Rolling Motion

Q: Why does a solid sphere reach the bottom of an incline before a hollow sphere of the same mass and radius?

Because a smaller fraction of the solid sphere's mass sits far from the axis, its shape factor is smaller (two-fifths against two-thirds for the hollow one). Acceleration on an incline is g sin theta over one plus c, so smaller c gives larger acceleration. Equivalently, less of the released potential energy goes into rotation and more into translation.

Q: Does mass or radius affect how quickly a body rolls down an incline?

No, neither. The acceleration g sin theta over one plus c depends only on the incline angle and the shape factor c. A tennis ball and a bowling ball, if both are solid spheres, roll down the same incline with the same acceleration and reach the bottom together.

Q: If friction does no work in pure rolling why do we still need friction?

Friction is essential to maintain the rolling condition: it supplies the torque that spins the body up to match the translational speed. Without friction a ball placed on an incline would just slide down without rolling. So friction is a passive constraint force, present in whatever amount the geometry demands but doing zero work because the contact point is instantaneously at rest.

Q: Can pure rolling happen on a smooth frictionless surface?

Only if the rolling has already been set up and no external tangential force acts. A body already rolling with v equals R omega on a smooth horizontal surface will continue rolling forever. But a ball placed at rest on a smooth incline will only slide down; without friction there is no torque to spin it, so it can never roll.

Q: What is the velocity of the topmost point of a rolling wheel?

Twice the velocity of the centre. In pure rolling the centre moves at v, the top moves at v plus R omega equals two v because translation and rotation add at the top, and the bottom moves at v minus R omega equals zero. On a moving car the top of each tire moves at twice the car's speed while the bottom is instantaneously at rest.

Q: What is the difference between sliding and toppling?

Sliding happens when the applied horizontal force exceeds the maximum static friction, so the body starts to translate. Toppling happens when the applied force produces a torque large enough to rotate the body about its leading edge, so it tips over without any translation of the base. Which happens first depends on the base width, the height at which the force is applied, and the friction coefficient.

Q: Why does using the instantaneous axis of rotation give the same total kinetic energy?

Because the two descriptions (translation of CM plus rotation about CM, or pure rotation about the IAOR) are physically identical, just seen from different reference decompositions. The parallel axes theorem is exactly the mathematical bridge that makes half M v squared plus half I about CM omega squared equal to half I about IAOR omega squared, using v equals d omega.

Q: In a body launched with initial speed but no spin on a rough surface does the final rolling speed depend on the friction coefficient?

No. Angular momentum about the bottom-most contact line is conserved because friction acts at that line and gives zero torque. This gives the final rolling speed purely from initial conditions and the shape factor: v equals v-nought over one plus c. The friction coefficient only affects how quickly pure rolling starts, not the final speed reached.

PhysicsSystem Of Particles And Rotational MotionFor NEET aspirants

Rolling motion combines translation of the centre of mass with rotation about it. For pure rolling (no slipping) on a stationary surface, the point of contact has zero velocity, giving the condition $v_{cm} = R ω$ and $a_{cm} = R α$ . The total kinetic energy of a rolling body is $K = \frac{1}{2} M v_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$ . On an incline without slipping, acceleration is $a = \frac{g sin θ}{1 + I _{cm} / M R ^{2}}$ , so a solid sphere reaches the bottom fastest, then a solid cylinder, then a hollow sphere, then a ring. Related concepts (toppling, instantaneous axis of rotation) round out the JEE and NEET rolling-motion syllabus.

