IB MYP 4-5 Physics- Circular motion and centripetal force- Study Notes - New Syllabus
IB MYP 4-5 Physics-Circular motion and centripetal force- Study Notes
Key Concepts
- Circular motion and centripetal force
Circular motion and centripetal force
Circular Motion
Circular motion occurs when an object moves in a circular path at a constant speed or varying speed. Even if speed is constant, the velocity changes because its direction changes continuously.
Key Terms
- Radius (\(r\)) The distance from the center of the circle to the object.
- Period (\(T\)) Time taken for one complete revolution, measured in seconds.
- Frequency (\(f\)) Number of revolutions per second: \( f = \dfrac{1}{T} \).
- Angular displacement (\(\theta\)) The angle swept by the radius vector, measured in radians.
- Angular velocity (\(\omega\)) Rate of change of angular displacement: \( \omega = \dfrac{\theta}{t} \) and \( \omega = \dfrac{2\pi}{T} \).
- Linear velocity (\(v\)) Speed along the circular path: \( v = \omega r \).
Centripetal Acceleration
Even with constant speed, the object accelerates toward the center due to change in velocity direction.
Formula: \( a_c = \dfrac{v^2}{r} = \omega^2 r \).
Centripetal Force
The net force acting towards the center to maintain circular motion.
Formula: \( F_c = \dfrac{mv^2}{r} = m\omega^2 r \).
Relationship Between Linear and Angular Quantities
- Linear velocity: \( v = \omega r \)
- Linear acceleration: \( a_c = \omega^2 r \)
Types of Circular Motion
- Uniform Circular Motion Speed is constant, only velocity direction changes, centripetal acceleration present.
- Non-uniform Circular Motion Speed and direction both change; has centripetal and tangential acceleration.
Example:
A satellite orbits the Earth in a circular path of radius \( 7.0 \times 10^6 \, \text{m} \) with a speed of \( 7.5 \times 10^3 \, \text{m/s} \). Find its orbital period.
▶️ Answer/Explanation
Step 1: Formula for period of circular motion:
\( T = \dfrac{2\pi r}{v} \)
Step 2: Substitute values:
\( T = \dfrac{2\pi (7.0 \times 10^6)}{7.5 \times 10^3} \)
\( T = \dfrac{4.398 \times 10^7}{7.5 \times 10^3} \)
\( T \approx 5866 \, \text{s} \)
Step 3: Convert to hours:
\( T \approx 1.63 \, \text{hours} \)
Final Answer:
\(\boxed{1.63 \, \text{hours}}\)
Example:
A wheel of radius \( 0.5 \, \text{m} \) rotates at \( 120 \, \text{rpm} \). Find the linear speed of a point on its rim.
▶️ Answer/Explanation
Step 1: First, find angular speed in rad/s:
\( \omega = \dfrac{2\pi N}{60} = \dfrac{2\pi (120)}{60} = 4\pi \, \text{rad/s} \)
Step 2: Linear speed formula:
\( v = \omega r \)
Step 3: Substitute values:
\( v = (4\pi)(0.5) = 2\pi \, \text{m/s} \)
\( v \approx 6.28 \, \text{m/s} \)
Final Answer:
\(\boxed{6.28 \, \text{m/s}}\)
Centripetal Force
Centripetal force is the resultant (net) force directed towards the center of a circular path that keeps an object in circular motion. It does not change the object’s speed but continuously changes its velocity direction.
Formula
\( F_c = \dfrac{mv^2}{r} = m\omega^2 r \)
- \(m\) – mass of the object (kg)
- \(v\) – linear speed (m/s)
- \(\omega\) – angular speed (rad/s)
- \(r\) – radius of the circular path (m)
Direction
Always directed towards the center of the circular path.
Key Notes
- Centripetal force is not a separate physical force it is provided by other forces such as tension, gravity, or friction.
- If centripetal force is removed, the object will move tangentially to the circle (Newton’s First Law).
Examples of Forces Providing Centripetal Force
- Tension in a string for a ball being whirled.
- Friction for a car turning a bend.
- Gravitational force for the Moon orbiting Earth.
Example:
A car of mass \( 1000 \, \text{kg} \) is taking a circular turn of radius \( 50 \, \text{m} \) at a speed of \( 20 \, \text{m/s} \). Find the centripetal force acting on it.
