Physics 1A: Circular Motion, Relative Motion, and Rocket Dynamics – Study Notes

Circular Motion in a Vertical Circle

Speed at Different Points in a Vertical Circle

When a particle moves in a vertical circle under the influence of gravity, its speed varies depending on its position due to the conservation of mechanical energy. The work-energy theorem provides a kinematic invariant for such motion.

• Kinematic Invariant: The quantity $\frac{1}{2}v(\phi)^2 - gR \cos \phi$ remains constant for a particle of mass m moving in a vertical circle of radius R under gravity g, where $\phi$ is the angle from the vertical downward axis.

• At the Top (Point A, $\phi = \pi$): $\frac{1}{2}v^2 + gR$

• At the Bottom (Point B, $\phi = 0$): $\frac{1}{2}v_{\text{bot}}^2 - gR$

• Equating values at top and bottom:

$\frac{1}{2}v^2 + gR = \frac{1}{2}v_{\text{bot}}^2 - gR$

$\Rightarrow v_{\text{bot}}^2 = v^2 + 4gR$

$\Rightarrow v_{\text{bot}} = \sqrt{v^2 + 4gR}$

• Example: If $v = 2\ \text{m/s}$, $g = 9.8\ \text{m/s}^2$, $R = 1\ \text{m}$, then $v_{\text{bot}} = \sqrt{4 + 39.2} \approx 6.5\ \text{m/s}$.

Radial (Centripetal) Acceleration

The radial acceleration at any point in the circle is given by $a_r(\phi) = \frac{v(\phi)^2}{R}$.

• At the Top ($\phi = \pi$): $a_r(\pi) = \frac{v^2}{R}$ (direction: downward, toward center)

• At the Bottom ($\phi = 0$): $a_r(0) = \frac{v_{\text{bot}}^2}{R} = \frac{v^2 + 4gR}{R} = \frac{v^2}{R} + 4g$ (direction: upward, toward center)

Normal Force and Radial Dynamics

Applying Newton's Second Law in the radial direction allows us to analyze the forces acting on the particle at any angle $\phi$.

• Forces: Normal force $N(\phi)$ (inward), radial component of gravity $-mg \cos \phi$ (outward or inward depending on $\phi$).

• Newton's Second Law (radial):

$N(\phi) - mg \cos \phi = m \frac{v(\phi)^2}{R}$

• Normal force acceleration: $a_N(\phi) = \frac{N(\phi)}{m} = \frac{v(\phi)^2}{R} + g \cos \phi$

• At the Top ($\phi = \pi$): $a_N(\pi) = \frac{v^2}{R} - g$

• At the Bottom ($\phi = 0$): $a_N(0) = \frac{v^2}{R} + 5g$

• Difference in normal force acceleration between bottom and top: $\Delta a_N = 6g$

Minimum Speed to Maintain Contact

The particle (or car) will lose contact with the track if the normal force becomes zero or negative. The critical point is at the top of the circle.

• Condition for contact at the top: $N(\pi) \geq 0$

• Setting $N(\pi) = 0$:

$0 = m \frac{v_{\text{min}}^2}{R} - mg$

$\Rightarrow v_{\text{min}}^2 = gR$

$\Rightarrow v_{\text{min}} = \sqrt{gR}$

• Example: For $R = 2\ \text{m}$, $g = 9.8\ \text{m/s}^2$, $v_{\text{min}} = \sqrt{19.6} \approx 4.43\ \text{m/s}$.

Relative Motion in River Crossing

River Width and Current Speed

When a boat crosses a river with a current, the effective velocity relative to the bank is the vector sum of the boat's velocity relative to the water and the current's velocity.