Key Formulas - Quick Reference

Pure rolling condition on a stationary surface: $v_{cm} = R ω$ , $a_{cm} = R α$
Velocity of top-most point: $v_{top} = 2 v_{cm} = 2 R ω$
Velocity of bottom-most point (contact): $v_{bottom} = 0$
Total kinetic energy: $K = \frac{1}{2} M v_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$
Using $I_{cm} = c M R^{2}$ : $K = \frac{1}{2} M v_{cm}^{2} (1 + c)$
Acceleration on incline (pure rolling): $a = \frac{g sin θ}{1 + c}$ where $c = I_{cm} / M R^{2}$
Minimum friction for rolling on incline: $μ_{m i n} = \frac{c tan θ}{1 + c}$
Speed after rolling down height $h$ : $v = \frac{2 g h}{1 + c}$
Instantaneous axis of rotation (IAOR): passes through the contact point; $K = \frac{1}{2} I_{IAOR} ω^{2}$
Toppling starts when normal reaction reaches the edge of the base

1. Combined Translation and Rotation

A rolling body executes two motions simultaneously: (i) its centre of mass moves with velocity $v_{cm}$ , and (ii) the body spins about an axis through its centre of mass with angular velocity $ω$ . The velocity of any point $P$ on the body is the vector sum:

$v_{P} = v_{cm} + ω \times r_{P, cm}$

where $r_{P, cm}$ is the position vector of $P$ from the centre of mass. The second term is the point's velocity relative to the CM, which is purely rotational.

Figure 1: Rolling wheel — all velocities shown are ground-frame (net). Centre C moves right at

v

; top point moves at

2 v

; contact point is instantaneously at rest (

v = 0

); left and right ends of the horizontal diameter move at

2 v

45°

(vector sum of centre translation

v

and rotational velocity

R ω = v

2. Pure Rolling Condition

Rolling is pure (no slipping) when the point of contact between the body and the surface is instantaneously at rest. Superposing translation and rotation at that point:

$v_{contact} = v_{cm} - R ω = 0 \Rightarrow v_{cm} = R ω$

Differentiating,

$a_{cm} = R α$

These two conditions must both be satisfied for pure rolling on a stationary surface. If $v_{cm} > R ω$ , the body slips forward (contact point moves in the direction of motion). If $v_{cm} < R ω$ , it slips backward (contact point moves opposite to motion).

Distance per revolution. In one full rotation, the centre of a rolling body advances by exactly the circumference of the rim,

2 π R

. Forward slipping gives

s > 2 π R

; backward slipping gives

s < 2 π R

3. Kinetic Energy of a Rolling Body

Total kinetic energy of a rigid body in combined translation and rotation splits cleanly:

$K = translational \frac{1}{2} M v_{cm}^{2} + rotational about CM \frac{1}{2} I_{cm} ω^{2}$

Writing $I_{cm} = c M R^{2}$ (where $c$ is a shape factor: $\frac{1}{2}$ for a solid cylinder, $\frac{2}{5}$ for a solid sphere, etc.) and using pure rolling $v_{cm} = R ω$ ,

$K = \frac{1}{2} M v_{cm}^{2} (1 + c)$

The bracket $(1 + c)$ tells you what fraction of the total KE is stored in rotation. For a ring ( $c = 1$ ), half the energy is rotational; for a solid sphere ( $c = 2/5$ ), only $2/7$ of the KE is rotational.

4. Rolling on an Inclined Plane

Figure 2: Free-body diagram of a body rolling down a rough incline. Friction

f

acts up the incline at the contact point to prevent slipping and provide the spinning torque.

Let a rigid body of mass $M$ , radius $R$ and $I_{cm} = c M R^{2}$ roll (no slipping) down an incline at angle $θ$ . Along the incline, taking down as positive:

Newton's second law (translation): $M g sin θ - f = M a$

Rotation about the CM: $f R = I_{cm} α = c M R^{2} \cdot \frac{a}{R} = c M R a$ , giving $f = c M a$

Substituting into the first equation:

$M g sin θ - c M a = M a \Rightarrow a = \frac{g sin θ}{1 + c}$

Required friction: $f = c M a = \frac{c M g sin θ}{1 + c}$ . For friction not to exceed the limiting static value $μ_{s} M g cos θ$ ,

