LEVELJEE Main

Animated Solution for Physics - Rotational Motion: 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

Q: Is there any tangential force acting on the particles?

No. Tangential force depends on angular acceleration ($F_t = m\alpha r$). Since the angular speed $\omega$ is constant, the angular acceleration $\alpha$ is zero, making the tangential force zero.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Inner radius = R_{1}$
$Outer radius = R_{2}$
$Angular speed = ω (constant)$

$Analyzing Accelerations$

$a_{c m} = 0$
$α = 0$
$a = - ω^{2} r$

$Force on Inner Particle$

$F_{1} = m ω^{2} R_{1}$

$Force on Outer Particle$

$F_{2} = m ω^{2} R_{2}$

$Calculating the Ratio$

$\frac{F _{1}}{F _{2}} = \frac{m ω ^{2} R _{1}}{m ω ^{2} R _{2}}$
$\frac{F _{1}}{F _{2}} = \frac{R _{1}}{R _{2}}$

$What if α \neq = 0 ?$

$If α \neq = 0 ⟹ a = a_{c m} + α \times r - ω^{2} r$

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The Sigma Insight: Rolling Motion

Solution Diagram

Visualizing the Rolling Ring

Imagine an annular ring, much like a thick hollow cylinder, rolling smoothly across a flat surface.

This ring has an inner radius

R_{1}

and an outer radius

R_{2}

The problem states that it is rolling without slipping with a uniform angular speed

ω

This single piece of information is the master key to unlocking the entire problem.

The Master Equation

Acceleration

Because the angular speed

ω

is constant, the linear velocity of the center of mass

v_{c m}

is also constant.

Why? Because for rolling without slipping,

v_{c m} = ω R_{2}

Since both

ω

and

R_{2}

are constant, the acceleration of the center of mass

a_{c m}

is exactly zero.

Furthermore, a constant angular speed means the angular acceleration

α

is also zero.

Therefore, the only acceleration experienced by any particle on this ring is the centripetal acceleration, which is always directed towards the center of the ring.

Calculating the Forces

Let's analyze a particle of mass

m

situated on the inner edge of the ring.

The centripetal force required to keep this particle moving in its circular path is given by:

F_{1} = m ω^{2} R_{1}

Now, let's look at a similar particle of mass

m

situated on the outer edge.

Even though it is further away, it rotates with the exact same angular speed

ω

. The force it experiences is:

F_{2} = m ω^{2} R_{2}

Notice how the force is directly proportional to the radial distance from the center. The further out you go, the greater the force required to maintain that circular motion!

The Final Ratio

The question asks for the ratio of these two forces,

F_{1} / F_{2}

Let's divide our two equations:

\frac{F _{1}}{F _{2}} = \frac{m ω ^{2} R _{1}}{m ω ^{2} R _{2}}

The mass

m

and the angular speed squared

ω^{2}

are identical for both particles, so they cancel out perfectly.

We are left with a beautifully simple result:

\frac{F _{1}}{F _{2}} = \frac{R _{1}}{R _{2}}

This elegant ratio tells us that in a uniformly rotating system, the centripetal force scales linearly with the radius.

Common Traps & FAQs

Why don't we consider the linear velocity of the rolling ring?

Force is determined by acceleration, not velocity. Since the ring rolls with a uniform angular speed, its linear velocity is also uniform, meaning its linear acceleration is zero.

Does the mass of the particles matter?

The question asks for the ratio of forces on 'two particles'. It implicitly assumes we are comparing identical test particles to understand the effect of the radial position. Even if they had different masses, the force is proportional to mass, but standard problems imply identical masses unless specified.

What if the ring was slipping?

If it was slipping but still maintained a uniform angular speed and uniform linear speed, the acceleration of the center of mass would still be zero, and the result would be exactly the same!

Why is the force directed towards the center?

Because the only acceleration present is the centripetal acceleration. This acceleration is required to continuously change the direction of the particle's velocity vector, keeping it in circular motion relative to the center.

Is there any tangential force acting on the particles?

No. Tangential force depends on angular acceleration (

F_{t} = m α r

). Since the angular speed

ω

is constant, the angular acceleration

α

is zero, making the tangential force zero.