• Definitions:

  • $v'$: Boat speed relative to water

  • $v$: Speed of current

  • $v_{\text{net}}$: Boat speed relative to bank

  • $d$: Width of river

• Effective crossing speed (perpendicular crossing): $v_{\text{net}} = \sqrt{(v')^2 - v^2}$

• Width equation: $d = t \sqrt{(v')^2 - v^2}$

• Given two trips with different $v'$ and $t$, but same $d$:

$t_1 \sqrt{(v'_1)^2 - v^2} = t_2 \sqrt{(v'_2)^2 - v^2}$

$\Rightarrow v = \sqrt{\frac{t_1^2 (v'_1)^2 - t_2^2 (v'_2)^2}{t_1^2 - t_2^2}}$

• Example Calculation: For $t_1 = 20.1\ \text{s}$, $v'_1 = 6.00\ \text{m/s}$, $t_2 = 11.2\ \text{s}$, $v'_2 = 9.00\ \text{m/s}$:

$v \approx 3.97\ \text{m/s}$

$d = 20.1 \sqrt{6.00^2 - 3.97^2} \approx 90.5\ \text{m}$

Shortest Time to Cross and Downstream Drift

• Shortest crossing time occurs when the boat heads directly perpendicular to the bank: $t_{\text{min}} = \frac{d}{|v'|}$

• Drift downstream: $x = v \cdot t_{\text{min}}$

• Example: $t_{\text{min}} = \frac{90.5}{6.00} = 15.1\ \text{s}$, $x = 3.97 \times 15.1 \approx 60.0\ \text{m}$

Rocket Soft Landing Dynamics

Forces and Acceleration During Landing

To achieve a soft landing, a rocket must apply an upward thrust to decelerate from its initial downward velocity to rest at ground level.

• Given: Initial speed $v = 30.0\ \text{m/s}$ (downward), initial height $h = 80.0\ \text{m}$, mass $m = 20.0\ \text{kg}$, final velocity $v_f = 0$ at $y = 0$.

• Newton's Second Law (vertical): $F - mg = ma \Rightarrow F = m(g + a)$

• Acceleration is constant since both $F$ and $mg$ are constant.

Kinematic Analysis for Required Acceleration

• Velocity condition at landing: $0 = -v + a t'$

• Position condition at landing: $0 = h - v t' + \frac{1}{2} a t'^2$

• Solving for $a$:

  • From velocity: $t' = \frac{v}{a}$

  • Substitute into position: $0 = h - \frac{v^2}{a} + \frac{1}{2} \frac{v^2}{a}$

  • Simplifies to: $a = \frac{v^2}{2h}$

• Alternative (time-independent) method: $v_f^2 - v_0^2 = 2a \Delta y$

• For $v_f = 0$, $v_0 = -v$, $\Delta y = -h$:

$0 - (-v)^2 = 2a(-h) \Rightarrow a = \frac{v^2}{2h}$

Calculating Required Thrust

• Substitute $a$ into force equation: $F = m \left( g + \frac{v^2}{2h} \right)$

• Numerical example: $F = 20.0 \times (9.80 + 5.625) = 308.5\ \text{N}$

Summary Table: Key Equations and Results

Topic	Key Equation	Result/Condition
Vertical Circle Speed (Bottom)	$v_{\text{bot}} = \sqrt{v^2 + 4gR}$	Speed at bottom given speed at top
Radial Acceleration (Bottom)	$a_r(0) = \frac{v^2}{R} + 4g$	Radial acceleration at bottom
Minimum Speed at Top	$v_{\text{min}} = \sqrt{gR}$	Minimum speed to maintain contact
River Current Speed	$v = \sqrt{\frac{t_1^2 (v'_1)^2 - t_2^2 (v'_2)^2}{t_1^2 - t_2^2}}$	Solving for current speed
Shortest Crossing Time	$t_{\text{min}} = \frac{d}{|v'|}$	Boat heads perpendicular to bank
Rocket Thrust for Soft Landing	$F = m \left( g + \frac{v^2}{2h} \right)$	Constant thrust required

Additional info: Academic context and step-by-step derivations have been added for clarity and completeness. All equations are provided in LaTeX format as per instructions.