University Physics Notes on Newton's Laws and Circular Motion

Fluid Resistance and Terminal Speed

• The fluid resistance acting on an object is influenced by the speed of the object.

• A falling object reaches its terminal speed when the resisting force equals the weight of the object.

Figures and Graphs

Motion of Falling Objects

• Figure 5.25: Illustrates the motion of an object falling with and without fluid resistance.

• With Fluid Resistance:

  • Position changes more slowly as the object falls.

  • Acceleration decreases over time, tending toward zero.

  • Velocity reaches an upper limit (terminal velocity).

• Without Fluid Resistance:

  • Velocity continues to increase indefinitely.

  • Position changes according to a parabolic curve due to constant acceleration.

  • Acceleration remains constant at gravitational acceleration g.

Example 5.18: Terminal Speed of a Skydiver

• For a human body falling through air in a spread-eagle position:

  • The numerical value of the drag constant D is approximately 0.25 kg/m.

  • To find the terminal speed v_t for a skydiver with mass 50 kg:

    1. The force of gravity F_g = mg = 50 \text{ kg} \times 9.8 \text{ m/s}^2 = 490 \text{ N}.

    2. At terminal speed, the weight equals the drag force: Fd = D \times vt^2.

    3. Set 490 \text{ N} = 0.25 \text{ kg/m} \times v_t^2.

    4. Solve for the terminal speed:
       vt = \sqrt{\left(\frac{490 \text{ N}}{0.25 \text{ kg/m}}\right)} = \sqrt{1960} \text{ m/s} \rightarrow vt \approx 44.5 \text{ m/s}.

Dynamics of Circular Motion

• In uniform circular motion:

  • Both the acceleration and the net force on an object are directed towards the center of the circle.

  • The net force acting on a particle in circular motion is given by:
    F_{\text{net}} = m \times a = m \times \frac{v^2}{r}

  • Where m is the mass of the object, v is the tangential velocity, and r is the radius of the circular path.

Example 5.19: Force in Uniform Circular Motion

• A sled, with a mass of 25.0 kg on frictionless ice, is attached to a post by a 5.00 m rope.

• After being pushed, the sled revolves uniformly in a circle:

  • It makes five complete revolutions every minute.

  • To find the force F exerted on it by the rope, the speed of the sled can be calculated by converting revolutions per minute to radians per second:
    v = \frac{5 \text{ rev/min} \times 2 \pi \text{ rad/rev}}{60 \text{ s/min}} = \frac{10 \pi}{60} \text{ m/s} \rightarrow v \approx 0.524 \text{ m/s}.

  • Applying centripetal force formula, find F:
    F = m \frac{v^2}{r} = 25.0 \text{ kg} \times \frac{(0.524)^2 \text{ m}^2/\text{s}^2}{5.00 \text{ m}} = 0.55 \text{ N}.

Example 5.20: A Conical Pendulum

• An inventor designs a pendulum clock with a mass m at the end of a wire of length L moving in a horizontal circle:

  • The angle \theta is formed with the vertical.

  • This pendulum is called a conical pendulum, as the wire traces out a cone.

  • To find the tension F and period T for one revolution:

    • Tension must account for both gravitational and centripetal forces:

    • FT \cos(\theta) = mg and FT \sin(\theta) = m \frac{v^2}{R} (where R = L \sin(\theta)).

Example 5.21: Rounding a Flat Curve

• A car rounds a flat, unbanked curve with radius R:

  • Coefficient of static friction is \mu_s.

  • Maximum speed v{\text{max}} can be calculated and is limited by friction: F{\text{friction}} = \mus \times FN \rightarrow m \frac{v^2}{R} = \mus mg \rightarrow v{\text{max}} = \sqrt{\mu_s g R}.

Example 5.22: Rounding a Banked Curve

• Banking a curve requires calculating the right angle \theta so a car can navigate without friction:

  • If entering a curve at speed u, it can safely turn on a banked curve.

  • Forces resolve into components along the incline of the banked road: normal force and gravitational force play roles in centripetal requirements.

Example 5.23: Uniform Circular Motion in a Vertical Circle

• A passenger on a Ferris wheel moves in a vertical circle of radius R at a constant speed v:

  • To find forces exerted by the seat when at the top and bottom positions of the circle:

    • At the top, centripetal force is contributed by both gravity and seat's normal force:

    • F_{\text{top}} = N + mg = m \frac{v^2}{R}.

    • At the bottom, the normal force must counteract gravity plus provide enough force for centripetal acceleration:

    • F_{\text{bottom}} = N - mg = m \frac{v^2}{R}.

The Fundamental Forces of Nature

• Current understanding categorizes forces into four fundamental interactions:

  1. Gravitational interactions – attraction between masses.

  2. Electromagnetic interactions – forces between charged particles.

  3. Strong interactions – binding protons and neutrons in atomic nuclei.

  4. Weak interactions – responsible for radioactive decay.

• Efforts are ongoing to unify these forces into a "theory of everything" that cohesively explains all fundamental forces in a single framework.