A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated u

NCERT Exemplar · Q2

Q. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

(a) its speed of rotation increases.

(b) its speed of rotation decreases.

(d) its speed increases because its moment of inertia increases.

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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}

That's why the figure skater spins faster when she pulls her arms in. Her moment of inertia $I$ decreases, so her angular velocity $\omega$ must increase to keep $L$ constant. No external torque — just redistribution of mass.

Watch out

A common mistake: thinking that angular momentum is always conserved. It's conserved only when the net external torque is zero. If you apply a torque (like friction on a spinning wheel), angular momentum changes.

Putting It All Together

When you encounter a rotational dynamics problem:

Identify the axis of rotation — everything depends on this
Find the net torque — sum all torques (with sign conventions)
Determine the moment of inertia — use the right formula for the shape and axis
Apply $\tau_{\text{net}} = I\alpha$ — this gives you angular acceleration
Use kinematic equations (if needed) — they're the same as linear ones, just with $\theta$ , $\omega$ , $\alpha$

The beauty of rotational dynamics is that once you understand the parallels, you already know most of the physics. The hard part is just the geometry — figuring out lever arms and mass distributions.

If you landed here looking for "Rotational Dynamics formula" or "Rotational Dynamics numericals class 11", it helps to know that Rotational Dynamics is a core, NCERT-aligned topic from the System of Particles and Rotational Motion portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET. Revisiting the NCERT Physics textbook exercises for this chapter alongside the walkthrough above is a solid way to convert this into exam-ready practice.

Heating the rod increases its length, which increases its moment of inertia. Since no external torque acts, angular momentum is conserved, so angular speed must decrease.

The Concept: Rotational Dynamics and Thermal Expansion

When a body rotates freely with no external torque, its angular momentum $L = I\omega$ stays constant. Here $I$ is the moment of inertia and $\omega$ the angular speed. If the body’s shape changes — as it does when heated — $I$ changes, and $\omega$ must adjust to keep $L$ unchanged.

The rod is uniform and rotates about its perpendicular bisector. Heating it uniformly causes linear expansion: every dimension increases slightly. For a rod, the length $L$ increases, and since the mass stays the same, the moment of inertia changes.

Step-by-Step Reasoning

Moment of inertia of a rod about its perpendicular bisector For a uniform rod of mass $M$ and length $\ell$ , rotating about an axis through its centre and perpendicular to its length, the moment of inertia is

I = \frac{1}{12} M \ell^2.

This is a standard result — the mass is distributed symmetrically, and the factor $\frac{1}{12}$ comes from integrating $r^2 \, dm$ .

Effect of heating on length When the temperature rises by $\Delta T$ , the rod expands linearly:

\ell' = \ell (1 + \alpha \Delta T),

where $\alpha$ is the coefficient of linear expansion. Since $\Delta T$ is small, $\alpha \Delta T \ll 1$ .

New moment of inertia The mass $M$ does not change. The new moment of inertia is

I' = \frac{1}{12} M (\ell')^2 = \frac{1}{12} M \ell^2 (1 + \alpha \Delta T)^2.

Expanding to first order (since $\alpha \Delta T$ is tiny):

I' \approx I (1 + 2\alpha \Delta T).

So $I' > I$ — the moment of inertia increases.

Conservation of angular momentum No external torque acts on the rod (it rotates freely, and heating does not apply a torque). Therefore

I \omega = I' \omega'.

Substituting $I' = I (1 + 2\alpha \Delta T)$ gives

\omega' = \frac{\omega}{1 + 2\alpha \Delta T} \approx \omega (1 - 2\alpha \Delta T).

Since $2\alpha \Delta T > 0$ , we have $\omega' < \omega$ — the angular speed decreases.

Watch out

A common mistake is to think that because the rod expands outward, its speed increases (like a spinning skater pulling arms in). But here the mass moves away from the axis, increasing $I$ , which slows the rotation — the opposite of the skater effect.

Checking the options
- (A) says speed increases — wrong.
- (B) says speed decreases — correct.
- (C) says speed remains same — wrong.
- (D) says speed increases because moment of inertia increases — the reason is backwards; increasing $I$ decreases speed.

Tip

You can remember this as: heating → expansion → larger $I$ → slower spin (for a free body). The skater’s trick works only when $I$ decreases.

✓Final answer

The correct option is (B) : its speed of rotation decreases.

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Q.A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

The Concept: Rotational Dynamics and Thermal Expansion

Step-by-Step Reasoning

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