Lesson 1: Angular Variables in Rotation

1. What is the difference between angular velocity and tangential velocity, and how do I convert between them?

Think of a record player. Every point on the record spins at the same Angular Velocity (ω), measured in radians per second—they all complete one circle in the same amount of time. However, a bug on the outer edge has to cover a huge distance compared to a bug near the center. That speed along the path is the Tangential Velocity (v).

• The Link: v = r · ω
• Note: You must be in radians for this equation to work!

2. Why do we have to use radians instead of degrees for all these rotational formulas?

Degrees are arbitrary (360 is just a number Babylonians liked). Radians are physical. One radian is defined as the angle where the arc length equals the radius. Because of this natural geometry, formulas like s = r · θ and the derivatives of sine and cosine only work cleanly if you use radians. If you use degrees, you'd have to carry a clumsy π/180 conversion factor in every single calculus step.

3. Does the direction of rotation (clockwise vs. counter-clockwise) actually change the sign of the answer?

Yes. By standard convention:

• Counter-Clockwise (CCW): Positive (+)
• Clockwise (CW): Negative (-) If you mix these up, your vector addition will fail. Always set your coordinate system at the start of the problem.

4. If an object has zero angular acceleration, does that mean it has zero angular velocity?

No. Just like a car can cruise at a steady 60 mph (zero acceleration, non-zero velocity), a spinning disk can rotate at a constant speed. Zero angular acceleration (α = 0) just means the rotation isn't getting faster or slower; it’s steady.

5. Is angular displacement actually a vector? Why doesn't it follow vector addition rules when rotations are large?

This is a deep nuance. Infinitesimal (tiny) rotations are vectors. However, large rotations do not commute. If you rotate a book 90° around the x-axis then 90° around the y-axis, it ends up in a different position than if you did y then x. Since vector addition must be commutative (A+B = B+A), large angular displacements fail the "Vector Test."

Lesson 2: Constant Angular Acceleration: Derivation of Equations

6. How do I derive the three rotational kinematic equations using integrals if I'm given constant alpha?

1. You start with the definition: α = dω/dt.
2. Rearrange: dω = α dt.
3. Integrate both sides: ∫ dω = ∫ α dt. Since α is constant, pull it out.
4. Result: ω - ω₀ = αt → ω = ω₀ + αt (Equation 1).
5. Next, use ω = dθ/dt. Replace ω with (ω₀ + αt) and integrate again to get the displacement equation.

7. How do I solve "Rolling Without Slipping" problems—specifically, what is the link between v and omega?

"Without slipping" is a constraint. It means the bottom of the wheel has zero velocity relative to the ground. This forces a strict relationship between how fast the center moves (v_cm) and how fast it spins (ω):

• The Golden Rule: v_cm = R · ω and a_cm = R · α.
• Warning: If the object is slipping (like a bowling ball sliding down a lane), this link breaks, and you must treat translation and rotation separately.

Lesson 3: Moment of Inertia and The Parallel Axis Theorem

8. In simple terms, what is Moment of Inertia, and is it just "mass for rotation"?

Yes, it is the "rotational laziness" of an object. In linear physics, Mass (m) tells you how hard it is to push something. In rotation, Moment of Inertia (I) tells you how hard it is to spin something. The catch? It depends on where the mass is. Mass further from the axis counts much more than mass near the axis.

9. Why is the Moment of Inertia higher for a hollow hoop than a solid cylinder of the same mass and radius?

In a solid cylinder, some mass is near the center (easy to spin) and some is at the edge. In a hollow hoop, 100% of the mass is located at the maximum distance (R) from the axis. Since I ~ mr², having all the mass at the furthest point maximizes the inertia.

10. In the Parallel Axis Theorem (I = I_cm + mh²), what exactly does the 'h' distance represent?

h is the perpendicular distance between the two parallel axes: the axis going through the Center of Mass (CM) and the new axis you are rotating around.

Example: If you spin a rod around its end, h is the distance from the center to the end (L/2).

11. Why does the Parallel Axis Theorem only work if one of the axes goes through the Center of Mass?

The derivation involves an integral where a "cross term" appears (relating to the position of the center of mass). This term only sums to zero if the origin is defined at the Center of Mass. If you try to shift from one arbitrary axis to another arbitrary axis without passing through the CM, the math fails.

Lesson 4: Torque and Direction

12. What actually is torque? Is it a force, energy, or something else entirely?

Torque is not energy (even though N·m looks like Joules!). It is a twist. It measures the "effectiveness" of a force in causing rotation. It combines the strength of the push with the leverage you have.

13. When calculating Torque (τ = r F sinθ), how do I know which angle to use for sin(θ)—the inside angle or the outside one?