$μ_{s} \geq \frac{c tan θ}{1 + c}$

4.1 Which Body Wins the Race?

Body	$c = I_{cm} / M R^{2}$	Acceleration $a$	Speed at bottom (height $h$ )
Solid sphere	$2/5$	$\frac{5 g sin θ}{7}$	$\frac{10 g h}{7}$
Solid cylinder / disc	$1/2$	$\frac{2 g sin θ}{3}$	$\frac{4 g h}{3}$
Hollow sphere	$2/3$	$\frac{3 g sin θ}{5}$	$\frac{6 g h}{5}$
Ring / hollow cylinder	$1$	$\frac{g sin θ}{2}$	$g h$

Smaller $c$ means larger $a$ and larger $v$ at the bottom, so the ranking (fastest first) is:

$solid sphere > solid cylinder > hollow sphere > ring$

Notice the result is independent of the mass $M$ and radius $R$ ; only the mass-distribution factor $c$ matters. This is why in the classic "which rolls down fastest" demonstration, size makes no difference.

Friction does no work in pure rolling on a stationary surface. The contact point has zero velocity, so the friction force (which acts at that point) has zero instantaneous power. Mechanical energy is therefore conserved, and

\frac{1}{2} M v^{2} (1 + c) = M g h

directly gives

v = 2 g h / (1 + c)

Solved Example 1

A solid sphere and a hollow sphere of the same mass and radius are released from rest at the top of the same incline. Which reaches the bottom first, and by what factor is its speed at the bottom greater?

Solution:

Solid sphere: $c = 2/5$ , so $v_{1} = \frac{2 g h}{1 + 2/5} = \frac{10 g h}{7}$ .

Hollow sphere: $c = 2/3$ , so $v_{2} = \frac{2 g h}{1 + 2/3} = \frac{6 g h}{5}$ .

Ratio:

$\frac{v _{1}}{v _{2}} = \frac{10/7}{6/5} = \frac{50}{42} = \frac{25}{21} \approx 1.09$

The solid sphere arrives first, with speed about $9%$ greater than the hollow sphere. Even though the hollow sphere may look heavier "at the rim", it has more of its energy locked into rotation, leaving less for translation.

Solved Example 2

A solid sphere is rolling without slipping on a horizontal surface with speed

v

. It then begins to move up (a) a rough incline steep enough to sustain pure rolling all the way up, and (b) a smooth incline on which it slides. In each case, find the maximum height reached.

Solution:

(a) Rough incline (pure rolling throughout). Friction does no work (contact point stationary), so mechanical energy is conserved. Both translational and rotational KE convert entirely to potential energy at the top, where $v = 0$ and hence $ω = v / R = 0$ :

$\frac{1}{2} M v^{2} + \frac{1}{2} \cdot \frac{2}{5} M R^{2} \cdot \frac{v ^{2}}{R ^{2}} = M g h_{1} \Rightarrow \frac{7}{10} M v^{2} = M g h_{1} \Rightarrow h_{1} = \frac{7 v ^{2}}{10 g}$

(b) Smooth incline (no friction). With no friction there is no torque, so the sphere's angular speed $ω = v / R$ is preserved as it slides up. Only the translational KE is available to lift the sphere:

$\frac{1}{2} M v^{2} = M g h_{2} \Rightarrow h_{2} = \frac{v ^{2}}{2 g}$

Ratio: $h_{1} / h_{2} = 7/5 = 1.4$ . The sphere climbs $40%$ higher on the rough incline because its rotational KE also gets converted to potential energy, whereas on the smooth incline that rotational KE stays trapped as spin at the top.

5. Rolling With an Applied Force

When a horizontal force $F$ acts on a rolling body, friction has a subtle role: it may act forward, backward, or vanish altogether, depending on where the force is applied. The safe method is to assume friction $f$ acts forward and solve. A negative answer means friction actually acts backward.

Consider a solid sphere ( $c = 2/5$ ) with a horizontal force $F$ applied at the centre. Newton's law: $F + f = M a$ . Rotational (about the CM): $- f R = I_{cm} α$ (friction is the only torque source since $F$ passes through the CM), so $f = - \frac{2}{5} M a$ . Substituting:

$F - \frac{2}{5} M a = M a \Rightarrow a = \frac{5 F}{7 M}, f = - \frac{2 F}{7} (backward)$

Friction points backward here to supply the torque that spins the sphere and keeps it rolling.

