Module Notes: Circular Motion and Combined 1D/2D Motion

Circular motion and acceleration components recap

  • The velocity vector at any point on a circular path is tangent to the path; the direction of motion is given by this tangent direction.
  • A velocity vector can be expressed as speed (magnitude) and direction via a unit vector along the motion.
  • Acceleration in circular motion can have two components:
    • Tangential component a_t (parallel to velocity) which changes the magnitude (speed) of the velocity vector.
    • Radial (centripetal) component a_r (perpendicular to velocity) which changes the direction of the velocity vector and points toward the center of the circle.
  • Notation for the components:
    • a_t = \dfrac{dv}{dt}
    • a_r = \dfrac{v^2}{r}
    • Total acceleration: \mathbf{a} = at \, \hat{t} + ar \, \hat{r}
  • Tangential component behavior:
    • If the speed is constant, a_t = 0 and there is no change in speed; the radial component still provides the centripetal acceleration toward the center.
  • Radial (centripetal) acceleration a_r:
    • Always directed toward the center of the circular path.
    • Depends on speed and radius: a_r = \dfrac{v^2}{r}.
  • Summary relationships:
    • Constant speed in circular motion: at = 0, a = ar toward center.
    • If speed changes, at ≠ 0 and contributes along the velocity direction, while ar remains toward the center.
  • How this ties into broader modules:
    • We’ll see these components appear when analyzing systems in terms of sources in Module 4.

Two-dimensional motion problem: combining 1D motion up a ramp with 2D projectile motion

  • Problem setup (as introduced in the lecture):
    • A cart starts from rest and accelerates up a ramp of height h = 8 m.
    • The ramp makes an angle θ with the horizontal and has an unknown length L.
    • The cart accelerates along the ramp according to a(t) (function of time).
    • It takes t = 2 s to reach the top of the ramp and then leaves the ramp to become a projectile.
    • We want to connect the 1D motion along the ramp to the subsequent 2D projectile motion after launch.
  • Key ideas used to bridge 1D and 2D parts:
    • Treat the motion up the ramp as a 1D problem along the ramp until launch.
    • Determine the launch speed and the launch angle from the ramp portion, then switch to a standard 2D projectile motion analysis after launch.
  • What we want to know from the 1D portion to set up the 2D problem:
    • The velocity at the top of the ramp (launch speed): v = v(2 s).
    • The ramp length L (distance along the ramp from start to the top): L = x(2 s).
    • The launch angle θ of the velocity relative to the horizontal, derived from the ramp geometry (right triangle with height h and ramp length L).
  • Numerical example (as described in the transcript):
    • Given the integrated ramp motion, the velocity at t = 2 s is v = 31.2 m/s.
    • The ramp length is L = 25.6 m (found by integrating x(t) from t = 0 to t = 2 s, with x(0) = 0).
    • The ramp’s height is h = 8 m, so the ramp angle satisfies sin θ = h / L = 8 / 25.6 = 0.3125, giving θ ≈ 18.21°.
    • Velocity components at launch:
    • v_x0 = v cos θ ≈ 31.2 cos(18.21°) ≈ 29.6 m/s
    • v_y0 = v sin θ ≈ 31.2 sin(18.21°) ≈ 9.75 m/s
  • Coordinate choices for the projectile portion:
    • Some solvers use origin at the ramp base (ground level) with x to the right and y up.
    • The lecture also discusses setting the launch point as the origin for the 2D portion, but then you must account for the initial height when applying the projectile equations.
    • In the setup presented, after the ramp portion is solved, the projectile motion is treated with standard 2D kinematics with acceleration ax = 0 and ay = -g, where g ≈ 9.8 m/s².
  • 2D projectile equations used for the landing location (starting from the launch point at the top of the ramp):
    • x(t) = vx0 t, since ax = 0.
    • y(t) = h + v_y0 t − (1/2) g t^2, where h is the launch height (here h = 8 m if origin is at ramp base).
  • Solving for the landing time and distance:
    • The landing condition is y(t) = 0.
    • Quadratic in t: −(1/2) g t^2 + v_y0 t + h = 0.
    • Coefficients: a = −(1/2) g, b = v_y0, c = h.
    • Numerical roots found: t ≈ 2.61 s (positive root) and t ≈ −0.61 s (negative root). We take t = 2.61 s as the physical flight time from launch to landing.
    • Horizontal landing distance: xfinal = vx0 t ≈ 29.6 m/s × 2.61 s ≈ 77.5 m.
  • Conceptual takeaway: this problem demonstrates breaking a mixed 1D and 2D problem into two parts (ramp with 1D motion, then projectile motion) and then stitching the results together by carrying over the launch speed and angle as initial conditions for the 2D portion.
  • Practical steps to solve problems like this (as illustrated):
    • Step 1: Solve the ramp (1D) portion to obtain the launch speed v and launch angle θ (from the ramp geometry).
    • Step 2: Use v and θ to find velocity components vx0 and vy0 for the 2D projectile portion.
    • Step 3: Set up the projectile equations with appropriate initial height h and initial velocity components.
    • Step 4: Solve y(t) = 0 for the flight time t (a quadratic in t).
    • Step 5: Compute the horizontal range xfinal = vx0 t for the landing spot.
  • Important mathematical relationships used:
    • Ramp geometry: sin θ = h / L, with h = 8 m and L = 25.6 m → θ ≈ \arcsin(0.3125) ≈ 18.21°.
    • Velocity components: vx0 = v cos θ, vy0 = v sin θ.
    • Projectile motion equations (constant gravity, no air resistance):
    • x(t) = v_x0 t
    • y(t) = h + v_y0 t - \dfrac{1}{2} g t^2
    • Flight time from y(t) = 0: t = \frac{-vy0 + \sqrt{vy0^2 + 2 g h}}{g} (the positive root).

