AP Physics C Mechanics- 2.10 Circular Motion- Study Notes- New Syllabus
AP Physics C Mechanics- 2.10 Circular Motion – Study Notes
AP Physics C Mechanics- 2.10 Circular Motion – Study Notes – per latest Syllabus.
Key Concepts:
- Centripetal Acceleration
- Forces Producing Centripetal Acceleration
- Tangential Acceleration
- Net Acceleration in Circular Motion
- Period and Frequency in Uniform Circular Motion
- Satellites in Circular Orbit
Centripetal Acceleration
Centripetal acceleration is the component of an object’s acceleration that is always directed toward the center of the circular path along which the object is moving. It is responsible for continuously changing the direction of the velocity vector, keeping the object in circular motion.
Even if the object’s speed remains constant, its velocity is still changing because direction is changing — hence, there is always an acceleration directed inward.
Magnitude of Centripetal Acceleration
The magnitude of centripetal acceleration is given by the ratio of the object’s tangential speed squared to the radius of its circular path:
\( \mathrm{a_c = \dfrac{v^2}{r}} \)
- \( \mathrm{a_c} \): centripetal acceleration (m/s²)
- \( \mathrm{v} \): tangential (linear) speed (m/s)
- \( \mathrm{r} \): radius of the circular path (m)
Key Idea: If the object’s speed increases, centripetal acceleration increases with the square of speed. If the radius increases, centripetal acceleration decreases.
Direction of Centripetal Acceleration
Centripetal acceleration is always directed radially inward — that is, toward the center of the circle. It is perpendicular to the tangential velocity vector at every point of the circular motion.
- It does not change the speed of the object, only the direction of motion.
- If this inward acceleration stops, the object will move off tangentially to the circle in a straight line.
Example
A satellite orbits Earth in a circular path of radius \( \mathrm{7.0\times10^6\,m.} \) It moves at a constant speed of \( \mathrm{7.5\times10^3\,m/s.} \) Determine the magnitude and direction of its centripetal acceleration.
▶️ Answer / Explanation
Step 1: Apply the formula for centripetal acceleration:
\( \mathrm{a_c = \dfrac{v^2}{r}} \)
Step 2: Substitute given values:
\( \mathrm{a_c = \dfrac{(7.5\times10^3)^2}{7.0\times10^6}} \)
\( \mathrm{a_c = \dfrac{56.25\times10^6}{7.0\times10^6} = 8.04\,m/s^2.} \)
Step 3: Interpret the direction:
The acceleration is directed toward the center of Earth, keeping the satellite in circular orbit.
Result: The satellite experiences a centripetal acceleration of \( \mathrm{8.0\,m/s^2} \), directed radially inward toward Earth’s center.
Forces Producing Centripetal Acceleration
An object in circular motion must always experience a net inward (centripetal) force to maintain its path. This centripetal force may arise from one or more physical forces — such as gravity, normal force, friction, or tension — or from the components of these forces acting toward the center of the circular path.
Motion at the Top of a Vertical Circular Loop
For an object (like a roller coaster car or a pendulum bob) at the top of a vertical circular loop, both gravity and the normal (or tension) force act toward the center of the loop. At the minimum speed required to maintain circular motion, the gravitational force alone provides the necessary centripetal force.
\( \mathrm{F_{centripetal} = F_g = m\dfrac{v^2}{r}} \)
Derived Equation:
\( \mathrm{v = \sqrt{gr}} \)
- \( \mathrm{v} \): minimum speed at the top of the loop
- \( \mathrm{r} \): radius of the circular path
If the object’s speed is less than \( \mathrm{\sqrt{gr}} \), it will not have enough centripetal acceleration to stay in circular motion and will lose contact with the path.
Banked Curves and Friction Components
When a car moves along a banked curve, both the normal force and the frictional force can provide the centripetal acceleration needed to maintain circular motion.
- The horizontal component of the normal force (\( \mathrm{F_N \sin\theta} \)) acts toward the center of the curve.