5.1 Force Applied at a Height $h$ Above the Centre

Now shift the point of application. Let a horizontal force $F$ act on a solid sphere at a point a distance $h$ vertically above (or below) the centre. About the CM, $F$ now supplies a torque $F h$ (in addition to the friction torque $f R$ ). Choose the sign so that positive $h$ makes $F$ tend to spin the sphere forward:

Translation: $F + f = M a$ .

Rotation about CM: $F h - f R = I_{cm} α = \frac{2}{5} M R^{2} \cdot \frac{a}{R} = \frac{2}{5} M R a$ .

Figure 3: A horizontal force

F

applied at height

h

above the centre of a solid sphere. When

h = 2 R /5

, the friction required for rolling drops to zero.

Solved Example 3

A horizontal force

F

acts on a solid sphere of mass

M

and radius

R

at a height

h

above its centre. The sphere rolls without slipping on a rough horizontal surface. (i) Find the acceleration of the centre. (ii) For what value of

h

does the required friction become zero?

Solution:

Take rightward positive and let friction $f$ act forward.

Translation: $F + f = M a$ .

Rotation about CM: $F h - f R = I_{cm} α = \frac{2}{5} M R^{2} \cdot (a / R) = \frac{2}{5} M R a$ , hence $f = \frac{F h}{R} - \frac{2 M a}{5}$ .

Substituting into the translation equation:

$F + \frac{F h}{R} - \frac{2 M a}{5} = M a \Rightarrow F (1 + \frac{h}{R}) = \frac{7 M a}{5} \Rightarrow a = \frac{5 F}{7 M} (1 + \frac{h}{R})$

Now compute $f$ :

$f = \frac{F h}{R} - \frac{2 M}{5} \cdot \frac{5 F}{7 M} (1 + \frac{h}{R}) = \frac{F h}{R} - \frac{2 F}{7} (1 + \frac{h}{R}) = \frac{F}{7 R} (5 h - 2 R)$

Setting $f = 0$ gives $h = \frac{2 R}{5}$ . For $h > 2 R /5$ friction acts backward (the applied force over-spins the sphere); for $h < 2 R /5$ friction acts forward. At the sweet spot $h = 2 R /5$ , the applied torque exactly matches what rotation demands and no friction is required.

5.2 The Yo-yo (Rolling on a String)

Figure 4: A yo-yo unwinding under gravity. The string tangent point on the disc has zero instantaneous velocity, so

a = R α

acts as the rolling constraint.

A yo-yo is a disc unwinding along a string, and the string acts exactly like the ground in ordinary rolling: the tangent point on the disc where the string leaves is instantaneously stationary, so the constraint $a = R α$ holds with $a$ the downward acceleration of the centre.

Solved Example 4

A uniform disc of mass

M

and radius

R

(moment of inertia

I_{cm} = \frac{1}{2} M R^{2}

) has a light inextensible string wound around its rim. The free end of the string is held fixed and the disc is released from rest. Find (i) the downward acceleration of the centre, (ii) the tension in the string.

Solution:

Take downward positive. Let $a$ be the downward acceleration of the centre and $T$ the tension in the string (acting upward on the disc at the rim).

Translation: $M g - T = M a$ .

Rotation about the centre: $T R = I_{cm} α = \frac{1}{2} M R^{2} \cdot \frac{a}{R} = \frac{1}{2} M R a$ , hence $T = \frac{1}{2} M a$ .

Substituting:

$M g - \frac{1}{2} M a = M a \Rightarrow a = \frac{2 g}{3}, T = \frac{M g}{3}$

The centre falls at only two-thirds of $g$ because the tension supports part of the weight while also supplying the torque that spins the disc up. For a hoop ( $I = M R^{2}$ ) the same argument gives $a = g /2$ , $T = M g /2$ ; for a solid sphere ( $I = \frac{2}{5} M R^{2}$ ) it gives $a = 5 g /7$ , $T = 2 M g /7$ .

