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Rotational Dynamics
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 $L$.
Application Strategy
Worked Examples
Problem Statement
A 1000 kg car travels at 10 m/s around a circular curve of radius 10 m. What force must the pavement exert on the tires?
$F_c = mv^2/r$
Strategic Toolkit
$$F_c = \frac{mv^2}{r}$$
Step 1
Identify the Centripetal Force Provider
On a flat road, static friction between tires and pavement provides the centripetal force.
Step 2
Substitute Values
Apply the centripetal force formula directly.
$F_c = \frac{1000 \times (10)^2}{10} = \frac{100{,}000}{10} = 10{,}000$ N
Step 3
Interpret Result
The pavement must exert $1.0 \times 10^4$ N of friction inward. If the road is icy (low friction), the car cannot maintain this curve at this speed.
Problem Statement
A ball on a string swings in a vertical circle of radius r. Find the tension at the top of the circle when the ball has speed v.
Centripetal force as net radial forceVertical circle analysis
Strategic Toolkit
$$T + mg = \frac{mv^2}{r}$$
Step 1
Draw Free-Body Diagram at the Top
At the top, both tension T and weight mg point downward (toward center).
Step 2
Apply Newton's 2nd Law Radially
The net inward force equals the centripetal force.
$T + mg = \frac{mv^2}{r}$
Step 3
Solve for Tension
Rearrange to isolate T.
$T = m\left(\frac{v^2}{r} - g\right)$
Step 4
Find Minimum Speed
When T = 0, the string goes slack. This gives the minimum speed to maintain the circular path.
$v_{min} = \sqrt{gr}$
Problem Statement
Find the moment of inertia of a system consisting of two 2 kg point masses connected by a light 1.0 m rod, rotating about the center of the rod.
$I = \sum mr^2$
Strategic Toolkit
$$I = m_1r_1^2 + m_2r_2^2$$
Step 1
Determine Radius
The axis is at the center of the 1.0 m rod, so each mass is halfway along the length.
$r = \frac{1.0}{2} = 0.5$ m
Step 2
Apply Summation Formula
Calculate I for both identical masses.
$I = 2(0.5)^2 + 2(0.5)^2 = 0.5 + 0.5 = 1.0$ kg·m²
Problem Statement
A constant torque of 10 N·m is applied to a wheel with a moment of inertia of 5 kg·m². How long does it take to reach 20 rad/s from rest?
$\tau = I\alpha$Angular kinematics
Strategic Toolkit
$$\tau = I\alpha, \quad \omega_f = \omega_i + \alpha t$$
Step 1
Find Angular Acceleration
Use the rotational Newton's 2nd Law.
$\alpha = \frac{\tau}{I} = \frac{10}{5} = 2$ rad/s²
Step 2
Solve for Time
Apply kinematic equation with $\omega_i = 0$.
$20 = 0 + 2t \implies t = 10$ s
Problem Statement
A skater spins at 2 rad/s with arms outstretched (I = 6 kg·m²). She pulls her arms in, reducing I to 2 kg·m². Find her new angular velocity and the change in kinetic energy.
Conservation of angular momentumRotational KE
Strategic Toolkit
$$I_i\omega_i = I_f\omega_f, \quad K = \frac{1}{2}I\omega^2$$
Step 1
Apply Conservation
No external torque acts, so L is conserved.
$6 \times 2 = 2 \times \omega_f \implies \omega_f = 6$ rad/s
Step 2
Calculate Initial KE
$K_i = \frac{1}{2}(6)(2)^2 = 12$ J
Step 3
Calculate Final KE
$K_f = \frac{1}{2}(2)(6)^2 = 36$ J
Step 4
Interpret
KE tripled from 12 J to 36 J. The extra 24 J came from the skater's muscles doing work to pull her arms inward against centripetal acceleration.
Problem Statement
A solid sphere rolls without slipping down an incline of height 5 m. Find its speed at the bottom.
Energy conservation with rolling$v = r\omega$
Strategic Toolkit
$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$
Step 1
Write Energy Conservation
PE at top converts to translational + rotational KE.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$
Step 2
Simplify
The m and r cancel out.
$gh = \frac{1}{2}v^2 + \frac{1}{5}v^2 = \frac{7}{10}v^2$
Step 3
Solve for v
$v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10 \times 9.8 \times 5}{7}} = \sqrt{70} \approx 8.37$ m/s
Memorize formulas
Centripetal Acceleration
Center-directed acceleration for circular motion.