- The frictional force may act toward or away from the center, depending on the speed.
The net inward force provides the required centripetal acceleration:
\( \mathrm{F_{net} = F_N \sin\theta + F_f = m\dfrac{v^2}{r}} \)
A properly banked curve allows the car to turn without relying solely on friction — reducing the chance of slipping.
Conical Pendulum
For a conical pendulum (a mass moving in a horizontal circle at the end of a string), the horizontal component of the tension in the string provides the centripetal acceleration.
\( \mathrm{T\sin\theta = m\dfrac{v^2}{r}} \)
and
\( \mathrm{T\cos\theta = mg.} \)
Combining these equations relates the angular speed and geometry of the pendulum.
Key Note: The string tension has two roles — balancing the weight vertically and providing the horizontal (centripetal) component that keeps the mass moving in a circle.
Example
A 2.0 kg object moves in a vertical circular path of radius \( \mathrm{r = 1.5\,m.} \) Determine the minimum speed the object must have at the top of the loop to maintain circular motion.
▶️ Answer / Explanation
Step 1: At minimum speed, \( \mathrm{F_g = F_{centripetal}} \).
\( \mathrm{mg = m\dfrac{v^2}{r}} \)
Step 2: Simplify:
\( \mathrm{v = \sqrt{gr} = \sqrt{(9.8)(1.5)} = 3.83\,m/s.} \)
Result: The object must have a minimum speed of \( \mathrm{3.83\,m/s} \) at the top of the loop to stay in circular motion.
Example
A small mass is attached to a string of length \( \mathrm{1.2\,m} \) and moves in a horizontal circle such that the string makes an angle of \( \mathrm{30°} \) with the vertical. Determine the speed of the mass.
▶️ Answer / Explanation
Step 1: From the geometry of circular motion:
\( \mathrm{r = L\sin\theta = 1.2\sin30° = 0.6\,m.} \)
Step 2: From the tension equations:
- \( \mathrm{T\cos\theta = mg} \)
- \( \mathrm{T\sin\theta = m\dfrac{v^2}{r}} \)
Divide the second by the first:
\( \mathrm{\tan\theta = \dfrac{v^2}{rg}} \Rightarrow \mathrm{v = \sqrt{r g \tan\theta}} \)
Step 3: Substitute values:
\( \mathrm{v = \sqrt{(0.6)(9.8)\tan30°} = \sqrt{(5.88)(0.577)} = \sqrt{3.39} = 1.84\,m/s.} \)
Result: The speed of the mass is \( \mathrm{1.84\,m/s.} \)
Tangential Acceleration
Tangential acceleration is the rate of change of the speed (magnitude of velocity) of an object moving along a circular path. Unlike centripetal acceleration, which changes the direction of velocity, tangential acceleration changes the magnitude of velocity.
Direction:
- It is always tangent to the circular path of motion.
- If the object speeds up, tangential acceleration is in the same direction as motion.
- If the object slows down, it is opposite to the motion.
Relevant Equation:
\( \mathrm{a_t = \dfrac{dv}{dt}} \)
or equivalently, since \( \mathrm{v = r\omega} \):
\( \mathrm{a_t = r\alpha} \)
- \( \mathrm{a_t} \): tangential acceleration (m/s²)
- \( \mathrm{r} \): radius of circular path (m)
- \( \mathrm{\alpha} \): angular acceleration (rad/s²)
Tangential acceleration only exists when the object’s angular velocity is changing; it represents linear acceleration along the direction of motion.
Example
A wheel of radius \( \mathrm{0.40\,m} \) starts from rest and accelerates uniformly with an angular acceleration of \( \mathrm{2.5\,rad/s^2.} \) Determine the tangential acceleration of a point on the wheel’s rim.
▶️ Answer / Explanation
Step 1: Use the relation \( \mathrm{a_t = r\alpha} \).