6. Transition from Slipping to Rolling

When a body is launched with initial linear speed $v_{0}$ but no spin (or with mismatched $v$ and $ω$ ), friction acts to bring the motion into pure rolling. During the slipping phase friction is kinetic; once pure rolling starts it becomes static (typically less than the maximum).

Solved Example 5

A solid sphere (

I_{cm} = \frac{2}{5} M R^{2}

) is placed on a rough horizontal surface with initial linear speed

v_{0}

forward and zero angular speed. Coefficient of kinetic friction is

μ_{k}

. Find (i) the time when pure rolling starts, (ii) the linear speed at that instant.

Solution:

During slipping, kinetic friction $μ_{k} M g$ acts backward on the CM, decelerating it, and provides the forward torque that spins the sphere up.

Linear: $v = v_{0} - μ_{k} g t$ .

Rotational: $τ = μ_{k} M g R$ , so $α = \frac{μ _{k} M g R}{\frac{2}{5} M R ^{2}} = \frac{5 μ _{k} g}{2 R}$ , and $ω = α t$ .

Pure rolling condition $v = R ω$ :

$v_{0} - μ_{k} g t = R \cdot \frac{5 μ _{k} g}{2 R} t = \frac{5 μ _{k} g}{2} t$

$v_{0} = \frac{7 μ _{k} g}{2} t \Rightarrow t = \frac{2 v _{0}}{7 μ _{k} g}$

At that instant, $v = v_{0} - μ_{k} g \cdot \frac{2 v _{0}}{7 μ _{k} g} = \frac{5 v _{0}}{7}$ .

Alternative (much faster) route: angular momentum about the bottom-most contact line is conserved because friction acts at that line and has zero torque about it. Before: $L = M v_{0} R$ . After (pure rolling): $L = M v R + I_{cm} ω = M v R + \frac{2}{5} M R^{2} \cdot v / R = \frac{7}{5} M v R$ . Equating: $v_{0} = \frac{7}{5} v$ , so $v = 5 v_{0} /7$ .

7. Toppling

Figure 5: Applied force

F

at height

b

on a block of base

a

. The normal reaction

N

shifts toward the leading edge; toppling begins when

N

reaches that edge.

When a horizontal force $F$ is applied at height $b$ above the base of a block of width $a$ , two failure modes compete:

Sliding starts when $F > μ_{s} M g$ .
Toppling starts when the torque about the leading edge just overturns the block: $F b = M g \cdot (a /2)$ , giving $F > \frac{M g a}{2 b}$ .

Whichever happens at the smaller $F$ decides the mode. A tall, narrow block topples easily (small $a$ , large $b$ ); a broad short block slides first.

Why blocks with a wide base are stable. The normal reaction has room to shift toward the leading edge without going past it, so even a large applied force can be balanced without producing a net toppling torque. Tall skinny blocks give the normal reaction almost no room to shift, so they topple at tiny forces.

8. Instantaneous Axis of Rotation (IAOR)

At any instant, the combined translation-plus-rotation motion of a rigid body can be viewed as a pure rotation about a special axis where the instantaneous velocity is zero. This axis is called the instantaneous axis of rotation.

Figure 6: For a wheel rolling without slipping, the instantaneous axis of rotation is the line of contact with the ground.

For a wheel in pure rolling, the contact point has zero velocity, so the IAOR is the line of contact. Every other point on the wheel moves as if it were rotating in a circle about this line. The speed of a point at distance $r$ from the IAOR is $v = r ω$ , and the total kinetic energy can be written as

$K = \frac{1}{2} I_{IAOR} ω^{2}$

where $I_{IAOR}$ is the moment of inertia about the IAOR. Using the parallel axes theorem for a rolling disc: $I_{IAOR} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$ . Both approaches give the same total KE, as they must.

