Centripetal Force and Circular Motion
Centripetal Force and Acceleration
Any object moving in a circular path requires a centripetal acceleration directed radially inward. This acceleration arises because the velocity vector continuously changes direction, even if the speed remains constant.
$$a_c = \frac{v^2}{r} = r\omega^2$$
The acceleration needed to maintain circular motion, directed toward the center of the circle.
$a_c$=Centripetal acceleration(m/s²)
$v$=Linear (tangential) speed(m/s)
$r$=Radius of circular path(m)
$\omega$=Angular velocity(rad/s)
Using $v = r\omega$
→$a_c = r\omega^2$ — at constant $\omega$, acceleration increases with radius
Two Forms: $v^2/r$ is useful when linear speed is known; $r\omega^2$ is useful when angular velocity is known. They give opposite proportionality with $r$.
Derivation Insight: The proof uses similar isosceles triangles — the velocity-change triangle mirrors the position triangle, giving $\Delta v / v = \Delta s / r$.
Centripetal force is not a new type of force — it is the net radial force (tension, friction, gravity, normal force, or a combination) that provides the required centripetal acceleration.
$$F_c = \frac{mv^2}{r} = mr\omega^2$$
The net inward force required to maintain circular motion.
$F_c$=Centripetal force(N)
$m$=Mass of the object(kg)
Horizontal circle on string
→$T = mv^2/r$ — tension alone provides centripetal force
Car on flat road
→Static friction provides $F_c$: $f_s = mv^2/r$
Not a Separate Force: Never draw centripetal force as a separate arrow in a free-body diagram. It is the resultant of real forces (tension, gravity, friction, normal).
Vertical Circle — Top: Both weight and tension point inward: $T + mg = mv^2/r$.
Vertical Circle — Bottom: Tension opposes weight: $T - mg = mv^2/r$, so $T = mv^2/r + mg$ (tension is maximum here).
Minimum Speed at Top: When $T = 0$, the critical condition gives $v_{min} = \sqrt{gr}$.
Moment of Inertia: Rotational Resistance
The moment of inertia ($I$) measures an object's resistance to Angular Acceleration, depending on both total mass and its distribution relative to the Axis of Rotation. It plays the same role in rotational motion that mass plays in linear motion.
$$I = \sum m_i r_i^2$$
The sum of all point masses multiplied by the square of their distance from the axis.
$I$=Moment of Inertia(kg·m²)
$m_i$=Mass of element $i$(kg)
$r_i$=Distance from axis(m)
Point Mass
→$I = mr^2$
All mass at same $r$
→$I = Mr^2$ (hoop/thin ring)
Mass Distribution: Mass further from the axis increases $I$ significantly due to the $r^2$ term.
Shape Factor: Objects like hoops have higher $I$ than solid disks of the same mass because all mass is at the maximum radius.
Axis Dependence: Unlike mass, $I$ changes when you change the axis of rotation. A rod has different $I$ about its center vs. its end.
Dimensional Check: $[I] = [M][L]^2 = $ kg·m². Any moment of inertia formula must have the form (number) × $mr^2$ or $mL^2$.
Standard Moment of Inertia values for common shapes about their central axis. These are derived from integration but must be memorized for quick problem-solving.
Thin Rod (center): $I = \frac{1}{12}mL^2$ — mass is distributed along the length.
Thin Ring / Hoop: $I = mr^2$ — all mass sits at distance $r$; this is the maximum possible $I$ for a given $m$ and $r$.
Solid Disk / Cylinder: $I = \frac{1}{2}mr^2$ — mass is spread from center to edge; average $r^2$ is half the maximum.
Solid Sphere: $I = \frac{2}{5}mr^2$ — mass concentrated more toward center than a disk.
Ranking Shortcut: For same $m$ and $r$: $I_{hoop} > I_{disk} > I_{sphere}$, so sphere wins rolling races.
Quick Coefficient Ranking (highest to lowest I)
1
Hoop: $k = 1$
2
Solid Disk/Cylinder: $k = 1/2$
3
Solid Sphere: $k = 2/5$
Here $I = kmr^2$. Lower $k$ means less rotational inertia and faster rolling.
Torque and Newton's Second Law for Rotation
Torque ($\tau$) is the rotational analogue of force, determined by the magnitude of force and the length of the Lever Arm. It is what causes Angular Acceleration.
$$\tau = r F \sin(\theta) = I\alpha$$
Relates the twisting force to the resulting angular acceleration and rotational inertia.
