JEE Advanced 2004

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Animated Solution for Physics - Rotational Motion: A disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points equidistant from $C$ . Let $v_{P}$ , $v_{Q}$ and $v_{C}$ be the magnitude of velocities of points $P$ , $Q$ and $C$ respectively, then

Q: Can we solve this using relative velocity instead of ICR?

Absolutely! The velocity of any point can be written as $\vec{v} = \vec{v}_{cm} + \vec{v}_{rotation}$. By adding the constant forward vector $\vec{v}_{cm}$ and the tangential rotational vector $\vec{\omega} \times \vec{r}$, you will find that points higher up have these vectors adding constructively, while points lower down have them partially cancelling out.

Q: Does the angular velocity $\omega$ change for different points?

No, the angular velocity $\omega$ is a property of the rigid body as a whole. Every point on the disc rotates with the exact same angular velocity $\omega$ about the Instantaneous Center of Rotation.

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Visualized Solution

Instantaneous Center of Rotation

For a disc rolling without slipping, the point of contact with the ground acts as the Instantaneous Center of Rotation (ICR).
Let's call this point $O$ . The velocity of this point is zero.

Velocity of Any Point

The entire disc can be assumed to be in pure rotation about the ICR ( $O$ ) with angular velocity $ω$ .
The velocity of any point is $v = r ω$ , where $r$ is its distance from $O$ .

Distance of Point $Q$

Point $Q$ is vertically above the center $C$ .
Its distance from $O$ is $r_{Q}$ .

Distance of Point $C$

Point $C$ is the center of the disc.
Its distance from $O$ is $r_{C} = R$ .

Distance of Point $P$

Point $P$ is equidistant from $C$ as $Q$ , but it lies below the horizontal level of $C$ .
Its distance from $O$ is $r_{P}$ .

Comparing Distances

From the geometry, we can clearly see that:
$r_{Q} > r_{C} > r_{P}$

Comparing Velocities

Since $v \propto r$ , the velocities will follow the same order:
$v_{Q} > v_{C} > v_{P}$

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The Sigma Insight: Rolling Motion

Solution Diagram

The Magic of Pure Rolling

When a rigid body like a disc rolls on a surface without slipping, it exhibits a fascinating kinematic property. The point of the disc that is in direct contact with the ground is momentarily at rest. This point is known as the Instantaneous Center of Rotation (ICR).

Imagine the entire disc is not just translating and rotating, but purely rotating about this single point of contact. This powerful visualization simplifies complex rolling problems into straightforward rotational mechanics.

The Master Equation

Because the disc is in pure rotation about the ICR, the velocity of any arbitrary point on the disc is directly proportional to its distance from the ICR. Mathematically, this is expressed as:

v = r ω

Here,

v

is the magnitude of the velocity,

r

is the straight-line distance from the ICR to the point, and

ω

is the angular velocity of the disc. This means the farther a point is from the ground contact, the faster it moves!

Analyzing the Distances

Let's denote the point of contact as

O

. We need to compare the velocities of three points:

Q

C

, and

P

Point

C

is the geometric center of the disc. Its distance from

O

is simply the radius of the disc,

r_{C} = R

Point

Q

is located vertically above the center

C

. Geometrically, this is the highest point on the disc. Its distance from

O

r_{Q} = 2 R

. Clearly,

r_{Q} > r_{C}

Point

P

is given to be equidistant from

C

Q

, meaning the distance

C P = C Q

. However, looking at the visual representation, point

P

lies below the horizontal axis passing through

C

. Because it is situated in the lower half of the disc, its distance from the bottom contact point

O

is less than the radius

R

. Therefore,

r_{P} < r_{C}

The Final Verdict

Putting it all together, we have a clear geometric inequality for the distances from the Instantaneous Center of Rotation:

r_{Q} > r_{C} > r_{P}

Since the velocity magnitude is directly proportional to this distance (

v \propto r

), the velocities must follow the exact same hierarchy:

$v_{Q} > v_{C} > v_{P}$

The top of the wheel flies forward, the center moves at a steady pace, and the points near the bottom sluggishly make their way around. Physics is beautiful, isn't it?

Common Traps & FAQs

Why is the point of contact at rest?

In pure rolling, the translational velocity of the center of mass (

v_{c m}

) is exactly balanced by the backward tangential velocity due to rotation (

R ω

) at the bottom-most point. Since

v_{c m} = R ω

, the net velocity at the contact point is

v_{c m} - R ω = 0

Can we solve this using relative velocity instead of ICR?

Absolutely! The velocity of any point can be written as

v = v_{c m} + v_{r o t a t i o n}

. By adding the constant forward vector

v_{c m}

and the tangential rotational vector

ω \times r

, you will find that points higher up have these vectors adding constructively, while points lower down have them partially cancelling out.

What if point P was horizontally aligned with C?

If P was exactly on the horizontal line passing through C, its distance from the bottom point O would be

R^{2} + R^{2} = 2 R

. In that case,

r_{Q} (2 R) > r_{P} (2 R) > r_{C} (R)

, so the velocity order would be

v_{Q} > v_{P} > v_{C}

Does the angular velocity

ω

change for different points?

No, the angular velocity

ω

is a property of the rigid body as a whole. Every point on the disc rotates with the exact same angular velocity

ω

about the Instantaneous Center of Rotation.

Why is Q's velocity the highest?

Point Q is at the very top of the disc. Here, the translational velocity of the disc and the tangential velocity due to rotation are in the exact same forward direction. They add up perfectly to give

v_{Q} = v_{c m} + R ω = 2 v_{c m}

, which is the maximum possible velocity on the disc.

Animated Solution for Physics - Rotational Motion: A disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points equidistant from $C$ . Let $v_{P}$ , $v_{Q}$ and $v_{C}$ be the magnitude of velocities of points $P$ , $Q$ and $C$ respectively, then

Select Answer:

Visualized Solution

Instantaneous Center of Rotation

Velocity of Any Point

Distance of Point $Q$

Distance of Point $C$

Distance of Point $P$

Comparing Distances

Comparing Velocities

The Sigma Insight: Rolling Motion

The Magic of Pure Rolling

The Master Equation

Analyzing the Distances

The Final Verdict

Similar Questions

A circular disc reaches from top to bottom of an inclined plane of length $L$ . When it slips down the plane, it takes time $t_{1}$ . When it rolls down the plane, it takes time $t_{2}$ . The value of $\frac{t _{2}}{t _{1}}$ is $\frac{3}{x}$ . The value of $x$ will be …… .

The centre of a wheel rolling on a plane surface moves with a speed $v_{0}$ . A particle on the rim of the wheel at the same level as the centre will be moving at a speed $x v_{0}$ . Then, the value of $x$ is ……… .

An annular ring with inner and outer radii $R_{1}$ and $R_{2}$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $F_{1} / F_{2}$ is

A uniform sphere of mass $500 g$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00 cm/s$ . Its kinetic energy is

Select Answer:

Visualized Solution

Instantaneous Center of Rotation

Velocity of Any Point

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

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

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

Comparing Distances

Comparing Velocities

The Sigma Insight: Rolling Motion

The Magic of Pure Rolling

The Master Equation

Analyzing the Distances

The Final Verdict

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is

Distance of Point $Q$

Distance of Point $C$

Distance of Point $P$

A uniform sphere of mass $500 g$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00 cm/s$ . Its kinetic energy is