Lab logistics and expectations for the 2D/3D motion activities

  • Schedule and lab structure:
    • Wednesday: first full-day lab (two hours, hands-on activity).
    • There is no video prep for Wednesday.
    • The instructor will open the lab document in the course portal for students to view; reading beforehand is optional.
    • In-class time will be provided to read and discuss the lab instructions as a group.
    • All handouts will be provided; students do not need to bring materials for the lab.
  • Weekly progression and expectations:
    • Finish Unit 1 by today; then over the next week you should complete Unit 3.
    • Module 4 starts on Friday.
  • Lab setup for the projectile motion activity (as described in the transcript):
    • Launchers may be vertically mounted or rotated to achieve a desired launch angle; to adjust, loosen the set screw and rotate the launcher.
    • The launcher indicator should be rotated so that the launch angle is set to a target value (in one example, a nearly horizontal launch corresponds to angle near 0°).
    • The string/plumb bob can be used to align the angle from the launcher to the planned trajectory; example target angle given around 55° in the described setup.
    • A medium-range setting (two clicks) is used with yellow plastic balls (as opposed to metal balls) for measurements.
    • A launch speed of about 4.8 m/s is used in an example experiment (based on prior data across classes);
    • This value comes from averaging previously measured values (e.g., 14 trials) to obtain a representative speed.
    • Measurements to perform before launching:
    • Determine the initial height of the launch (use table clamps and a meter stick or ruler for accuracy).
    • Align the launch so that the initial position sits over the edge of the table for easier height measurement.
    • After launching, predict where the projectile will land and mark a landing target on the floor with a cone or tape;
    • Typical process: set a predicted landing distance xpredict, then observe and compare to the actual landing distance xactual.
    • If needed, repeat up to three tries to improve accuracy.
  • Lab safety and coordination:
    • Launching into the room requires awareness of other tables and people; communicate with nearby groups to avoid collisions.
    • Secure the launcher and measurement tools with clamps to minimize movement during firing.
    • The instructor emphasizes that the goal is to analyze the motion and improve prediction accuracy rather than achieving a perfect hit on the first try.
  • Connection to practical physics concepts and real-world relevance:
    • The exercise demonstrates how to decompose a complex motion problem into simpler components (1D ramp motion and 2D projectile motion).
    • It highlights the use of kinematic equations and triangle geometry to compute launch conditions and landing ranges.
    • The activity links theory (circular motion, centripetal acceleration, tangential vs radial components) with an applied experimental setup.
  • Quick references to formulas and steps to memorize for exams:
    • Circular motion decompositions:
    • at = \dfrac{dv}{dt}, \quad ar = \dfrac{v^2}{r}
    • \mathbf{a} = at \hat{t} + ar \hat{r}
    • Projectile motion (two-dimensional):
    • x(t) = vx0 t, \quad y(t) = h + vy0 t - \dfrac{1}{2} g t^2
    • Launch angle from ramp geometry:
    • sin θ = \dfrac{h}{L}, \quad θ = \arcsin\left( \dfrac{h}{L} \right)
    • Key derived values (example numbers):
    • h = 8 m, L = 25.6 m, thus θ ≈ 18.21°
    • v = 31.2 m/s, vx0 ≈ 29.6 m/s, vy0 ≈ 9.75 m/s
    • Time of flight (from y(t) = 0): t ≈ 2.61 s
    • Landing distance: xfinal ≈ vx0 t ≈ 77.5 m
  • Connections to prior lectures and future topics:
    • This example introduces the approach of combining one-dimensional and two-dimensional analyses, which will recur when studying systems with constraints and forces.
    • The enzyme-like or energy-based perspectives (to be covered later) will intersect with the kinematic analyses done here when discussing energy and work in motion.
  • Ethical, philosophical, and practical implications discussed:
    • Emphasis on safe experimental practice, adherence to lab rules, and responsible use of equipment.
    • Encourages a mindset of prediction, measurement, verification, and iteration, which mirrors scientific inquiry.
  • Student tips for exam prep based on this material:
    • Be comfortable switching coordinate systems (ramp-aligned vs standard x-y axes) and know how to translate results between them.
    • Practice breaking composite problems into 1D and 2D components, then reassembling results into a full solution.
    • Remember the physical meaning behind at and ar, and how each component affects speed and direction.
    • Be able to derive and apply the time-of-flight formula for y(t) = 0 with initial height h and vertical speed v_y0:
    • t = \dfrac{vy0}{g} + \sqrt{\left(\dfrac{vy0}{g}\right)^2 + \dfrac{2h}{g}}
    • Practice numerical substitutions with the given numbers (g = 9.8 m/s², h = 8 m, v = 31.2 m/s, θ ≈ 18.21°) to reproduce the example results.