A uniform solid sphere of mass m and radius R rests on a rough horizontal floor; its centre is therefore at he

NCERT Exemplar · Q18

m

(a)

h = R/2

;

(b)

h = R

;

(c)

h = 3R/2

;

(d)

h = 7R/5

.
Column II:

(i) Sphere rolls without slipping at constant velocity, with no loss of energy.

(ii) Sphere spins in the forward (clockwise) sense and loses energy by friction.

(iii) Sphere spins in the backward (anti-clockwise) sense and loses energy by friction.

(iv) Sphere has only translational motion (no spin) and loses energy by friction.

Punjab PsebShort· 3mImportance★★★★★est

82% · 47/57 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 →

The blow $J$ gives translation $v=J/m$ and, because it acts a height $(h-R)$ above the centre, a spin $\omega = J(h-R)/I$ with $I=\tfrac25 mR^2$ . Requiring $v=\omega R$ (pure rolling, no friction needed) gives the special height $h=7R/5$ . Other heights leave a mismatch between $v$ and $\omega R$ , so friction acts and energy is lost; the spin is backward for $h<R$ , zero for $h=R$ , forward for $h>R$ .

Concept

Impulse $J$ at height $h$ from the floor is at height $(h-R)$ relative to the centre. It produces:

linear speed of the centre: $v = J/m$ ;
angular speed about the centre: $\omega = \dfrac{J(h-R)}{I}$ , with $I = \tfrac{2}{5}mR^2$ .

The rolling height

Pure rolling with no friction requires the contact point to be instantaneously at rest: $v = \omega R$ .

\frac{J}{m} = \frac{J(h-R)}{\tfrac25 mR^2}\,R \ \Rightarrow\ 1 = \frac{h-R}{\tfrac25 R} \ \Rightarrow\ h - R = \frac{2R}{5} \ \Rightarrow\ h = \frac{7R}{5}.

So (d) $h=7R/5$ gives rolling without slipping at constant velocity, no energy loss → (i).

The other cases

The sign of $(h-R)$ fixes the spin sense (taking the blow toward the right, forward rolling is clockwise): …

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

←Previous questionQ17 — A uniform cube of mass m and side a rests on a frictionless horizontal…→Next questionQ19 — The vector sum of a system of non-collinear forces acting on a rigid b…

Concept

The rolling height

The other cases

🎓Unlock everything free for 14 days

Unlock everything free for 14 days