk is held stationary. The time $	au$ at which the smaller disks are again in contact is :
[Use $\sin(\Delta	heta) = \Delta	heta$ and ignore gravity.]

Options:

(A) $	au = 51 	imes \left(2\pi - \frac{4}{51}ight) / \omega$

(B) $	au = 51 	imes \left(2\pi - \frac{2}{51}ight) / 3\omega$

(C) $	au = 51 	imes \left(2\pi - \frac{4}{51}ight) / 3\omega$

(D) $	au = 51 	imes \left(2\pi - \frac{2}{51}ight) / \omega$

Sigma Percentile · Accepted Answer

This problem is a beautiful interplay of rolling kinematics and relative motion. It challenges us to carefully track the centers of two rolling disks and understand exactly what it means for them to "meet again."

### The Setup: Visualizing the Disks
Imagine a large stationary disk of radius $R$. Resting on its circumference are two tiny disks, each with a radius $r = R/50$. Initially, they are in perfect contact with each other. Because they have a finite size, their centers are not at the exact same point; they are separated by a small angle, which we'll call $\Delta	heta$.

### Finding the Initial Angular Separation
Let's zoom in on the initial state. The distance between the centers of the two small disks is simply $2r$ because they are touching. The distance from the center of the large disk to the center of either small disk is $R + r$.

We can form a triangle connecting the origin to the two small disk centers. Using the small angle approximation (as suggested by the problem statement $\sin(\Delta	heta) \approx \Delta	heta$), the arc length connecting their centers is approximately equal to the straight-line distance $2r$. Therefore, the angular separation is:

$$\Delta	heta = \frac{2r}{R+r}$$

Substituting $r = R/50$, we get:

$$\Delta	heta = \frac{2(R/50)}{R + R/50} = \frac{2/50}{51/50} = \frac{2}{51} 	ext{ rad}$$

### Kinematics of Rolling
Now, the disks start moving. The left disk rolls without slipping with an angular velocity $\omega$. This means the velocity of its center is $v_1 = \omega r$. Since this center is moving in a circle of radius $R+r$, its angular velocity around the origin is:

$$\Omega_1 = \frac{v_1}{R+r} = \frac{\omega r}{R+r}$$

Similarly, the right disk rolls in the opposite direction with an angular velocity $2\omega$. The velocity of its center is $v_2 = 2\omega r$, and its angular velocity around the origin is:

$$\Omega_2 = \frac{v_2}{R+r} = \frac{2\omega r}{R+r}$$

Because they are moving in opposite directions, their relative angular velocity is the sum of their individual angular velocities:

$$\Omega_{rel} = \Omega_1 + \Omega_2 = \frac{3\omega r}{R+r}$$

Substituting $r = R/50$ again, we find:

$$\Omega_{rel} = \frac{3\omega (R/50)}{51R/50} = \frac{3\omega}{51}$$

### The Meeting Condition: A Tricky Catch
Here is where many students make a subtle mistake. You might think that to meet again, the disks need to cover a total relative angle of $2\pi$. But remember, they are not point particles!

They start with their centers separated by $\Delta	heta$. When they meet on the other side of the large disk, they will again be in contact, meaning their centers will once again be separated by $\Delta	heta$.

Imagine the gap between them growing from $\Delta	heta$ to a maximum, and then shrinking back down to $\Delta	heta$. The total relative angular distance they must cover to close this gap is the full circle minus the initial separation and the final separation:

$$	heta_{rel} = 2\pi - 2\Delta	heta$$

### The Final Calculation
Now, we have everything we need. The time $	au$ it takes for them to meet is simply the relative angular distance divided by the relative angular velocity:

$$	au = \frac{	heta_{rel}}{\Omega_{rel}} = \frac{2\pi - 2\Delta	heta}{\frac{3\omega}{51}}$$

Substitute $\Delta	heta = \frac{2}{51}$:

$$	au = \frac{2\pi - 2\left(\frac{2}{51}ight)}{\frac{3\omega}{51}} = \frac{51}{3\omega}\left(2\pi - \frac{4}{51}ight)$$

This perfectly matches option (C). The elegance of this problem lies in carefully defining the geometry of the meeting condition rather than just blindly applying kinematic formulas.

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Visualized Solution

Initial Setup

Angular Separation $Δ θ$

Kinematics of Left Disk

Kinematics of Right Disk

Relative Angular Velocity

The Meeting Condition

Final Time Calculation

The Sigma Insight: Rolling Motion

The Setup

Finding the Initial Angular Separation Let's zoom in on the initial state

Kinematics of Rolling Now, the disks start moving

The Meeting Condition

The Final Calculation Now, we have everything we need

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Visualized Solution

Initial Setup

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

Kinematics of Left Disk

Kinematics of Right Disk

Relative Angular Velocity

The Meeting Condition

Final Time Calculation

The Sigma Insight: Rolling Motion

The Setup

Finding the Initial Angular Separation Let's zoom in on the initial state

Kinematics of Rolling Now, the disks start moving

The Meeting Condition

The Final Calculation Now, we have everything we need

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Angular Separation $Δ θ$