The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slip

Q: Why is the velocity of O $3R\omega$ and not $2R\omega$?

The rolling without slipping condition applies to the *outer ring* which is in contact with the ground. The outer ring has a radius of $3R$ and angular speed $\omega$, so $v_O = \omega \times 3R$.

Q: Why do we use $\omega/2$ for point P?

Point P is located on the *inner disc*, which is rotating independently of the outer ring due to the frictionless ball bearings. Its given angular speed is $\omega/2$.

Q: How do we determine the direction of $\mathbf{v}_{PO}$?

The relative velocity $\mathbf{v}_{PO}$ is always perpendicular to the position vector $\vec{r}_{P/O}$. Since the disc rotates anti-clockwise, the velocity vector points $90^\circ$ anti-clockwise from the position vector.

Q: Why do we add $\mathbf{v}_O$ and $\mathbf{v}_{PO}$?

Velocity is relative. The absolute velocity of a point (relative to the ground) is the vector sum of the velocity of the reference frame (center O) and the velocity of the point relative to that frame.

Q: What if the inner disc was also rolling on something?

If the inner disc were constrained by another surface, its angular velocity would be determined by that constraint. Here, it's freely rotating at a given $\omega/2$, so we just use that given value.

Question

The figure shows a system consisting of (i) a ring of outer radius $3R$ rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and (ii) an inner disc of radius $2R$ rotating anti-clockwise with angular speed $\omega/2$. The ring and disc are separated by frictionless ball bearings. The system is in the $x-z$ plane. The point $P$ on the inner disc is at a distance $R$ from the origin, where $OP$ makes an angle of $30^\circ$ with the horizontal. Then with respect to the horizontal surface,

Options:

(A) the point $O$ has a linear velocity $3R\omega \hat{\mathbf{i}}$

(B) the point $P$ has a linear velocity $\frac{11}{4} R\omega \hat{\mathbf{i}} + \frac{\sqrt{3}}{4} R\omega \hat{\mathbf{k}}$

(C) the point $P$ has a linear velocity $\frac{13}{4} R\omega \hat{\mathbf{i}} - \frac{\sqrt{3}}{4} R\omega \hat{\mathbf{k}}$

(D) the point $P$ has a linear velocity $\left(3 - \frac{\sqrt{3}}{4}\right) R\omega \hat{\mathbf{i}} + \frac{1}{4} R\omega \hat{\mathbf{k}}$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Rotations

Imagine a complex mechanical system where two distinct rotational motions are happening simultaneously. We have a large outer ring of radius $3R$ rolling smoothly (without slipping) on a horizontal surface. Inside this ring, separated by frictionless ball bearings, sits a smaller disc of radius $2R$.

While the outer ring rolls clockwise with an angular speed $\omega$, the inner disc is doing its own thing—rotating anti-clockwise with an angular speed of $\omega/2$. Our goal is to find the absolute velocity of a specific point $P$ located on the inner disc.

### Velocity of the Center: The Rolling Condition

Let's start by finding the velocity of the center of the entire system, point $O$. Because the outer ring is rolling without slipping, the velocity of its center is directly tied to its angular velocity and its radius.

The formula for pure rolling is $v = \omega r$. For the outer ring, the radius is $3R$. Since it's rolling clockwise (moving to the right), the velocity of the center $O$ is:

$$\mathbf{v}_O = 3R\omega \hat{\mathbf{i}}$$

This velocity $\mathbf{v}_O$ acts as the translational velocity for the entire system, including the inner disc.

### Relative Velocity: The Inner Disc's Secret

Now, let's shift our focus to the inner disc. It's rotating anti-clockwise with an angular speed of $\omega/2$. Point $P$ is located at a distance $R$ from the center $O$, making an angle of $30^\circ$ with the horizontal $x$-axis.

To find the velocity of $P$ relative to the center $O$, we use the relation $\mathbf{v}_{PO} = \vec{\omega}_{	ext{disc}} 	imes \vec{r}_{P/O}$. The magnitude of this relative velocity is simply the angular speed times the distance from the center:

$$|\mathbf{v}_{PO}| = \left(\frac{\omega}{2}ight) R = \frac{R\omega}{2}$$

Because the disc rotates anti-clockwise, the velocity vector $\mathbf{v}_{PO}$ will point $90^\circ$ anti-clockwise from the position vector $\vec{r}_{P/O}$. Using basic geometry, we can resolve this vector into its $x$ and $z$ components:

$$\mathbf{v}_{PO} = \frac{R\omega}{2} (-\sin 30^\circ \hat{\mathbf{i}} + \cos 30^\circ \hat{\mathbf{k}})$$

Substituting the trigonometric values, we get:

$$\mathbf{v}_{PO} = -\frac{R\omega}{4} \hat{\mathbf{i}} + \frac{\sqrt{3}R\omega}{4} \hat{\mathbf{k}}$$

### The Grand Finale: Absolute Velocity

Velocity is relative. The absolute velocity of point $P$ (its velocity with respect to the ground) is the vector sum of the velocity of the center $O$ and the velocity of $P$ relative to $O$. This is a classic application of Galilean relativity:

$$\mathbf{v}_P = \mathbf{v}_O + \mathbf{v}_{PO}$$

Let's plug in the vectors we've calculated:

$$\mathbf{v}_P = 3R\omega \hat{\mathbf{i}} + \left( -\frac{R\omega}{4} \hat{\mathbf{i}} + \frac{\sqrt{3}R\omega}{4} \hat{\mathbf{k}} ight)$$

Combining the $\hat{\mathbf{i}}$ components:

$$\mathbf{v}_P = \left(3 - \frac{1}{4}ight) R\omega \hat{\mathbf{i}} + \frac{\sqrt{3}}{4} R\omega \hat{\mathbf{k}}$$

$$\mathbf{v}_P = \frac{11}{4} R\omega \hat{\mathbf{i}} + \frac{\sqrt{3}}{4} R\omega \hat{\mathbf{k}}$$

And there we have it! By carefully breaking down the motion into translation of the center of mass and rotation about the center of mass, we've successfully navigated this multi-layered kinematics problem.

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Visualized Solution

$Visualizing the System$

$Velocity of Center O$

$Kinematics of Inner Disc$

$Relative Velocity of P$

$Resolving v_{P O}$

$Absolute Velocity of P$

$Final Calculation$

The Sigma Insight: Rolling Motion

The Setup

Velocity of the Center

Relative Velocity

The Grand Finale

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of Center O

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Velocity of P

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Absolute Velocity of P

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