AP Physics 1- 2.9 Circular Motion- Study Notes- New Syllabus
AP Physics 1-2.9 Circular Motion – Study Notes
AP Physics 1-2.9 Circular Motion – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Circular Motion
- Circular Orbits and Kepler’s Third Law
Circular Motion
Circular motion occurs when an object moves along a circular path. The velocity vector keeps changing direction, which means the object is accelerating even if its speed is constant.
Key Quantities:
Angular Displacement (\( \theta \)): The angle (in radians) through which an object moves around the circle, measured from a reference line.
Angular Velocity (\( \omega \)): Rate of change of angular displacement.
\( \omega = \dfrac{\Delta \theta}{\Delta t} \)
It tells how fast the object rotates.
Linear Velocity (\( v \)): The actual speed of the object along the circular path.
\( v = \omega r \)
Directly proportional to both radius and angular velocity.
Centripetal Acceleration (\( a_c \)): Acceleration directed toward the center of the circle, responsible for changing velocity direction.
\( a_c = \dfrac{v^2}{r} = \omega^2 r \)
Larger speed or smaller radius means stronger inward pull is required.
Centripetal Force (\( F_c \)): Net inward force required to maintain circular motion.
\( F_c = \dfrac{m v^2}{r} = m \omega^2 r \)
This is not a new force—it comes from tension, gravity, friction, etc.
Time Period (\( T \)): Time taken to complete one revolution.
\( T = \dfrac{2 \pi r}{v} = \dfrac{2 \pi}{\omega} \)
Inversely proportional to angular velocity.
Frequency (\( f \)): Number of revolutions per second.
\( f = \dfrac{1}{T} \)
Directly tells how many circles per second the object makes.
Vertical Circular Motion
In vertical circles (like a pendulum or a stone tied to a string moving vertically), gravity plays a role along with tension. The speed of the object is not constant because of the change in gravitational potential energy.
Minimum Speed at Top: To just maintain circular motion at the top, centripetal force is provided entirely by weight:
\( v_{\text{min}} = \sqrt{g r} \)
Maximum Speed at Bottom: By conservation of energy, if the particle just has \( v_{\text{min}} \) at the top, then at the bottom:
\( v_{\text{bottom}} = \sqrt{5 g r} \)
Tension in String:
- At Top: \( T_{\text{top}} = \dfrac{m v^2}{r} – mg \)
- At Bottom: \( T_{\text{bottom}} = \dfrac{m v^2}{r} + mg \)
Tension is greatest at the bottom and least at the top.
Example :
A stone of mass 0.5 kg is whirled in a vertical circle of radius 2 m. Find the minimum speed at the top of the circle to maintain motion.
▶️Answer/Explanation
\( v_{\text{min}} = \sqrt{g r} = \sqrt{9.8 \times 2} = \sqrt{19.6} \approx 4.43 \, \text{m/s} \)
Answer: Minimum speed at top = 4.43 m/s
Example :
If the same stone (radius 2 m) just maintains circular motion, find its speed and tension at the bottom of the path.
▶️Answer/Explanation
\( v_{\text{bottom}} = \sqrt{5 g r} = \sqrt{5 \times 9.8 \times 2} = \sqrt{98} \approx 9.9 \, \text{m/s} \)
Tension: \( T_{\text{bottom}} = \dfrac{m v^2}{r} + mg = \dfrac{0.5 \times (9.9^2)}{2} + (0.5)(9.8) \)
= \( \dfrac{0.5 \times 98}{2} + 4.9 = 24.5 + 4.9 = 29.4 \, \text{N} \)
Answer: Speed ≈ 9.9 m/s, Tension ≈ 29.4 N
Circular Orbits and Kepler’s Third Law
When a smaller body (like a planet or satellite) moves around a much larger body (like the Sun or Earth) in a nearly circular orbit, its motion is governed by the balance of gravitational force and centripetal force.
Force Balance in Circular Orbit:
The gravitational force provides the centripetal force needed to keep the object in orbit:
\( \dfrac{G M m}{r^2} = \dfrac{m v^2}{r} \)
- \( M \) = mass of central body (e.g., Earth, Sun)
- \( m \) = mass of orbiting body
- \( r \) = orbital radius
- \( v \) = orbital speed
Orbital Speed:
\( v = \sqrt{\dfrac{G M}{r}} \)
- Closer orbits → higher speed, farther orbits → slower speed.
Kepler’s Third Law :
The square of the orbital period of a planet is directly proportional to the cube of the radius of its orbit around the Sun (or any central mass).
\( T^2 \propto r^3 \)
- This law holds for all planets and satellites in circular orbits.
- The ratio \( \dfrac{T^2}{r^3} \) is the same for all objects orbiting the same central mass.
Orbital Period:
Time to complete one revolution:
\( T = \dfrac{2 \pi r}{v} = 2 \pi \sqrt{\dfrac{r^3}{G M}} \)
Example :
The Moon orbits Earth at an average radius of \( r = 3.84 \times 10^8 \, \text{m} \). Calculate its orbital period.
▶️Answer/Explanation
Use \( T = 2 \pi \sqrt{\dfrac{r^3}{GM}} \).
Take \( G = 6.67 \times 10^{-11}, \, M_{\text{Earth}} = 5.97 \times 10^{24} \, \text{kg} \).
\( T = 2 \pi \sqrt{\dfrac{(3.84 \times 10^8)^3}{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}} \)
≈ \( 2.36 \times 10^6 \, \text{s} \) ≈ 27.3 days
Answer: The Moon’s orbital period ≈ 27.3 days (matches reality).
Example :
A satellite is launched into a circular orbit 36000 km above Earth’s surface (geostationary orbit). Show that its period is 24 hours.
▶️Answer/Explanation
Total orbital radius: \( r = R_{\text{Earth}} + h = 6.37 \times 10^6 + 3.6 \times 10^7 = 4.24 \times 10^7 \, \text{m} \).
\( T = 2 \pi \sqrt{\dfrac{r^3}{GM}} \)
= \( 2 \pi \sqrt{\dfrac{(4.24 \times 10^7)^3}{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}} \)
≈ \( 8.64 \times 10^4 \, \text{s} \) = 24 h
Answer: Period = 24 hours → hence geostationary.