IB DP Physics Circular motion Study Notes - 2025 Syllabus

IB DP Physics Circular motion Study Notes

IB DP Physics Circular motion Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with guiding questions of

that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle—known as a centripetal acceleration as given by $a=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2}$
that circular motion is caused by a centripetal force acting perpendicular to the velocity
that a centripetal force causes the body to change direction even if its magnitude of velocity may remain constant
that the motion along a circular trajectory can be described in terms of the angular velocity ω which is related to the linear speed v by the equation as given by $v=\frac{2\pi r}{T}=\omega r$

Standard level and higher level: 10 hours
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IB DP Physics 2025 -Study Notes -All Topics

Centripetal acceleration:

When an object moves in a circle at constant speed, its velocity is constantly changing direction. Although the speed remains the same, the changing direction of velocity means the object is accelerating.

This acceleration is always directed toward the center of the circular path. It’s called centripetal acceleration.

The word “centripetal” comes from Latin: centrum = center, petere = to seek — i.e., “center-seeking” acceleration.

Key formulas for centripetal acceleration $ a $:

$ a = \frac{v^2}{r} $, where:
- $ v $ = linear speed (tangential speed)
- $ r $ = radius of the circular path
$ a = \omega^2 r $, where:
- $ \omega $ = angular velocity in rad/s
$ a = \frac{4\pi^2 r}{T^2} $, where:
- $ T $ = time period (time for one full revolution)

Conceptual Points:

Even though the object is moving at constant speed, it is accelerating because velocity is a vector (direction matters).
The acceleration is always perpendicular to the velocity vector and points inward.
This centripetal acceleration must be caused by a net inward force (centripetal force), e.g., tension, gravity, friction, or normal force.

Example:

A car of mass $ 1000 \, \text{kg} $ is moving at a constant speed of $ 20 \, \text{m/s} $ around a circular track of radius $ 50 \, \text{m} $. Find the centripetal acceleration of the car and the net force required to maintain this circular motion.

▶️Answer/Explanation

Step 1: Use the centripetal acceleration formula

$ a = \frac{v^2}{r} = \frac{(20)^2}{50} = \frac{400}{50} = \boxed{8 \, \text{m/s}^2} $

Step 2: Find the net force using $ F = ma $

$ F = 1000 \times 8 = \boxed{8000 \, \text{N}} $

Interpretation: The car requires an inward (centripetal) force of 8000 N to maintain its circular path. This could be provided by friction between the tires and the road.

Example:

A small object is attached to a string and swung in a horizontal circle of radius $ 0.5 \, \text{m} $ with a time period of $ 2 \, \text{s} $. Calculate the centripetal acceleration of the object using the formula involving time period.

▶️Answer/Explanation

Step 1: Use the formula involving time period:

$ a = \frac{4\pi^2 r}{T^2} $

Substitute: $ r = 0.5 \, \text{m}, \, T = 2 \, \text{s} $

$ a = \frac{4\pi^2 (0.5)}{(2)^2} = \frac{4\pi^2 \times 0.5}{4} = \pi^2 \times 0.5 \approx 9.87 \times 0.5 = \boxed{4.93 \, \text{m/s}^2} $

Interpretation: The object experiences a centripetal acceleration of approximately $ 4.93 \, \text{m/s}^2 $, directed towards the center of the circle.

• That circular motion is caused by a centripetal force acting perpendicular to the velocity

When an object moves in a circular path, its direction constantly changes — even if the speed (magnitude of velocity) remains the same.
This change in direction implies a change in velocity, which means there is acceleration.
This acceleration is called centripetal acceleration, always directed towards the center of the circle.
To cause this acceleration, there must be a net force directed towards the center. This is known as the centripetal force.
The centripetal force is always perpendicular to the instantaneous velocity vector of the object.
Since the force is perpendicular, it doesn’t do work on the object — it changes direction, not speed.