\( \mathrm{a_t = (0.40)(2.5) = 1.0\,m/s^2.} \)
Step 2: Direction:
The tangential acceleration is tangent to the circular path in the direction of rotation, since the wheel is speeding up.
Result: The point on the rim has a tangential acceleration of \( \mathrm{1.0\,m/s^2} \), directed along the tangent to its circular path.
Net Acceleration in Circular Motion
When an object moves along a circular path with a changing speed, it experiences two components of acceleration:
- Centripetal acceleration ( \( \mathrm{a_c} \) ) — directed toward the center, changes direction of velocity.
- Tangential acceleration ( \( \mathrm{a_t} \) ) — directed tangent to the path, changes magnitude of velocity.
The net acceleration is the vector sum of these two perpendicular components:
\( \mathrm{a_{net} = \sqrt{a_c^2 + a_t^2}} \)
If the speed is constant, \( \mathrm{a_t = 0} \), and the net acceleration is purely centripetal. If both components are present, the acceleration vector points between the radial and tangential directions.
Example
A motorcycle travels around a circular track of radius \( \mathrm{40\,m} \) and is speeding up. At a given instant, its tangential acceleration is \( \mathrm{1.5\,m/s^2} \), and its speed is \( \mathrm{16\,m/s.} \) Find the magnitude of the net acceleration.
▶️ Answer / Explanation
Step 1: Calculate the centripetal acceleration:
\( \mathrm{a_c = \dfrac{v^2}{r} = \dfrac{(16)^2}{40} = 6.4\,m/s^2.} \)
Step 2: Compute the net acceleration:
\( \mathrm{a_{net} = \sqrt{a_c^2 + a_t^2} = \sqrt{(6.4)^2 + (1.5)^2}} \)
\( \mathrm{a_{net} = \sqrt{40.96 + 2.25} = \sqrt{43.21} = 6.57\,m/s^2.} \)
Step 3: Direction:
The net acceleration points slightly inward from the tangent to the circular path.
Result: The motorcycle’s total acceleration is \( \mathrm{6.57\,m/s^2} \), directed between the radial and tangential directions.
Period and Frequency in Uniform Circular Motion
When an object moves in a circular path at a constant speed, it undergoes uniform circular motion. Although the object’s speed remains constant, its direction continuously changes, resulting in continuous centripetal acceleration. This motion can be described in terms of two important quantities — period (T) and frequency (f).
Period (T)
The period (T) is defined as the time taken for the object to complete one full revolution around the circular path.
Units: seconds (s)
\( \mathrm{T = \dfrac{\text{time}}{\text{number of revolutions}}} \)
Interpretation: If a car completes one lap every 10 seconds, its period is \( \mathrm{T = 10\,s.} \)
Frequency (f)
The frequency (f) is the number of complete revolutions per unit time.
Units: hertz (Hz) = revolutions per second
\( \mathrm{f = \dfrac{1}{T}} \)
Key Relationship:
\( \mathrm{T f = 1} \)
Interpretation: If a wheel completes 5 revolutions per second, its frequency is \( \mathrm{5\,Hz,} \) and its period is \( \mathrm{T = \dfrac{1}{5} = 0.2\,s.} \)
Period and Speed Relationship
For uniform circular motion, the tangential (linear) speed \( \mathrm{v} \) is related to the circumference of the circle and the period of revolution:
\( \mathrm{v = \dfrac{2\pi r}{T}} \)
Rearranging gives the derived equation for period:
\( \mathrm{T = \dfrac{2\pi r}{v}} \)
- \( \mathrm{r} \): radius of the circular path (m)
- \( \mathrm{v} \): tangential speed (m/s)
- \( \mathrm{T} \): period (s)
A faster object (higher \( \mathrm{v} \)) completes each revolution in a shorter time, so \( \mathrm{T} \) decreases as \( \mathrm{v} \) increases.