8.1 Locating the IAOR When It Is Not Obvious

If the body is not simply rolling on a surface, you can still locate the IAOR from two facts:

The IAOR lies on the line perpendicular to the velocity of any point, passing through that point.
If the velocities of two points are known, draw perpendiculars to each velocity at its location; the IAOR is where the perpendiculars intersect.

Solved Example 6

A ladder of length

L

slides down a smooth wall while its foot slips along a smooth horizontal floor. At an instant when the ladder makes angle

θ

with the horizontal, find the ratio of the speed of the top to the speed of the foot.

Solution:

The top of the ladder slides straight down along the wall (velocity vertical); the foot slides horizontally along the floor (velocity horizontal). The IAOR is at the intersection of the horizontal line through the top and the vertical line through the foot, which forms a rectangle with the ladder as diagonal.

Distance from IAOR to the top: $L sin θ$ . Distance from IAOR to the foot: $L cos θ$ .

Both points rotate about the IAOR with the same angular speed $ω$ , so

$\frac{v _{top}}{v _{foot}} = \frac{L sin θ}{L cos θ} = tan θ$

Notice: the top moves faster than the foot when $θ > 45°$ and slower when $θ < 45°$ .

Solved Example 7

A uniform disc of mass

M

and radius

R

rolls without slipping on a horizontal surface with speed

v

of its centre. Find its total kinetic energy.

Solution:

Two equivalent ways.

Method 1: Split into translation and rotation.

$K = \frac{1}{2} M v^{2} + \frac{1}{2} \cdot \frac{M R ^{2}}{2} \cdot (\frac{v}{R})^{2} = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2}$

Method 2: Treat as pure rotation about the IAOR (bottom-most contact line).

$I_{IAOR} = I_{cm} + M R^{2} = \frac{M R ^{2}}{2} + M R^{2} = \frac{3 M R ^{2}}{2}$

$K = \frac{1}{2} I_{IAOR} ω^{2} = \frac{1}{2} \cdot \frac{3 M R ^{2}}{2} \cdot \frac{v ^{2}}{R ^{2}} = \frac{3}{4} M v^{2}$

Same answer, as expected.

Common Mistakes to Avoid

Watch out

Assuming friction does work on a rolling body. On a stationary rough surface with pure rolling, friction acts at a point with zero velocity, so it does zero work and mechanical energy is conserved.
Forgetting to include the rotational kinetic energy $\frac{1}{2} I ω^{2}$ when applying energy conservation to a rolling body.
Using $v_{cm} = R ω$ when the body is slipping. This condition holds only in pure rolling.
Assuming a sphere and a cylinder reach the bottom of an incline together. They do not; their shape factors $c$ differ, so accelerations differ.
Treating the friction on an incline as always $μ_{k} N$ . In pure rolling it is static friction, at whatever value the equations require, up to the limit $μ_{s} N$ .
Assuming the answer depends on mass or radius when rolling down an incline. It does not; only the shape factor $c$ matters.
Forgetting to shift the moment of inertia to the IAOR (via the parallel axes theorem) when using $K = \frac{1}{2} I_{IAOR} ω^{2}$ .
In toppling problems, forgetting that the normal reaction can shift along the base but not go beyond the edge.
Confusing forward slipping ( $v > R ω$ ) with backward slipping ( $v < R ω$ ). Draw the arrows for the velocity of the contact point to check which way it moves relative to the ground.

Frequently Asked Questions

Why does a solid sphere reach the bottom of an incline before a hollow sphere of the same mass and radius?

Because a smaller fraction of the solid sphere's mass sits far from the axis, its shape factor is smaller ( $c = 2/5$ against $2/3$ for the hollow one). Acceleration on an incline is $g sin θ / (1 + c)$ , so smaller $c$ gives larger acceleration. Equivalently, less of the released potential energy goes into rotation and more into translation.

Does mass or radius affect how quickly a body rolls down an incline?

No, neither. The acceleration $g sin θ / (1 + c)$ depends only on the incline angle and the shape factor $c$ . A tennis ball and a bowling ball, if both are solid spheres, roll down the same incline with the same acceleration and reach the bottom together.

