Is it correct to say: Drawing 'centripetal force' as a separate arrow alongside tension, friction and weight?

Centripetal force is the net radial component of the real forces. Draw only the real forces, then set their net component towards the centre equal to mv²/r.

Is it correct to say: Substituting rpm directly as ω?

ω must be in rad/s: ω = 2π × (rpm/60). In I₁ω₁ = I₂ω₂ you may keep rpm, but only if both sides use the same unit.

Is it correct to say: Taking √(rg) as the minimum speed at the lowest point of a vertical circle?

√(rg) is the minimum speed at the top. At the bottom the body needs √(5rg) to complete the circle.

Is it correct to say: Using the parallel-axes theorem between two axes, neither of which passes through the centre of mass?

I_O = I_C + Mh² needs one axis through the centre of mass. To go between two other axes, go through I_C first.

Is it correct to say: Applying the perpendicular-axes theorem to a sphere or a solid cylinder?

I_Z = I_X + I_Y holds only for a plane lamina, with X and Y in its plane.

Is it correct to say: Writing mgh = ½mv² for a rolling body?

Include rotational KE: mgh = ½mv²(1 + K²/R²). The rolling body is slower than a body sliding down a frictionless incline.

📄 Formula sheet (PDF)

Rotational Dynamics

Circular motion, banking of roads, conical pendulum, vertical circular motion, moment of inertia, angular momentum and rolling motion — Maharashtra HSC Physics Ch 1

📐 20 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Start every banking-of-road or conical-pendulum answer with a free-body diagram (weight mg, normal reaction or string tension, friction if present). The derivation follows directly from resolving these forces.
→Centripetal force is not an extra force. On the diagram show only real forces; their net component towards the centre equals mv²/r.
→In vertical circular motion, combine the force equation at a point (top or bottom) with conservation of energy between the top and the bottom. This gives √(rg), √(3rg) and √(5rg) quickly.
→Moment of inertia always refers to an axis. Write the axis next to every I you use, e.g. 'disc, about its own axis' or 'rod, about a perpendicular axis through one end'.
→Convert rpm to rad/s before substituting: ω = 2πn, with n in revolutions per second (rpm ÷ 60).
→For rolling bodies, the total kinetic energy is translational plus rotational. Using mgh = ½mv² alone is a common way to lose the numerical.

📐 Formulas(20)

Linear and Angular Speed

v = r ω, ω = \frac{2 π}{T} = 2 π n

Symbol	Meaning
$v$	Linear speed (m s⁻¹)
$ω$	Angular speed (rad s⁻¹)
$r$	Radius of the circular path (m)
$T$	Period of revolution (s)
$n$	Frequency of revolution (Hz = rev s⁻¹)

Centripetal (Radial) Acceleration

a_{r} = \frac{v ^{2}}{r} = ω^{2} r

Symbol	Meaning
$a_{r}$	Centripetal acceleration (m s⁻²)
$v$	Linear speed (m s⁻¹)
$ω$	Angular speed (rad s⁻¹)
$r$	Radius (m)

Tangential and Resultant Acceleration (Non-uniform Circular Motion)

a_{t} = r α, a = a_{r}^{2} + a_{t}^{2}

Symbol	Meaning
$a_{t}$	Tangential acceleration (m s⁻²)
$α$	Angular acceleration (rad s⁻²)
$a_{r}$	Radial (centripetal) acceleration (m s⁻²)

Centripetal Force

F = \frac{m v ^{2}}{r} = m ω^{2} r

Symbol	Meaning
$F$	Centripetal force (N)
$m$	Mass of the body (kg)
$v$	Linear speed (m s⁻¹)
$r$	Radius (m)

Maximum Speed on a Level Curved Road

v_{ma x} = μ_{s} r g

Symbol	Meaning
$μ_{s}$	Coefficient of static friction (no unit)
$r$	Radius of the curve (m)
$g$	Acceleration due to gravity (m s⁻²)

Angle of Banking (Most Safe Speed, No Friction)★ Board fav

tan θ = \frac{v ^{2}}{r g}

Symbol	Meaning
$θ$	Angle of banking (with the horizontal)
$v$	Most safe speed (m s⁻¹)
$r$	Radius of the curve (m)

Maximum Safe Speed on a Banked Road with Friction

v_{ma x} = r g \frac{μ _{s} + t a n θ}{1 - μ _{s} t a n θ}

Symbol	Meaning
$μ_{s}$	Coefficient of static friction
$θ$	Angle of banking

Conical Pendulum — Period★ Board fav

T = 2 π \frac{L c o s θ}{g}, tan θ = \frac{v ^{2}}{r g}

Symbol	Meaning
$T$	Period of revolution (s)
$L$	Length of the string (m)
$θ$	Angle made by the string with the vertical
$r$	Radius of the horizontal circle = L sin θ (m)

