A small roller of diameter 20 cm has an axle of diameter 10 cm. It is on a horizontal floor and a me

Q: Why does the scale move faster than the roller?

Because the scale is resting on the top of the axle! The top of the axle is moving forward due to the translation of the center, AND it is moving forward due to the clockwise rotation of the roller. These two velocities add up, making the scale move faster than the roller itself.

Q: Why is the initial contact point at $x = 100$ cm?

Look closely at the diagram in the question. The scale is a meter scale (100 cm long). The left edge is marked as $x = 0$. The diagram shows the scale extending to the left of the roller, with its rightmost edge resting exactly on the axle. Therefore, the initial contact point is at $x = 100$ cm.

Q: Why do we subtract the relative displacement from 100?

The scale moves 75 cm forward, while the roller only moves 50 cm forward. This means the scale slides 25 cm forward *relative* to the roller. If the scale slides forward, the roller effectively moves backward along the scale. Since it started at the $x = 100$ mark, moving 25 cm backward puts it at $100 - 25 = 75$ cm.

Q: What if the scale was placed at the bottom of the axle?

If it were at the bottom, the tangential velocity due to rotation would point backward (opposite to the center's velocity). The scale's velocity would be $V_c - \omega r$, making it move slower than the roller!

Q: Does the mass of the roller or the scale matter here?

Not at all! This is a pure kinematics problem. As long as the 'no slipping' conditions are maintained, the motions are strictly governed by geometry and the radii of the cylinders.

Question

A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the position of the scale will look like (figures are schematic and not drawn to scale)-

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Two Cylinders

Imagine a fascinating mechanical dance: a large roller resting on the ground, with a smaller axle protruding from its center. On top of this axle rests a meter scale. The problem asks us to determine the final position of the scale after the roller has moved a certain distance.

Before we dive into the math, we must carefully observe the initial state. The diagram shows the scale extending to the left of the roller, with its rightmost edge resting exactly on top of the axle. Since it is a standard meter scale and the left edge is marked as $x = 0$, this initial contact point is exactly at $x = 100 	ext{ cm}$. This spatial visualization is the crucial first step to unlocking the solution.

### The Master Equation: Kinematics of Rolling

When the scale is pushed, the roller begins to roll without slipping on the ground. The magic of pure rolling is that the translational velocity of the center, $V_c$, is perfectly synchronized with the angular velocity, $\omega$. The relationship is given by the classic equation:

$$ V_c = \omega R $$

where $R$ is the outer radius of the roller. Because the diameter is $20 	ext{ cm}$, we know $R = 10 	ext{ cm}$.

### The Velocity of the Scale

Now, let's shift our focus to the scale. It is resting on the *top* of the inner axle. Therefore, the scale must move with the exact same velocity as the top point of that axle.

Because the roller is moving forward and rotating clockwise, the velocity at the top of the axle is the sum of the center's translational velocity and the tangential velocity due to rotation. We can write this as:

$$ V_s = V_c + \omega r $$

where $r$ is the radius of the inner axle. Since the axle's diameter is $10 	ext{ cm}$, $r = 5 	ext{ cm}$.

By substituting $\omega = \frac{V_c}{R}$ into our equation, we get a beautiful, mass-independent relation:

$$ V_s = V_c \left(1 + \frac{r}{R}ight) $$

Plugging in our radii, $\frac{r}{R} = \frac{5}{10} = 0.5$. This means the velocity of the scale is exactly $1.5$ times the velocity of the roller's center:

$$ V_s = 1.5 V_c $$

### From Velocities to Displacements

Because the ratio of their velocities is constant throughout the motion, the ratio of their displacements will be exactly the same. If the scale is always moving $1.5$ times faster than the roller, it will cover $1.5$ times the distance in the same amount of time.

$$ \Delta x_s = 1.5 \Delta x_c $$

The problem states that the roller has moved $50 	ext{ cm}$ forward. Substituting this into our equation:

$$ \Delta x_s = 1.5 	imes 50 = 75 	ext{ cm} $$

The scale has moved a total of $75 	ext{ cm}$ forward relative to the ground.

### The Final Catch: Relative Motion

Here is where many students make a silly mistake. The scale moved $75 	ext{ cm}$, but the roller also moved $50 	ext{ cm}$ in the same direction. This means the scale slid forward *relative* to the roller by the difference of their displacements:

$$ \Delta x_{rel} = 75 - 50 = 25 	ext{ cm} $$

If the scale slides $25 	ext{ cm}$ forward relative to the roller, the contact point on the scale must shift $25 	ext{ cm}$ backward.

Remember our initial observation? The contact point started at the right edge, at $x = 100 	ext{ cm}$. Shifting this point backward by $25 	ext{ cm}$ gives us our final answer:

$$ x_{final} = 100 - 25 = 75 	ext{ cm} $$

This perfectly matches the visual representation in Option (B). The beauty of this problem lies not in complex calculus, but in the elegant application of relative kinematics and spatial reasoning.

Select Answer:

Visualized Solution

$x_{ini t ia l} = 100 cm$

$V_{c} = ω R$

$V_{s} = V_{c} + ω r$

$V_{s} = V_{c} (1 + \frac{r}{R})$

$V_{s} = 1.5 V_{c}$

$Δ x_{s} = 1.5Δ x_{c}$

$Δ x_{s} = 75 cm$

$x_{f ina l} = 75 cm$

$Δ x_{r e l} = 25 cm$

The Sigma Insight: Rolling Motion

The Setup

The Master Equation

The Velocity of the Scale

From Velocities to Displacements

The Final Catch

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xinitial​=100 cm

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vs​=Vc​(1+Rr​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vs​=1.5Vc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δxs​=1.5Δxc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δxs​=75 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xfinal​=75 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δxrel​=25 cm