Similar Questions

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The figure shows a system consisting of (i) a ring of outer radius $3 R$ rolling clockwise without slipping on a horizontal surface with angular speed $ω$ and (ii) an inner disc of radius $2 R$ rotating anti-clockwise with angular speed $ω /2$ . 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 $O P$ makes an angle of $3 0^{\circ}$ with the horizontal. Then with respect to the horizontal surface,

* Multiple Correct Options

(A)

the point

O

has a linear velocity

3 R ω \hat{i}

(B)

the point

P

has a linear velocity

\frac{11}{4} R ω \hat{i} + \frac{3}{4} R ω \hat{k}

(C)

the point

P

has a linear velocity

\frac{13}{4} R ω \hat{i} - \frac{3}{4} R ω \hat{k}

(D)

the point

P

has a linear velocity

(3 - \frac{3}{4}) R ω \hat{i} + \frac{1}{4} R ω \hat{k}

JEE Main 2021 (20 July Shift-II)

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Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $\frac{x}{2}$ . Then, the value of $x$ is …… .

JEE Main 2019, 10 Jan Shift-I

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A homogeneous solid cylindrical roller of radius $R$ and mass $m$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is

(A)

\frac{F}{2 m R}

(B)

\frac{2 F}{3 m R}

(C)

\frac{3 F}{2 m R}

(D)

\frac{F}{3 m R}

LEVELJEE Main

A round uniform body of radius $R$ , mass $M$ and moment of inertia $I$ , rolls down (without slipping) an inclined plane making an angle $θ$ with the horizontal. Then, its acceleration is

(A)

\frac{g s i n θ}{1 + I / M R ^{2}}

(B)

\frac{g s i n θ}{1 + M R ^{2} / I}

(C)

\frac{g s i n θ}{1 - I / M R ^{2}}

(D)

\frac{g s i n θ}{1 - M R ^{2} / I}

JEE Advanced 2004

LEVELJEE Main

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

(A)

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

(B)

v_{Q} < v_{C} < v_{P}

(C)

v_{Q} = v_{P}, v_{C} = \frac{1}{2} v_{P}

(D)

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

JEE Advanced 2023

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An annular disk of mass M, inner radius a and outer radius b is placed on a horizontal surface with coefficient of friction $μ$ , as shown in the figure. At some time, an impulse $J_{0} \overset{x}{^}$ is applied at a height h above the center of the disk. If $h = h_{m}$ then the disk rolls without slipping along the x-axis. Which of the following statement(s) is(are) correct ?

* Multiple Correct Options

(A)

For

μ \neq = 0

and

a \to 0

h_{m} = b /2

(B)

For

μ \neq = 0

and

a \to b

h_{m} = b

(C)

For

h = h_{m}

, the initial angular velocity does not depend on the inner radius a.

(D)

For

μ = 0

and

h = 0

, the wheel always slides without rolling.

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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 …… .

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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 $Δ θ$ between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities $ω$ and $2 ω$ while the large disk is held stationary. The time $τ$ at which the smaller disks are again in contact is : [Use $sin (Δ θ) = Δ θ$ and ignore gravity.]

(A)

τ = 51 \times (2 π - \frac{4}{51}) / ω

(B)

τ = 51 \times (2 π - \frac{2}{51}) /3 ω

(C)

τ = 51 \times (2 π - \frac{4}{51}) /3 ω

(D)

τ = 51 \times (2 π - \frac{2}{51}) / ω

JEE Advanced 2020

LEVELJEE Advanced

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)-

JEE Main 2019, 8 April Shift-II

LEVELJEE Advanced

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\frac{h _{sph}}{h _{cyl}}$ is given by

(A)

\frac{2}{5}

(B)

\frac{14}{15}

(C)

1

(D)

\frac{4}{5}

Select Answer:

Visualized Solution

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if α=0?

The Sigma Insight: Rolling Motion

Visualizing the Rolling Ring

The Master Equation

Calculating the Forces

The Final Ratio

Similar Questions

$Visualizing the Setup$

$Analyzing Accelerations$

$Force on Inner Particle$

$Force on Outer Particle$

$Calculating the Ratio$

$What if α \neq = 0 ?$