Vertical Circular Motion — Critical (Minimum) Speeds★ Board fav

v_{t o p} = r g, v_{mi d} = 3 r g, v_{b o tt o m} = 5 r g

Symbol	Meaning
$r$	Radius of the vertical circle (m)
$g$	Acceleration due to gravity (m s⁻²)

Tension Difference in a Vertical Circle

T_{b o tt o m} - T_{t o p} = 6 m g

Symbol	Meaning
$T_{b o tt o m}$	Tension at the lowest point (N)
$T_{t o p}$	Tension at the highest point (N)
$m$	Mass of the body (kg)

Moment of Inertia and Radius of Gyration

I = \sum_{i} m_{i} r_{i}^{2} = M K^{2}, K = \frac{I}{M}

Symbol	Meaning
$I$	Moment of inertia about the given axis (kg m²)
$m_{i}$	Mass of the i-th particle (kg)
$r_{i}$	Perpendicular distance of the i-th particle from the axis (m)
$M$	Total mass of the body (kg)
$K$	Radius of gyration (m)

Standard Moments of Inertia

I_{r in g} = M R^{2}, I_{d i sc} = \frac{1}{2} M R^{2}, I_{r o d} = \frac{1}{12} M L^{2}, I_{s p h er e} = \frac{2}{5} M R^{2}

Symbol	Meaning
$M$	Mass of the body (kg)
$R$	Radius of ring, disc or sphere (m)
$L$	Length of the rod (m)

Theorem of Parallel Axes★ Board fav

I_{O} = I_{C} + M h^{2}

Symbol	Meaning
$I_{O}$	MI about the given axis through O (kg m²)
$I_{C}$	MI about the parallel axis through the centre of mass (kg m²)
$h$	Distance between the two parallel axes (m)

Theorem of Perpendicular Axes

I_{Z} = I_{X} + I_{Y}

Symbol	Meaning
$I_{Z}$	MI about the axis perpendicular to the plane (kg m²)
$I_{X}$	MI about the X axis in the plane (kg m²)
$I_{Y}$	MI about the Y axis in the plane (kg m²)

Kinetic Energy of Rotation

E_{r o t} = \frac{1}{2} I ω^{2}

Symbol	Meaning
$E_{r o t}$	Rotational kinetic energy (J)
$I$	Moment of inertia (kg m²)
$ω$	Angular speed (rad s⁻¹)

Angular Momentum

L = r \times p, L = I ω

Symbol	Meaning
$L$	Angular momentum (kg m² s⁻¹)
$r$	Position vector from the axis (m)
$p$	Linear momentum (kg m s⁻¹)

Torque and Angular Acceleration

τ = I α, τ = \frac{d L}{d t}

Symbol	Meaning
$τ$	Torque (N m)
$I$	Moment of inertia (kg m²)
$α$	Angular acceleration (rad s⁻²)

Law of Conservation of Angular Momentum★ Board fav

I_{1} ω_{1} = I_{2} ω_{2} (when τ_{e x t} = 0)

Symbol	Meaning
$I_{1}, I_{2}$	Initial and final moment of inertia (kg m²)
$ω_{1}, ω_{2}$	Initial and final angular speed (rad s⁻¹ or rpm, same unit on both sides)

Kinetic Energy of a Rolling Body★ Board fav

E = \frac{1}{2} m v^{2} (1 + \frac{K ^{2}}{R ^{2}})

Symbol	Meaning
$m$	Mass (kg)
$v$	Speed of the centre of mass (m s⁻¹)
$K$	Radius of gyration about the axis through the centre of mass (m)
$R$	Radius of the rolling body (m)

Rolling Down an Inclined Plane★ Board fav

a = \frac{g s i n θ}{1 + K ^{2} / R ^{2}}, v = \frac{2 g h}{1 + K ^{2} / R ^{2}}

Symbol	Meaning
$a$	Linear acceleration along the incline (m s⁻²)
$θ$	Angle of inclination
$h$	Vertical height descended (m)

✏️ Solved Examples

1Solved Exampleeasy3 steps

A car takes a turn on a level (unbanked) circular road of radius 50 m. The coefficient of static friction between the tyres and the road is 0.4. Find the maximum speed at which the car can take the turn without skidding. (g = 9.8 m/s²)

On a level road, static friction supplies the centripetal force, so v_max = √(μ_s r g).

v_{ma x} = μ_{s} r g

2Solved Exampleboard3 steps

A curved road of radius 100 m is to be banked so that a vehicle moving at 54 km/h needs no friction to take the turn. Find the angle of banking. (g = 9.8 m/s²)

Convert the speed to SI units: 54 km/h = 54 × 1000/3600 = 15 m/s.

v = 15 m s^{- 1}

3Solved Exampleboard4 steps

A stone of mass 0.5 kg tied to a string 1 m long is whirled in a vertical circle. Find (a) the minimum speed at the highest point, (b) the minimum speed at the lowest point needed to complete the circle, and (c) the tension in the string at the lowest point in that case. (g = 9.8 m/s²)

At the top, the minimum speed corresponds to zero tension.

v_{t o p} = r g = 1 \times 9.8 \approx 3.13 m s^{- 1}

4Solved ExampleHOTS4 steps

A solid sphere starts from rest and rolls without slipping down an inclined plane, descending a vertical height of 1.4 m. Find (a) its speed at the bottom and (b) the fraction of its kinetic energy that is rotational. (g = 9.8 m/s²)

For a solid sphere about a diameter, I = ⅖MR², so K²/R² = 2/5 = 0.4.

