A disc of radius R is rotating with an angular speed ωo about a horizontal axis. It is placed on a horizontal

NCERT Exemplar · Q25

R

(a) What was the velocity of its centre of mass before being brought in contact with the table?

(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?

(d) Which force is responsible for the effects in

(b) and (c).

(e) What condition should be satisfied for rolling to begin?

(f) Calculate the time taken for the rolling to begin.

Punjab PsebLong· 5mImportance★★★★★est

95% · 54/57 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Rotational Dynamics

Rotational Dynamics: The Physics of Spinning Things

Imagine you're trying to open a heavy door. You push near the hinge — it barely moves. Push near the handle — it swings open easily. Same force, different result. That's the first clue: rotation isn't just about how much you push, but where and in what direction.

Now think about a spinning bicycle wheel. Why is it so hard to tilt it sideways when it's spinning fast? And why does a figure skater spin faster when she pulls her arms in? These are the questions rotational dynamics answers.

The Core Idea

Rotational dynamics is the study of why things rotate and how their rotation changes. It's the spinning-world equivalent of Newton's laws for straight-line motion.

In linear motion, you have:

Force ( $F$ ) causes acceleration ( $a$ )
Mass ( $m$ ) resists acceleration

In rotational motion, you have:

Torque ( $\tau$ ) causes angular acceleration ( $\alpha$ )
Moment of inertia ( $I$ ) resists angular acceleration

The master equation is:

\tau_{\text{net}} = I \alpha

This is the rotational version of $F = ma$ . Every term has a direct parallel.

Breaking It Down

Torque — The Rotational "Push"

Torque isn't just force — it's force multiplied by the distance from the pivot point (the lever arm). That's why the door handle works better than the hinge.

\tau = r F \sin\theta

Where $r$ is the distance from the axis, $F$ is the force, and $\theta$ is the angle between them. Maximum torque happens when you push perpendicular to the lever arm ( $\theta = 90^\circ$ ).

Tip

Think of torque as "twisting effectiveness." A wrench works because the handle gives you a long lever arm. A short wrench needs more force to do the same job.

Moment of Inertia — The Rotational "Mass"

Mass resists linear acceleration. Moment of inertia resists angular acceleration. But unlike mass, moment of inertia depends on how the mass is distributed relative to the axis of rotation.

For a point mass $m$ at distance $r$ from the axis:

I = mr^2

For extended objects, you sum (or integrate) over all mass elements:

I = \sum m_i r_i^2

Note

| Object | Axis | Moment of Inertia |

|--------|------|-------------------|

| Thin hoop | Through center, perpendicular to plane | $MR^2$ |

| Solid disk | Through center, perpendicular to plane | $\frac{1}{2}MR^2$ |

| Solid sphere | Through center | $\frac{2}{5}MR^2$ |

| Thin rod | Through center, perpendicular to rod | $\frac{1}{12}ML^2$ |

Notice: a hoop has more moment of inertia than a disk of the same mass and radius because its mass is farther from the axis. That's why a hoop is harder to start spinning.

Angular Acceleration — How Fast Rotation Changes

Just as acceleration is the rate of change of velocity, angular acceleration $\alpha$ is the rate of change of angular velocity $\omega$ :

\alpha = \frac{d\omega}{dt}

And angular velocity is the rate of change of angular displacement $\theta$ :

\omega = \frac{d\theta}{dt}

The Complete Picture: Rotational Analogues

Linear Quantity	Rotational Analogue
Displacement $x$	Angular displacement $\theta$
Velocity $v$	Angular velocity $\omega$
Acceleration $a$	Angular acceleration $\alpha$
Mass $m$	Moment of inertia $I$
Force $F$	Torque $\tau$
Newton's 2nd law: $F = ma$	$\tau = I\alpha$
Kinetic energy: $\frac{1}{2}mv^2$	$\frac{1}{2}I\omega^2$
Momentum: $p = mv$	Angular momentum: $L = I\omega$

The Key Insight: Conservation of Angular Momentum

This is where rotational dynamics gets beautiful. Just as linear momentum is conserved when no external force acts, angular momentum is conserved when no external torque acts:

$L = I\omega = \text{constant}$ …

Start your 14-day free trial to unlock the full solution →

A spinning disc dropped onto a rough table experiences friction that simultaneously accelerates its centre of mass forward and decelerates its rotation until pure rolling begins. Rolling starts when $v_{cm} = \omega R$ , which takes time $t = \frac{\omega_0 R}{3\mu_k g}$ .

