Lesson 1: Static Friction and Kinetic Friction

1. What is the actual difference between static and kinetic friction, and why do we need two different coefficients?

Static friction acts when surfaces are gripping each other (not moving relative to one another). Kinetic friction acts when surfaces are sliding past one another. You need two coefficients because it is physically harder to start an object moving than to keep it moving. At the microscopic level, the "teeth" of the surfaces are interlocked when static; once moving, they glide over the top of each other, resulting in a lower coefficient for kinetic friction (μk < μs).

2. Why isn't the force of static friction always equal to μ times Normal Force (fs = μN)? My answer key says I'm wrong

This is the most common trap. The formula fs_max = μs · N only calculates the maximum possible static friction—the breaking point right before it slips. Below that limit, static friction is a "smart force"; it simply adjusts to equal whatever force is pushing the object, just enough to keep it in equilibrium. So, generally, fs ≤ μs · N.

3. Does friction always oppose motion? What about when a car accelerates forward?

No. Friction opposes slipping, not necessarily motion. When a car accelerates, the tires try to slip backward against the road. Static friction opposes this slip by pushing the tire forward. Therefore, in this case, static friction is the propulsive force causing the motion.

4. Why is the coefficient of static friction usually higher than the coefficient of kinetic friction? What is happening at the molecular level?

When two surfaces are stationary, "cold welding" occurs—molecular bonds form between the contact points, and the microscopic peaks and valleys settle into each other. To start motion, you must snap these bonds and lift the surfaces out of the valleys. Once moving, the surfaces float above the valleys and bonds don't have time to form, resulting in less resistance.

Lesson 2: Terminal Velocity: Deriving the Formula with Drag Force

5. What exactly is 'terminal velocity' and why does an object stop accelerating when it falls?

Terminal velocity is the top speed a falling object reaches. As an object falls faster, the drag force (air resistance) pushing up increases. Eventually, the upward drag force equals the downward force of gravity (Fd = mg). At this point, the net force is zero. Newton’s Second Law says if F_net = 0, then acceleration is zero, and the velocity becomes constant.

6. How do I actually derive the terminal velocity formula using calculus? I get stuck integrating 1/(mg - kv)

Start with Newton's Second Law: mg - kv = ma. Rewrite acceleration a as dv/dt. This gives you a differential equation. You must separate the variables, putting all "v" terms on one side and "t" terms on the other. This usually results in an integral of the form ∫ dv / (mg - kv). You solve this using u-substitution (let u = mg - kv), which results in a natural log (ln) function.

7. When calculating drag, when should I use the formula with 'v' (linear) vs 'v-squared' (quadratic)?

It depends on the Reynolds number (interaction with the fluid). For very small, slow objects (like dust or oil droplets), use Linear Drag (F = -bv). For human-scale objects moving at normal speeds (like a baseball or a skydiver), use Quadratic Drag (F = ½CρAv²). In AP Physics C, the problem statement will usually specify which one to use.

8. How does the terminal velocity change if the object is denser or has a larger surface area?

Looking at the formula v_term = √(2mg / CρA), you can see the relationships. If mass (m) increases (making it denser), the numerator grows, so terminal velocity increases (it falls faster). If Area (A) increases (like opening a parachute), the denominator grows, so terminal velocity decreases (it slows down).

9. In the drag force equation, why does the drag coefficient (Cd) change for different shapes?

Cd is a measure of aerodynamic efficiency. A flat plate has a high Cd (~1.0) because it creates a chaotic, turbulent wake of air behind it. A teardrop shape has a very low Cd (~0.04) because the air flows smoothly around it (laminar flow) without separating, creating very little drag.

Lesson 3: What Causes Centripetal Acceleration? Key Concepts of Circular Motion

10. If I'm moving in a circle at a constant speed, why do you say I have acceleration?

Acceleration is defined as a change in velocity, not just speed. Velocity is a vector, meaning it has both magnitude (speed) and direction. In a circle, your direction is changing at every single instant. Since the direction changes, the velocity changes. If velocity changes, you are accelerating.

11. Is 'centripetal force' a real force like gravity or tension, or is it just a label we put on other forces?

It is a label (a "job title"). You will never see "Centripetal Force" on a Free Body Diagram. Instead, you will see real forces like Gravity, Tension, Friction, or Normal Force. The sum of those forces acting toward the center is what we call the centripetal force.