\frac{K ^{2}}{R ^{2}} = \frac{2}{5}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Drawing 'centripetal force' as a separate arrow alongside tension, friction and weight
✓Centripetal force is the net radial component of the real forces. Draw only the real forces, then set their net component towards the centre equal to mv²/r.
2
Substituting rpm directly as ω
✓ω must be in rad/s: ω = 2π × (rpm/60). In I₁ω₁ = I₂ω₂ you may keep rpm, but only if both sides use the same unit.
3
Taking √(rg) as the minimum speed at the lowest point of a vertical circle
✓√(rg) is the minimum speed at the top. At the bottom the body needs √(5rg) to complete the circle.
4
Using the parallel-axes theorem between two axes, neither of which passes through the centre of mass
✓I_O = I_C + Mh² needs one axis through the centre of mass. To go between two other axes, go through I_C first.
5
Applying the perpendicular-axes theorem to a sphere or a solid cylinder
✓I_Z = I_X + I_Y holds only for a plane lamina, with X and Y in its plane.
6
Writing mgh = ½mv² for a rolling body
✓Include rotational KE: mgh = ½mv²(1 + K²/R²). The rolling body is slower than a body sliding down a frictionless incline.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the moment of inertia of a uniform disc of mass 2 kg and radius 0.1 m about a tangent perpendicular to its plane.
Q2
A conical pendulum has a string of length 1 m that makes an angle of 30° with the vertical. Find its period. (g = 9.8 m/s²)
Q3
A person stands on a rotating platform with arms stretched out. The moment of inertia is 6 kg m² and the platform turns at 30 rpm. On folding the arms, the moment of inertia becomes 2 kg m². Find the new rate of rotation (ignore friction).
Q4
A constant torque of 20 N m acts on a wheel of moment of inertia 4 kg m², initially at rest. Find the angular acceleration and the angular speed after 4 s.
Q5
Find the radius of gyration of a solid sphere of radius 5 cm about a diameter.
Q6
A curve of radius 100 m is banked so that tan θ = 0.2. If μ_s = 0.3, find the maximum safe speed on the curve. (g = 9.8 m/s²)

📝 Notes

Rotational Dynamics — Maharashtra HSC Overview

Chapter 1 of the Maharashtra Board (Balbharati) Std XII Physics textbook has two halves. The first covers circular motion and its uses: banked roads, the conical pendulum and vertical circles. The second covers the rigid body: moment of inertia, the two axis theorems, angular momentum and rolling.

Circular motion: always start from forces

Every circular-motion problem has the same skeleton. List the real forces, resolve them along the radius, and set the net radial force equal to $mv^2/r$ . On a level road friction does this job, which gives $v_{max} = \sqrt{\mu_s rg}$ . On a banked road without friction, a component of the normal reaction does it, which gives $\tan\theta = v^2/rg$ . In a conical pendulum the horizontal component of the string tension does it, which gives $T = 2\pi\sqrt{L\cos\theta/g}$ . None of these results depends on the mass, which makes a good check.

Vertical circles: force at one point, energy between points

Use $mg = mv^2/r$ at the top for the critical speed. Then use energy conservation over a fall of $2r$ to reach the bottom. This gives $\sqrt{rg}$ at the top, $\sqrt{3rg}$ at the midway point and $\sqrt{5rg}$ at the bottom, and a tension difference of $6mg$ .

Moment of inertia: the axis decides everything

The textbook derives the MI of a uniform ring ( $MR^2$ ) and a uniform disc ( $\tfrac{1}{2}MR^2$ ) about their own axes. With the parallel-axes theorem ( $I_O = I_C + Mh^2$ ) and the perpendicular-axes theorem ( $I_Z = I_X + I_Y$ , laminae only), you can get most other axes. Keep the analogy table in mind: mass ↔ moment of inertia, force ↔ torque, $p = mv$ ↔ $L = I\omega$ , $F = ma$ ↔ $\tau = I\alpha$ .

Rolling

A rolling body has kinetic energy $\tfrac{1}{2}mv^2(1 + K^2/R^2)$ . The factor $K^2/R^2$ alone decides how fast it rolls down an incline: sphere first, then disc, then ring.