Formula
$a_c = \frac{v^2}{r} = r\omega^2$
Centripetal Force
Net inward force maintaining circular motion.
Formula
$F_c = \frac{mv^2}{r} = mr\omega^2$
Moment of Inertia
Rotational resistance; depends on mass distribution.
Formula
$I = \sum m_i r_i^2$
Newton's 2nd Law (Rotational)
Net torque causes angular acceleration.
Formula
$\tau_{net} = I \alpha$
Rotational Kinetic Energy
Energy stored in the spin of a rigid body.
Formula
$K_{rot} = \frac{1}{2} I \omega^2$
Angular Momentum
Conserved when net external torque is zero.
Formula
$L = I \omega$
Conservation of Angular Momentum
Internal redistribution of I changes ω inversely.
Formula
$I_i \omega_i = I_f \omega_f$
Review
1
A solid cylinder and a hollow cylinder of the same mass and radius roll down an incline from the same height. Which one has more total kinetic energy at the bottom?
Ans
Both have the same total kinetic energy
Review
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Use energy conservation for both cylinders
Each cylinder starts from rest at the same height, so the entire gravitational potential energy becomes total kinetic energy at the bottom. The shape changes how that kinetic energy is split between translation and rotation, not the total amount.
$$mgh = K_{total}$$
2
A 0.8 m long beam is pivoted at its left end. A 50 N force acts downward at 0.3 m from the pivot, and a 20 N force acts upward at the far end. What is the net torque about the pivot?
Ans
1 N·m counter-clockwise
Review
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Compute each torque with its direction
Use $\tau = rF$ because each force is perpendicular to the beam. The downward force on the right side tends to rotate the beam clockwise, while the upward force at the end tends to rotate it counter-clockwise.
$$\tau_1 = 50(0.3) = 15\ \text{N·m clockwise},\quad \tau_2 = 20(0.8) = 16\ \text{N·m counter-clockwise}$$
Subtract the opposite torques
The counter-clockwise torque is slightly larger, so the net torque points counter-clockwise.
$$\tau_{net} = 16 - 15 = 1\ \text{N·m counter-clockwise}$$
3
An ice skater spinning with arms outstretched ($I = 5$ kg·m², $\omega = 2$ rad/s) pulls her arms in ($I = 2$ kg·m²). What happens to her angular momentum and kinetic energy?
Ans
L stays the same, KE increases
Review
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Conserve angular momentum
With no external torque, angular momentum stays constant as the skater changes body position. That lets us solve for the new angular speed after the moment of inertia decreases.
$$L = I\omega = 5(2) = 10\ \text{kg·m}^2/\text{s},\qquad \omega_f = \frac{10}{2} = 5\ \text{rad/s}$$
Compare rotational kinetic energy before and after
Rotational kinetic energy increases because the skater does internal work while pulling the arms inward. The angular momentum stays fixed, but the smaller moment of inertia gives a larger speed and therefore a larger KE.
$$K_i = \tfrac12(5)(2^2) = 10\ \text{J},\qquad K_f = \tfrac12(2)(5^2) = 25\ \text{J}$$
4
If the net torque on a system is zero, which of the following must be true?
Ans
The angular acceleration is zero
Review
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Apply the rotational form of Newton's second law
The net torque and angular acceleration are linked by $\tau_{net} = I\alpha$. If the net torque is zero and the moment of inertia is not changing in this statement, the angular acceleration must also be zero.
$$\tau_{net} = 0 \Rightarrow I\alpha = 0 \Rightarrow \alpha = 0$$
5
A 0.4 kg ball on a string moves in a vertical circle of radius 1.5 m. At the top of the circle, its speed is 5 m/s. What is the tension in the string at that point?
Ans
2.75 N
Review
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Write the centripetal-force balance at the top
At the top of the circle, the center is downward, so both tension and weight point inward. Their sum must equal the required centripetal force.
$$T + mg = \frac{mv^2}{r}$$
Solve for the tension
Substitute the given values and isolate $T$.
$$T = \frac{0.4(5^2)}{1.5} - 0.4(9.8) = 2.75\ \text{N}$$
6
A car travels at 20 m/s around a flat circular curve. If the speed is doubled to 40 m/s, by what factor does the required centripetal force change?