Understanding the Physics

When a disc spinning about a horizontal axis is placed on a table, it initially has angular velocity but no translational motion. The contact point tries to slide backward relative to the table, so kinetic friction acts forward on the disc. This friction does two things simultaneously: it accelerates the centre of mass forward (translation) and creates a torque that opposes the rotation (reducing angular velocity). Eventually, the two motions synchronize into pure rolling.

The beauty of this problem lies in recognizing that the same force—friction—produces opposite effects on translation and rotation, and these effects conspire to establish rolling motion.

Step-by-Step Solution

1. Initial velocity of the centre of mass

Before contact with the table, the disc is simply rotating about a horizontal axis through its centre. The centre of mass itself is not moving translationally; it's the axis of rotation.

The velocity of the centre of mass is $v_{cm} = 0$ .

2. Linear velocity at the rim when contact is made

Initially, a point on the rim has tangential velocity $v = \omega_0 R$ due to rotation. The moment the disc touches the table, the contact point has this velocity directed horizontally (tangent to the disc). Since the table is stationary, there is relative sliding, and kinetic friction $f_k = \mu_k N = \mu_k mg$ acts on the disc.

The linear velocity of the rim point in contact with the table is initially $\omega_0 R$ (backward relative to the disc's intended rolling direction), but friction immediately begins to reduce the angular velocity while simultaneously accelerating the centre of mass forward. The rim point's velocity relative to the ground changes continuously.

3. Effect on the centre of mass velocity

The friction force $f_k = \mu_k mg$ acts forward (in the direction opposite to the sliding). This produces a linear acceleration:

a_{cm} = \frac{f_k}{m} = \mu_k g

The centre of mass, initially at rest, begins to accelerate forward. Its velocity increases linearly with time:

v_{cm}(t) = \mu_k g \cdot t

4. The responsible force

The force responsible for both effects is kinetic friction between the disc and the table. It acts forward on the disc at the contact point, creating both a linear acceleration of the centre of mass and a torque about the centre of mass that opposes the initial rotation.

Important

Friction here is not dissipative in the usual sense—it converts rotational kinetic energy into translational kinetic energy until rolling begins. Only after rolling starts does the disc roll without slipping (if friction is sufficient).

5. Condition for rolling to begin

Pure rolling occurs when the velocity of the contact point relative to the ground is zero. This requires:

v_{cm} = \omega R

At the contact point, the velocity due to translation is $v_{cm}$ forward, and the velocity due to rotation is $\omega R$ backward. For no slipping:

v_{cm} = \omega R

6. Time taken for rolling to begin …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ24 — Two discs of moments of inertia I₁ and I₂ about their respective axes …→Next questionQ26 — Two cylindrical hollow drums of radii R and 2R, and of a common height…

Q.A disc of radius $R$ is rotating with an angular speed $\omega_o$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $\mu_k$ .

Concept understanding — Rotational Dynamics

Rotational Dynamics: The Physics of Spinning Things

The Core Idea

Breaking It Down

Torque — The Rotational "Push"

Moment of Inertia — The Rotational "Mass"

Angular Acceleration — How Fast Rotation Changes

The Complete Picture: Rotational Analogues

The Key Insight: Conservation of Angular Momentum

Concept: Rotational Dynamics with Friction

Understanding the Physics

Step-by-Step Solution

Concept: Friction Simultaneously Accelerates Translation and Decelerates Spin Until Rolling Begins

Q.A disc of radius RRR is rotating with an angular speed ωo\omega_oωo​ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is μk\mu_kμk​.

Understanding the Physics

Step-by-Step Solution

🎓Unlock everything free for 14 days

Q.A disc of radius $R$ is rotating with an angular speed $\omega_o$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $\mu_k$ .

Unlock everything free for 14 days