Example
A 0.5 kg mass moves at a constant speed of \( \mathrm{4.0\,m/s} \) in a horizontal circular path of radius \( \mathrm{0.80\,m.} \) Determine the period and frequency of its motion.
▶️ Answer / Explanation
Step 1: Use the derived equation for period:
\( \mathrm{T = \dfrac{2\pi r}{v}} \)
Step 2: Substitute the given values:
\( \mathrm{T = \dfrac{2\pi(0.80)}{4.0} = \dfrac{5.03}{4.0} = 1.26\,s.} \)
Step 3: Find frequency:
\( \mathrm{f = \dfrac{1}{T} = \dfrac{1}{1.26} = 0.79\,Hz.} \)
Step 4: Interpretation:
- The object completes one revolution every \( \mathrm{1.26\,s.} \)
- It makes \( \mathrm{0.79} \) revolutions per second.
Result: \( \mathrm{T = 1.26\,s, \quad f = 0.79\,Hz.} \)
Satellites in Circular Orbit
For a satellite in a stable circular orbit around a much more massive central body (such as Earth or the Sun), the only force acting on the satellite that provides the necessary centripetal acceleration is the gravitational force between the two bodies.
Condition for Circular Motion:
\( \mathrm{F_{gravity} = F_{centripetal}} \)
That is,
\( \mathrm{\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}} \)
where:
- \( \mathrm{G} \): universal gravitational constant (\( \mathrm{6.67 \times 10^{-11}\,N·m^2/kg^2} \))
- \( \mathrm{M} \): mass of the central body (e.g., Earth)
- \( \mathrm{m} \): mass of the satellite
- \( \mathrm{r} \): radius of the satellite’s circular orbit (m)
The mass of the satellite cancels out, showing that the orbital speed depends only on the central mass and orbital radius:
\( \mathrm{v = \sqrt{\dfrac{GM}{r}}} \)
Relation Between Period and Radius
The orbital period \( \mathrm{T} \) is the time it takes for the satellite to make one complete revolution:
\( \mathrm{T = \dfrac{2\pi r}{v}} \)
Substitute \( \mathrm{v = \sqrt{\dfrac{GM}{r}}} \):
\( \mathrm{T = 2\pi \sqrt{\dfrac{r^3}{GM}}} \)
Derived Equation:
\( \mathrm{T^2 = \dfrac{4\pi^2}{GM} r^3} \)
This relationship shows that \( \mathrm{T^2 \propto r^3} \) — which is Kepler’s Third Law of planetary motion. It holds true for any object in circular orbit around a massive central body.
The greater the orbital radius, the longer the period of revolution — meaning satellites farther from Earth move more slowly and take more time to complete one orbit.
Example
A satellite orbits Earth at a height where its orbital radius (from Earth’s center) is \( \mathrm{r = 7.0\times10^6\,m.} \) Given \( \mathrm{M_E = 5.97\times10^{24}\,kg} \) and \( \mathrm{G = 6.67\times10^{-11}\,N·m^2/kg^2,} \) calculate the satellite’s orbital period.
▶️ Answer / Explanation
Step 1: Use the orbital period formula:
\( \mathrm{T = 2\pi \sqrt{\dfrac{r^3}{GM}}} \)
Step 2: Substitute known values:
\( \mathrm{T = 2\pi \sqrt{\dfrac{(7.0\times10^6)^3}{(6.67\times10^{-11})(5.97\times10^{24})}}} \)
\( \mathrm{T = 2\pi \sqrt{\dfrac{3.43\times10^{20}}{3.98\times10^{14}}}} \)
\( \mathrm{T = 2\pi \sqrt{8.62\times10^5} = 2\pi(928) = 5.83\times10^3\,s.} \)
Step 3: Convert to hours:
\( \mathrm{T = \dfrac{5.83\times10^3}{3600} = 1.62\,h.} \)
Result: The satellite completes one orbit around Earth every \( \mathrm{1.62\,hours} \) (approximately 1 hour 37 minutes).