A roller is made by joining together two corners at their vertices O. It is kept on two rails AB and

Q: Why does the radius $r_1$ decrease as the roller moves?

Because rail $AB$ is slanted inwards, the physical point of contact between the rail and the cone moves closer to the vertex $O$ of the cone. Since a cone gets narrower towards its vertex, the radius at the contact point decreases.

Q: Why does the contact point slip forward?

For pure rolling, the forward velocity $v$ must equal the backward rotational velocity $\omega r_1$. When $r_1$ decreases, $\omega r_1$ becomes smaller than $v$. The net velocity of the contact point is $v - \omega r_1$, which becomes positive, meaning it slips forward.

Q: Doesn't the right side also slip?

Initially, rail $CD$ is parallel to the direction of motion, so the contact point remains at a constant distance from the vertex $O$. Thus, $r_2$ is constant, and the right side remains in pure rolling until the roller actually begins to turn.

Q: How does backward friction cause a left turn?

Imagine pushing a shopping cart. If you pull backward on the left side of the handle, the cart turns left. Similarly, the backward friction on the left side of the roller creates a torque about the center of mass that steers it to the left.

Q: What if both rails were slanted inwards symmetrically?

If both rails were symmetric, the radius would decrease equally on both sides. Both sides would slip forward, and backward friction would act equally on both sides. This would slow the roller down, but the torques would cancel out, so it would not turn.

Question

A roller is made by joining together two corners at their vertices $O$. It is kept on two rails $AB$ and $CD$ which are placed asymmetrically (see the figure), with its axis perpendicular to $CD$ and its centre $O$ at the centre of line joining $AB$ and $CD$ (see the figure). It is given a light push, so that it starts rolling with its centre $O$ moving parallel to $CD$ in the direction shown. As it moves, the roller will tend to

Options:

(A) turn left

(B) turn right

(C) go straight

(D) turn left and right alternately

Sigma Percentile · Accepted Answer

### The Geometry of the Setup

Imagine a unique roller constructed by joining two identical cones at their vertices, forming a shape that resembles an hourglass pinched in the middle. This roller is placed on two rails, $AB$ and $CD$. The geometry of these rails is the key to unlocking this problem. Rail $CD$ is perfectly straight and parallel to the initial direction of motion. However, rail $AB$ is slanted inwards, meaning the distance between the two rails gradually decreases as you move forward.

When the roller is given a light push, it begins to move forward. Because the axis of the roller is initially perpendicular to rail $CD$, the center of the roller, $O$, moves in a straight line parallel to $CD$.

### The Physics of Pure Rolling

Initially, the roller experiences pure rolling at both contact points. In pure rolling, the point of the roller in contact with the rail has zero velocity relative to the rail. Mathematically, the forward velocity of the center of mass, $v$, is perfectly balanced by the backward tangential velocity due to rotation, $\omega r$.

So, at the left contact point $P$ (on rail $AB$) and the right contact point $Q$ (on rail $CD$), we have:

$v = \omega r_1 = \omega r_2$

Here, $r_1$ and $r_2$ are the radii of the conical cross-sections at the respective contact points.

### The Onset of Slipping

As the roller progresses forward, the slanted nature of rail $AB$ comes into play. Because the rail angles inwards, the contact point $P$ is forced to move closer to the central vertex $O$. Since the roller is conical, moving closer to the vertex means the radius of the cross-section at the contact point, $r_1$, begins to decrease.

This is where the pure rolling condition breaks down. The center of the roller is still translating forward with velocity $v$, but the rotational velocity at the left contact point, $\omega r_1$, has decreased because $r_1$ is smaller.

The net velocity of the contact point $P$ is given by:

$v_P = v - \omega r_1$

Since $v$ is now greater than $\omega r_1$, the velocity $v_P$ becomes positive. This means the left side of the roller is no longer gripping the rail perfectly; it is slipping forward.

### Friction and Torque: The Steering Mechanism

Nature always opposes relative motion between surfaces in contact. To fight this forward slipping, kinetic friction $f_L$ immediately acts on the left contact point in the **backward** direction.

Now, consider the effect of this backward frictional force on the entire roller. A backward force applied to the left side of the center of mass creates a torque. If you look at the roller from above, this torque acts in a counter-clockwise direction.

This torque acts exactly like a steering wheel. By pulling the left side of the roller backward, it forces the entire assembly to pivot and **turn towards the left**.

### Conclusion

Once the roller begins to turn left, its axis is no longer perpendicular to rail $CD$. This misalignment causes the right side to also begin slipping, which introduces additional frictional forces that further amplify the turning effect. However, the initial trigger is the decreasing radius on the slanted rail, which definitively causes the roller to turn left.

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Visualized Solution

$Initial Setup$

$Pure Rolling Condition$

$Change in Radius$

$Onset of Slipping$

$Kinetic Friction$

$Torque and Steering$

$Conclusion$

The Sigma Insight: Rolling Motion

The Geometry of the Setup

The Physics of Pure Rolling

The Onset of Slipping

Friction and Torque

Conclusion

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Visualized Solution

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

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

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

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

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

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

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