12. If I swing a ball on a string and the string breaks, why doesn't the ball fly outwards? Doesn't centrifugal force push it out?

No. There is no force pushing it out. The ball has inertia—it wants to travel in a straight line. The string forces it to turn. If the string breaks, that turning force disappears, and the ball simply follows its inertia, flying off in a straight line tangent to the circle, not directly outward.

13. How do you derive the formula for centripetal acceleration (v²/r) without using calculus?

You can use similar triangles. If you draw the position vectors for two points on a circle and the velocity vectors for those same points, you form two isosceles triangles. These triangles are similar (same angles). By setting up a ratio of their sides (Δv / v = Δr / r) and doing a little algebra dividing by Δt, you arrive at a = v²/r.

14. If I have a block on a rotating turntable, how far out can I place it before it slips?

The block slips when the required centripetal force exceeds the maximum static friction. Set the friction force equal to the necessary centripetal force: μs · mg = mv²/r. The masses cancel out. Solving for r, you get r = v² / (μs · g). Any radius larger than this requires more friction than the surface can provide.

Lesson 4: Understanding Banked Tracks and Circular Motion

15. On a banked curve, does the normal force point straight up or at an angle? I keep getting the components backward.

The Normal Force is always perpendicular to the surface. Since the track is banked (tilted), the Normal Force is tilted too. It points diagonally up and toward the center of the turn. This is crucial because the horizontal component of the Normal Force is what provides the centripetal acceleration to turn the car.

16. In banked track problems, how do I know if I should resolve gravity or resolve the Normal force into components?

This is the opposite of a "block on a ramp" problem. In circular motion, the acceleration is strictly horizontal (toward the center of the circle). You should always align your axes with the acceleration. Therefore, keep the x-axis horizontal. This means you must resolve the Normal Force into x and y components. Do not resolve gravity; gravity points straight down the y-axis.

17. What is the 'design speed' of a banked curve, and why is there no friction required at that specific velocity?

The design speed is the specific velocity where the horizontal component of the Normal Force exactly equals the required centripetal force (N·sinθ = mv²/r). At this speed, the car turns naturally due to the bank angle alone; the tires don't need to use any side-to-side friction to hold the car in the lane.

Lesson 5: The Conical Pendulum: Tension, Period & Frequency

18. What is a conical pendulum and how is the physics different from a normal swinging pendulum?

A normal pendulum swings back and forth in a vertical plane (stopping at the top of the swing). A conical pendulum traces a horizontal circle at a constant speed. The physics is different because the conical pendulum is in dynamic equilibrium vertically (forces cancel) but has constant centripetal acceleration horizontally.

19. When deriving the conical pendulum period, why do we use tan(θ) instead of just sin or cos?

It comes from dividing the equations. Vertically, T·cosθ = mg. Horizontally, T·sinθ = mv²/r. If you divide the horizontal equation by the vertical one, the Tension (T) cancels out, and sinθ / cosθ becomes tanθ. This gives you tanθ = v² / rg, which is much easier to work with.

20. How do I solve for the tension in the string of a conical pendulum if I only know the length and the speed?

You usually need to find the angle first. Knowing speed (v) and length (L), you can relate radius r = L·sinθ. Substituting this into tanθ = v²/rg allows you to solve for the angle θ. Once you have the angle, use the vertical equation: Tension = mg / cosθ.

21. If a conical pendulum spins faster and faster, can the string ever become perfectly horizontal (90 degrees)?

No. To be horizontal (θ = 90°), the vertical component of tension would have to be zero. But gravity is still pulling down with force mg. Something must balance gravity to keep the bob from falling. Since cos(90°) = 0, the equation T = mg / cosθ would require infinite Tension. The string will break before it ever gets to horizontal.

Lesson 6: Deep Dive: Banked Track and Maximum/ Minimum Velocity of Motion

24. How do I calculate the maximum speed a car can go around a banked curve before it slides up the track?

At maximum speed, the car wants to slide up and out of the turn. Therefore, static friction points down the slope to grab it and hold it in. Your centripetal force equation becomes: (Horizontal Normal Force) + (Horizontal Friction) = mv²/r.

25. Conversely, what is the minimum speed required on a banked track so the car doesn't slide down?

At minimum speed (imagine going very slow on a steep Daytona curve), gravity wants to pull the car down the slope. Therefore, static friction points up the slope to hold it up. Your centripetal force equation becomes: (Horizontal Normal Force) - (Horizontal Friction) = mv²/r.