If friction does no work in pure rolling, why do we still need friction?

Friction is essential to maintain the rolling condition: it supplies the torque that spins the body up to match the translational speed. Without friction, a ball placed on an incline would just slide down without rolling. So friction is a passive constraint force, present in whatever amount the geometry demands, but doing zero work because the contact point is instantaneously at rest.

Can pure rolling happen on a smooth (frictionless) surface?

Only if the rolling has already been set up and no external tangential force acts. A body already rolling with $v = R ω$ on a smooth horizontal surface will continue rolling forever (no torque changes $ω$ , no force changes $v$ ). But a ball placed at rest on a smooth incline will only slide down; without friction there is no torque to spin it, so it can never roll.

What is the velocity of the topmost point of a rolling wheel?

Twice the velocity of the centre. In pure rolling, the centre moves at $v_{cm}$ ; the top moves at $v_{cm} + R ω = 2 v_{cm}$ (translation plus rotation both point the same way at the top). The bottom moves at $v_{cm} - R ω = 0$ . So on a moving car, the top of each tire is moving at twice the car's speed while the bottom is instantaneously at rest.

What is the difference between sliding and toppling?

Sliding happens when the applied horizontal force exceeds the maximum static friction, so the body starts to translate. Toppling happens when the applied force produces a torque large enough to rotate the body about its leading edge, so it tips over without any translation of the base. Which happens first depends on the base width, the height at which the force is applied, and the friction coefficient.

Why does using the instantaneous axis of rotation give the same total kinetic energy?

Because the two descriptions (translation of CM plus rotation about CM, or pure rotation about the IAOR) are physically identical, just seen from different reference decompositions. The parallel axes theorem $I_{IAOR} = I_{cm} + M d^{2}$ is exactly the mathematical bridge that makes $\frac{1}{2} M v^{2} + \frac{1}{2} I_{cm} ω^{2}$ equal to $\frac{1}{2} I_{IAOR} ω^{2}$ (using $v = d ω$ ).

In a body launched with $v_{0}$ but no spin on a rough surface, does the final rolling speed depend on the friction coefficient?

No. Angular momentum about the bottom-most contact line is conserved because friction acts at that line (zero torque). This gives the final rolling speed purely from initial conditions and the shape factor: $v = v_{0} / (1 + c)$ . The friction coefficient only affects how quickly pure rolling starts, not the final speed reached.

Previous year questions on Rolling Motion

17 questions from past papers, each with a step-by-step solution.

Show all 17 questions

Ready to master System Of Particles And Rotational Motion?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. Combined Translation and Rotation

2. Pure Rolling Condition

3. Kinetic Energy of a Rolling Body

4. Rolling on an Inclined Plane

4.1 Which Body Wins the Race?

5. Rolling With an Applied Force

5.1 Force Applied at a Height h Above the Centre

5.2 The Yo-yo (Rolling on a String)

6. Transition from Slipping to Rolling

7. Toppling

8. Instantaneous Axis of Rotation (IAOR)

8.1 Locating the IAOR When It Is Not Obvious

Common Mistakes to Avoid

Frequently Asked Questions

Why does a solid sphere reach the bottom of an incline before a hollow sphere of the same mass and radius?

Does mass or radius affect how quickly a body rolls down an incline?

If friction does no work in pure rolling, why do we still need friction?

Can pure rolling happen on a smooth (frictionless) surface?

What is the velocity of the topmost point of a rolling wheel?

What is the difference between sliding and toppling?

Why does using the instantaneous axis of rotation give the same total kinetic energy?

In a body launched with v0​ but no spin on a rough surface, does the final rolling speed depend on the friction coefficient?

Previous year questions on Rolling Motion

Ready to master System Of Particles And Rotational Motion?

5.1 Force Applied at a Height $h$ Above the Centre

In a body launched with $v_{0}$ but no spin on a rough surface, does the final rolling speed depend on the friction coefficient?