▶️ Answer/Explanation
Step 1: Formula for centripetal force:
\( F_c = \dfrac{mv^2}{r} \)
Step 2: Substitute values:
\( F_c = \dfrac{(1000)(20^2)}{50} \)
\( F_c = \dfrac{(1000)(400)}{50} \)
\( F_c = 8000 \, \text{N} \)
Final Answer:
\(\boxed{8000 \, \text{N}}\) directed towards the center.
Example:
A stone of mass \( 0.5 \, \text{kg} \) is tied to a string of length \( 0.8 \, \text{m} \) and whirled in a horizontal circle at an angular speed of \( 5 \, \text{rad/s} \). Find the tension in the string.
▶️ Answer/Explanation
Step 1: Here, the tension provides the centripetal force:
\( T = m\omega^2 r \)
Step 2: Substitute values:
\( T = (0.5)(5^2)(0.8) \)
\( T = (0.5)(25)(0.8) \)
\( T = 10 \, \text{N} \)
Final Answer:
\(\boxed{10 \, \text{N}}\) directed towards the center.
Centrifugal Force
The centrifugal force is an apparent or fictitious force experienced in a rotating (non-inertial) reference frame. It is not a real force in the Newtonian sense but rather a perceived effect due to the inertia of a body when observed from a rotating frame.
- It appears to push an object away from the center of the circular path.
- It is equal in magnitude but opposite in direction to the centripetal force.
- It arises only in a rotating frame of reference, not in an inertial frame.
Mathematical Form
\( F_{\text{centrifugal}} = \dfrac{mv^2}{r} \)
- \( m \) = mass of the object (kg)
- \( v \) = tangential speed (m/s)
- \( r \) = radius of circular path (m)
Key Notes
- Exists only from the perspective of an observer inside the rotating frame.
- Always directed radially outward from the axis of rotation.
- Used in engineering problems such as vehicle banking and rotating rides to explain apparent outward push on passengers.
| Aspect | Centripetal Force | Centrifugal Force |
|---|---|---|
| Definition | The real force that acts towards the center of a circular path, keeping an object in circular motion. | A fictitious or apparent force that appears to act outward on an object moving in a circle, observed in a rotating reference frame. |
| Direction | Always directed towards the center of the circular path. | Always directed away from the center of the circular path. |
| Nature | Real force (can be gravitational, tension, friction, etc.). | Pseudo force (exists only in non-inertial, rotating frames). |
| Cause | Caused by actual physical interactions (e.g., tension in a string, gravitational pull). | Perceived due to the inertia of the object resisting a change in direction. |
| Formula | \( F_{\text{cp}} = \dfrac{mv^2}{r} \) | \( F_{\text{cf}} = \dfrac{mv^2}{r} \) |
| Reference Frame | Observed in an inertial (non-accelerating) frame of reference. | Observed in a rotating (non-inertial) frame of reference. |
| Example | The tension in a string when an object is whirled in a circle. | The outward push you feel when a car takes a sharp turn. |
Example:
A car of mass \( m = 800 \, \text{kg} \) travels at \( v = 20 \, \text{m/s} \) around a circular curve of radius \( r = 50 \, \text{m} \). Find the centrifugal force experienced by the passengers from the car’s frame of reference.
▶️ Answer/Explanation
Step 1: Formula for centrifugal force:
\( F_{\text{cf}} = \dfrac{mv^2}{r} \)
Step 2: Substituting values:
\( F_{\text{cf}} = \dfrac{800 \times (20)^2}{50} \)
\( F_{\text{cf}} = \dfrac{800 \times 400}{50} \)
\( F_{\text{cf}} = 6400 \, \text{N} \)
Interpretation: This is the apparent outward force felt by passengers.
\(\boxed{6400 \, \text{N}}\)
Example:
A washing machine drum has a radius \( r = 0.25 \, \text{m} \) and spins at \( v = 15 \, \text{m/s} \). If a wet cloth inside has a mass of \( m = 0.5 \, \text{kg} \), calculate the centrifugal force acting on it in the rotating drum’s frame of reference.
▶️ Answer/Explanation
Step 1: Formula:
\( F_{\text{cf}} = \dfrac{mv^2}{r} \)
Step 2: Substituting values:
\( F_{\text{cf}} = \dfrac{0.5 \times (15)^2}{0.25} \)
\( F_{\text{cf}} = \dfrac{0.5 \times 225}{0.25} \)
\( F_{\text{cf}} = \dfrac{112.5}{0.25} \)
\( F_{\text{cf}} = 450 \, \text{N} \)
Interpretation: This force pushes the cloth outward against the drum wall, helping to squeeze water out.
\(\boxed{450 \, \text{N}}\)