It technically doesn't matter because sin(θ) = sin(180 - θ). However, the standard definition uses the angle between the tails of the position vector r and the force vector F. If you extend the r vector out past the object, it's the angle between that extension and the force.

14. How do I compute the cross product for Torque if I'm given the vectors in i, j, k notation?

1. You use the determinant method:
2. Top row: i, j, k
3. Middle row: components of r (rx, ry, rz)
4. Bottom row: components of F (Fx, Fy, Fz)
5. Then solve: i(ryFz - rzFy) - j(...) + k(...).

15. How do I determine the direction of the torque vector using the Right-Hand Rule without twisting my wrist off?

1. Point your fingers in the direction of r (from pivot to force).
2. Curl your fingers in the direction the Force is pushing.
3. Your thumb points in the direction of the Torque vector.

Tip: If your thumb points OUT of the page, torque is Positive (CCW). If INTO the page, it's Negative (CW).

16. Why is Torque considered a "pseudovector" or "axial vector" rather than a real vector?

Real vectors (like velocity) flip direction if you invert the coordinate system (mirror reflection). Torque, being the product of two vectors (r × F), behaves differently. If you look at a spinning system in a mirror, the rotation direction flips, but the definition of the cross product makes the torque vector behave strangely compared to "polar" vectors. It’s a mathematical artifact of 3D space.

Lesson 5: Torque & Angular Momentum

17. What is the mathematical step to get from F=ma to Torque = I · alpha?

1. Start with Newton's 2nd law for a particle: F_tangential = m · a_tangential.
2. Multiply both sides by radius r: r · F_t = r · m · a_t.
3. Recognize r · F_t is Torque (τ).
4. Substitute a_t = r · α.
5. Result: τ = m · r · (r · α) = (mr²) · α. Since mr² = I, we get τ = I · α.

18. Why isn't the tension in the rope equal on both sides of a pulley when the pulley has mass?

If the tension were equal (T1 = T2), the net torque on the pulley would be zero (T1·R - T2·R = 0). If torque is zero, the pulley cannot angularly accelerate (α = 0). For a massive pulley to start spinning, T1 must be different from T2 to create the net torque required to overcome the pulley's inertia.

19. Can a single particle moving in a straight line have angular momentum, or does it have to be spinning?

Yes, it can! Angular Momentum is L = r × p. Even if a particle is moving in a straight line, if that line does not pass through the origin, it has an "angular" position relative to that origin.

Value: L = mvr, where r is the perpendicular distance (impact parameter) from the origin to the line of motion.

20. For the "Ice Skater" problem (Conservation of Angular Momentum), why does the angular velocity increase when she pulls her arms in?

Angular Momentum (L = I · ω) is conserved (stays constant) because there is no external torque. When she pulls her arms in, she drastically reduces her Moment of Inertia (I). To keep L constant, if I drops, ω must increase proportionally.

21. Does the equation Torque = dL/dt work if the axis of rotation is accelerating?

Generally, no. This equation holds strictly for inertial reference frames or if the axis is the Center of Mass. If you choose an arbitrary axis that is accelerating (and not the CM), you introduce "fictitious forces" that break the standard τ = dL/dt relationship unless you add complex correction terms.

22. How does the reference point change the Angular Momentum of a system?

Angular Momentum is not intrinsic; it is relative. If you calculate L relative to the pivot, it might be non-zero. If you calculate L relative to the particle itself, it is zero. You must always declare "Angular Momentum about point P."

Lesson 6: Kinetic Energy of Rotating Bodies

23. What does it mean for a body to have "Rotational Kinetic Energy" if the object isn't actually moving across the room?

The object as a whole might not be going anywhere, but the billions of atoms inside it are zooming in circles. Rotational KE is simply the sum of the linear Kinetic Energies (½mv²) of every single particle making up the object. We simplify this sum to ½Iω².

24. If I have a rod rotating around its end versus its center, how does that change the Kinetic Energy calculation?

• It changes the I (Moment of Inertia).
• Rod about center: I = 1/12 mL²
• Rod about end: I = 1/3 mL² (4x larger!)
• Therefore, spinning a rod about its end at the same speed requires 4 times more energy than spinning it about its center.

25. Which object wins the race down a ramp: A solid sphere, a hollow sphere, or a cylinder, and why?

• Solid Sphere (Wins)
• Solid Cylinder
• Hollow Sphere/Hoop (Loses)

Why? Conservation of Energy. Potential Energy (mgh) turns into Translation (½mv²) + Rotation (½Iω²). Objects with higher I (like hollow shapes) hoard more energy in rotation, leaving less energy for translation (v), making them slower. The solid sphere has the lowest I coefficient, so it puts the most energy into speed.