$\tau$=Net Torque(N·m)
$r$=Distance from pivot(m)
$F$=Applied Force(N)
$\theta$=Angle between $\vec{r}$ and $\vec{F}$
$\alpha$=Angular acceleration(rad/s²)
$\theta = 90°$
→Maximum torque: $\tau = rF$
$\theta = 0°$ or $180°$
→Zero torque — force acts along the lever arm
Force through pivot ($r = 0$)
→Zero torque regardless of force magnitude
Rotational Newton's Law: $\tau_{net} = I\alpha$ is the direct analogue of $F = ma$. For constant torque, higher $I$ means lower $\alpha$.
Directionality: Torque is a vector; counter-clockwise is traditionally positive.
Equilibrium Condition: For static balance, $\sum \tau_{CW} = \sum \tau_{CCW}$.
Rotational Kinetic Energy
A spinning body stores energy in its rotation. A rolling object possesses both translational and rotational kinetic energy, partitioning its total energy based on its shape.
$$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$
Total energy is the sum of linear motion of the center of mass and spinning motion around the axis.
$K_{total}$=Total Kinetic Energy(J)
$\frac{1}{2}mv^2$=Translational KE(J)
$\frac{1}{2}I\omega^2$=Rotational KE(J)
Rolling Without Slipping
→$v = r\omega$ links translational and rotational motion
Pure spinning (fixed axis)
→$K = \frac{1}{2}I\omega^2$ only — no translational term
Energy Partition: The fraction of total KE that is rotational equals $k/(1+k)$. Hoop: 1/2. Disk: 1/3. Sphere: 2/7.
The Rolling Race: All shapes released from height $h$ have $K_{total} = mgh$, but objects with lower $k$ convert more PE into translational KE, reaching higher speed: $v = \sqrt{2gh/(1+k)}$.
Rotational Collision (Clutch): When two discs engage, angular momentum is conserved but KE is lost — analogous to perfectly inelastic collisions in linear motion.
Angular Momentum and Conservation
Angular momentum ($L$) for a particle is defined as the cross product of position vector and linear momentum. For a rigid body rotating about a fixed axis, it simplifies to $L = I\omega$.
$$L = r \times p = mrv\sin\theta$$
Measures the 'quantity of rotation' of a particle or system.
$L$=Angular Momentum(kg·m²/s (or J·s))
$r$=Position vector from reference point(m)
$p$=Linear momentum ($mv$)(kg·m/s)
$\theta$=Angle between $\vec{r}$ and $\vec{p}$
Circular motion ($\theta = 90°$)
→$L = mvr = mr^2\omega$
Rigid body about fixed axis
→$L = I\omega = (\sum m_i r_i^2)\omega$
Vector Direction: $L$ is perpendicular to the plane of $\vec{r}$ and $\vec{p}$, found by the right-hand rule.
Spin vs. Orbital: A planet has orbital $L$ (around the Sun) and spin $L$ (around its own axis). Total angular momentum is the sum of both.
Dimensional Check: $[L] = [M][L]^2[T]^{-1}$ — same dimensions as energy × time (J·s).
The law of conservation of angular momentum states: if the Net Torque from external forces on a system is zero, the total angular momentum remains constant in both magnitude and direction.
$$I_i\omega_i = I_f\omega_f$$
When no external torque acts, any decrease in $I$ must be compensated by an increase in $\omega$, and vice versa.
$I_i, I_f$=Initial and final moments of inertia(kg·m²)
$\omega_i, \omega_f$=Initial and final angular velocities(rad/s)
$I$ decreases
→$\omega$ increases (skater pulls arms in)
$I$ increases
→$\omega$ decreases (diver opens up before water entry)
KE Changes: Although $L$ is conserved, $K_{rot} = L^2/(2I)$ changes when $I$ changes — the skater does internal work with muscles.
Axis Stability: The direction of $\vec{L}$ also remains fixed (gyroscopic effect). This is why the Earth's axis points toward Polaris year-round.
Diver's Somersault: The diver tucks to reduce $I$, spinning faster to complete extra rotations, then extends before entry to slow down.
Water Drip Puzzle: Water dripping into a spinning beaker increases $I$ (added mass at some $r$), so $\omega$ must decrease to conserve $L$.