MHT-CET Physics · Rotational Dynamics

Rotational Kinetic Energy and Rolling Motion

A spinning body stores ½Iω²; a rolling one stores that plus ½mv², tied together by v = Rω, so the factor 1 + k²/R² decides how fast each shape rolls, how far it climbs and how its energy splits.

All Rotational Dynamics notes MHT-CET Physics strategy

Why this matters

24 PYQs, two HARD. Three shapes: rotational kinetic energy and what a swinging or falling rod turns it into, the split of a rolling body's energy between translation and rotation, and the race down an incline — acceleration, speed at the bottom, distance up a ramp. One number, k²/R², carries all three.

Concept 1 of 3: Rotational Kinetic Energy, and Rods That Swing or Fall

½Iω² is the rotational twin of ½mv². A rod pivoted at one end trades that energy with the height of its centre of mass — which sits at its middle, so it rises or falls only half the rod's length.

Definition

$K_{\text{rot}} = \dfrac{1}{2}I\omega^2$ ; $\omega$ up 20% ⇒ $K$ up 44%.
Sphere and cylinder of equal M, R, the cylinder at twice the angular speed: $\dfrac{K_s}{K_c} = \dfrac{\frac{2}{5}}{\frac{1}{2} \times 4} = \dfrac{1}{5}$ .
Rod pivoted at an end with max angular speed $\omega$ : $\dfrac{1}{2}\cdot\dfrac{ML^2}{3}\omega^2 = Mgh \Rightarrow h = \dfrac{L^2\omega^2}{6g}$ .
Rod standing on its hinged end, falling flat: $mg\dfrac{L}{2} = \dfrac{1}{2}\cdot\dfrac{mL^2}{3}\omega^2 \Rightarrow \omega = \sqrt{\dfrac{3g}{L}}$ .

Rotational kinetic energy

K_{\text{rot}} = \tfrac{1}{2}I\omega^2

Worked example

A flywheel of I = 4 kg m² turns at 10 rad/s. Its kinetic energy? By what percentage does it rise if ω rises 10%?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q36Moderate

Example 1 · Rotational Dynamics · Rotational Kinetic Energy and Rolling Motion

A solid cylinder of mass M and radius R is rotating about its geometrical axis. A solid sphere of the same mass and same radius is also rotating about its diameter with an angular speed half that of the cylinder. The ratio of the kinetic energy of rotation of the sphere to that of the cylinder will be

Raising the centre of mass by the whole length

A rod's weight acts at its middle, so falling from upright lowers it by

\frac{L}{2}

, not

L

. Using

L

gives

\sqrt{6g/L}

, a printed distractor.

Drill 8 more on rotational kinetic energy, and rods that swing or fall

Concept 2 of 3: How a Rolling Body's Energy Splits

A body rolling without slipping moves and spins at once, with v = Rω. Its total energy is ½mv²(1 + k²/R²). A ring puts as much into spinning as into moving; a solid sphere puts only 2 parts in 7 into spinning.

Definition

$K = \dfrac{1}{2}mv^2\left(1 + \dfrac{k^2}{R^2}\right)$ ; $\dfrac{k^2}{R^2}$ : ring 1, disc/cylinder $\dfrac{1}{2}$ , solid sphere $\dfrac{2}{5}$ , shell $\dfrac{2}{3}$ .
Rotational share $= \dfrac{k^2/R^2}{1 + k^2/R^2}$ : ring $\dfrac{1}{2}$ , disc $\dfrac{1}{3}$ , sphere $\dfrac{2}{7}$ — so total : rotational = $\dfrac{7}{2}$ for a sphere.
Ring and disc of equal mass at the same speed: $K_{\text{disc}} = \dfrac{3}{4}K_{\text{ring}}$ .
A string unwinding from a wheel: the falling mass's lost energy feeds both its own motion and the wheel's spin.

Rolling energy

K = \tfrac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)

Worked example

A 3 kg disc rolls at 2 m/s. Its total kinetic energy, and how much of it is rotational?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q31Moderate

Example 2 · Rotational Dynamics · Rotational Kinetic Energy and Rolling Motion

A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 6 J, then total kinetic energy of the disc is

Forgetting the rotational part

A rolling body at speed v has MORE than

\frac{1}{2}mv^2

. Setting

mgh = \frac{1}{2}mv^2

for a rolling body is the sliding answer.

Drill 5 more on how a rolling body's energy splits

Concept 3 of 3: Rolling Down (and Up) an Incline

Rolling down a slope, some of the lost height goes into spin, so a rolling body is slower than a sliding one — and the more of its mass is at the rim, the slower. The solid sphere always wins the race, then the disc, then the ring.

Definition

$a = \dfrac{g\sin\theta}{1 + k^2/R^2}$ : solid sphere $\dfrac{5}{7}g\sin\theta$ , disc $\dfrac{2}{3}g\sin\theta$ , ring $\dfrac{1}{2}g\sin\theta$ . On 30°, a sphere: $\dfrac{5g}{14}$ .
$v = \sqrt{\dfrac{2gh}{1 + k^2/R^2}}$ : sphere $\sqrt{\dfrac{10gh}{7}}$ , disc $\sqrt{\dfrac{4gh}{3}}$ , ring $\sqrt{gh}$ .
Compared with sliding ( $V = \sqrt{2gh}$ ): disc $V\sqrt{\dfrac{2}{3}}$ , ring $\dfrac{V}{\sqrt{2}}$ .
Rolling UP a ramp from speed $v$ : $\dfrac{1}{2}mv^2\left(1 + \dfrac{k^2}{R^2}\right) = mgs\sin\theta$ .
Cylinder to sphere acceleration ratio: $\dfrac{2/3}{5/7} = \dfrac{14}{15}$ .

Rolling on an incline

a = \frac{g\sin\theta}{1 + k^2/R^2}, \qquad v = \sqrt{\frac{2gh}{1 + k^2/R^2}}

Worked example

A ring rolls from rest down a 30° incline of height 1.8 m. Its acceleration and speed at the bottom? (g = 10)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q33Moderate

Example 3 · Rotational Dynamics · Rotational Kinetic Energy and Rolling Motion

A solid cylinder and a solid sphere having same mass and same radius roll down on the same inclined plane. The ratio of the acceleration of the cylinder

a_c

to that of sphere

a_s

Using sin θ twice, or not at all

a

carries

\sin\theta

;

v

at the bottom depends only on the height

h

. '30°, solid sphere' is

\frac{5g}{14}

, and

\frac{5g}{7}

is what you get by forgetting the

\sin 30^\circ

Drill 8 more on rolling down (and up) an incline

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Rotational Kinetic Energy, and Rods That Swing or Fall
Rotational kinetic energy
$K_{\text{rot}} = \tfrac{1}{2}I\omega^2$
How a Rolling Body's Energy Splits
Rolling energy
$K = \tfrac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$
Rolling Down (and Up) an Incline
Rolling on an incline
$a = \frac{g\sin\theta}{1 + k^2/R^2}, \qquad v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$

Watch out for (3)

Raising the centre of mass by the whole length
→ Rotational Kinetic Energy, and Rods That Swing or Fall
Forgetting the rotational part
→ How a Rolling Body's Energy Splits
Using sin θ twice, or not at all
→ Rolling Down (and Up) an Incline

Test yourself on Rotational Dynamics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Rotational Dynamics chapter test Drill all 24 questions

Related notes

Moment of Inertia — the k²/R² values used here →