Ans
4 times
Review
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Use the speed-squared dependence
Centripetal force scales with the square of speed. When the speed doubles, the force does not double linearly; it increases by the square of the factor.
$$F_c \propto v^2 \Rightarrow \frac{F_{new}}{F_{old}} = \frac{(2v)^2}{v^2} = 4$$
7
A person stands at the center of a freely rotating platform ($I_{platform} = 200$ kg·m², $\omega = 3$ rad/s). The person ($m = 50$ kg) then walks to the rim, 2 m from the center. What is the new angular velocity of the system?
Ans
1.5 rad/s
Review
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Conserve angular momentum
There is no external torque, so the total angular momentum remains constant as the person moves outward. The platform slows only because the total moment of inertia becomes larger.
$$L_i = I_i\omega_i = 200(3) = 600\ \text{kg·m}^2/\text{s}$$
Add the person's moment of inertia at the rim
At the rim, the person contributes $mr^2$ to the system's moment of inertia. Use the new total inertia to solve for the final angular speed.
$$I_f = 200 + 50(2^2) = 400\ \text{kg·m}^2,\qquad \omega_f = \frac{600}{400} = 1.5\ \text{rad/s}$$
8
A solid sphere rolls without slipping down a 3 m high incline. What is its linear speed at the bottom?
Ans
6.48 m/s
Review
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Use rolling energy conservation
For rolling without slipping, the loss in gravitational potential energy becomes both translational and rotational kinetic energy. For a solid sphere, $I = \frac{2}{5}mr^2$, which gives the standard rolling factor.
$$mgh = \frac12 mv^2 + \frac12 I\omega^2 = \frac12 mv^2\left(1 + \frac{2}{5}\right)$$
9
A uniform rod of mass $m$ and length $L$ has a moment of inertia of $\frac{1}{12}mL^2$ about its center. A torque of 6 N·m produces an angular acceleration of 3 rad/s² about the center. What is $mL^2$?
Ans
24 kg·m²
Review
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Find the moment of inertia from torque and angular acceleration
Use $\tau = I\alpha$ to solve for $I$ directly. Then substitute that value into the rod formula to recover $mL^2$.
$$I = \frac{\tau}{\alpha} = \frac{6}{3} = 2\ \text{kg·m}^2$$
Convert from $I$ to $mL^2$
The given rod relation is $I = \frac{1}{12}mL^2$, so multiply both sides by 12.
$$mL^2 = 12I = 12(2) = 24\ \text{kg·m}^2$$
10
A disc rotates at 6 rad/s. A piece of putty (moment of inertia contribution = $I/2$) is dropped onto it and sticks. What fraction of the original kinetic energy is lost?
Ans
1/3
Review
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Conserve angular momentum during sticking
When the putty lands and sticks, external torque is negligible, so angular momentum stays constant. The total moment of inertia increases from $I$ to $I + I/2$.
$$I(6) = \left(I + \frac{I}{2}\right)\omega_f \Rightarrow \omega_f = 4\ \text{rad/s}$$
Compare initial and final rotational kinetic energy
The kinetic energy drops because the collision is inelastic. Compute the fraction lost from the difference between the initial and final energies.
$$K_i = \frac12 I(6^2) = 18I,\qquad K_f = \frac12 \left(\frac{3I}{2}\right)(4^2) = 12I,\qquad \frac{K_i-K_f}{K_i} = \frac{1}{3}$$
11
Two particles of equal mass orbit the same point. Particle A orbits at radius $r$ and particle B at radius $2r$, both at the same angular velocity $\omega$. What is the ratio of their centripetal accelerations $a_A / a_B$?
Ans
1/2
Review
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Use the angular-speed form of centripetal acceleration
Since both particles share the same angular velocity, the relevant relation is $a_c = r\omega^2$. That makes the acceleration directly proportional to radius.
$$\frac{a_A}{a_B} = \frac{r\omega^2}{(2r)\omega^2} = \frac{1}{2}$$
12
A solid disk and a hoop of the same mass and radius both have the same angular momentum. Which has more rotational kinetic energy?
Ans
The disk
Review
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Compare rotational energy at fixed angular momentum
For the same angular momentum, rotational kinetic energy is larger when the moment of inertia is smaller. The disk has the smaller inertia, so it must have the larger kinetic energy.
$$K_{rot} = \frac{L^2}{2I},\qquad I_{disk} = \frac12 mr^2 < I_{hoop} = mr^2$$
Question 1 of 12
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1
A solid cylinder and a hollow cylinder of the same mass and radius roll down an incline from the same height. Which one has more total kinetic energy at the bottom?