• That a centripetal force causes the body to change direction even if its magnitude of velocity may remain constant

The centripetal force does not increase or decrease the object’s speed.
Instead, it continuously changes the direction of the velocity vector, which results in circular motion.
This is a fundamental concept: an unbalanced force perpendicular to motion changes the direction of velocity, not its magnitude.
The absence of a centripetal force would cause the object to move off in a straight line, according to Newton’s first law (inertia).
Examples of centripetal forces include:
- Tension in a string for a ball swung in a circle
- Gravitational force for planets in orbit
- Frictional force for a car turning on a flat road

Example:

A car of mass $ 1000 \, \text{kg} $ is traveling around a circular track of radius $ 50 \, \text{m} $ at a constant speed of $ 10 \, \text{m/s} $. Calculate:

The centripetal acceleration
The required frictional force acting as the centripetal force
Why the car doesn’t speed up even though a net force is acting on it

▶️Answer/Explanation

Step 1: Centripetal acceleration

$ a = \frac{v^2}{r} = \frac{(10)^2}{50} = \frac{100}{50} = \boxed{2 \, \text{m/s}^2} $

Step 2: Centripetal force (here, friction provides it)

$F = ma = 1000 \times 2 = \boxed{2000 \, \text{N} \, \text{(towards the center)}} $

Step 3: Why the speed doesn’t change

The net force is perpendicular to the velocity vector.
It changes the direction of velocity (i.e. car’s heading), not the speed.
Hence, the car moves in a circle with constant speed — circular motion with unchanging kinetic energy.

Vertical Circular Motion

Vertical circular motion occurs when an object moves in a circular path under the influence of gravity, with forces acting both radially and tangentially.
The tension or normal force changes depending on the position in the circle (e.g., top, bottom, or side).
Unlike horizontal motion, the speed is not constant due to the effect of gravity doing work on the object.

1. Conical Pendulum

A small mass is suspended by a string and moves in a horizontal circle, forming a cone shape with the string.
Here, gravity and tension balance in the vertical direction, while the horizontal component of tension provides the centripetal force.

Let $ \theta $ be the angle the string makes with the vertical, $ T $ the tension, $ m $ the mass, and $ r $ the radius of the horizontal circle.

Vertical: $ T \cos \theta = mg $
Horizontal (centripetal): $ T \sin \theta = \frac{mv^2}{r} $

From this, we can derive the relationship between the speed and the angle of the pendulum:

$ \tan \theta = \frac{v^2}{rg} $

2. Vertical Circle (e.g., mass on string or roller coaster loop)

At different points in the motion, the net force (and thus required tension or normal force) changes.
At the top: both gravity and tension act downwards.
At the bottom: gravity acts downward, tension acts upward.

At top of the loop:
$ T_{\text{top}} + mg = \frac{mv^2}{r} $

At bottom of the loop:
$ T_{\text{bottom}} – mg = \frac{mv^2}{r} $

At minimum speed to stay in contact at the top of the loop, $ T = 0 $, so:
$ v_{\text{min}} = \sqrt{gr} $

3. Orbital Motion (e.g., satellites)

Gravitational force provides the centripetal force.
Assume circular orbit of radius $ r $, mass of orbiting body $ m $, and mass of central body $ M $:

$ \frac{mv^2}{r} = \frac{GMm}{r^2} \Rightarrow v = \sqrt{\frac{GM}{r}} $

Orbital period: $ T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r^3}{GM}} $

Example :

A 0.40 kg ball is attached to a string 1.5 m long and rotates in a horizontal circle at a constant speed. The string makes an angle $ \theta $ with the vertical. If the speed of the ball is $ 2.0 \, \text{m/s} $, calculate the angle $ \theta $.

▶️ Answer/Explanation

We use the relation for a conical pendulum: $ \tan \theta = \frac{v^2}{r g} $

First, determine $ r $:

$ r = L \sin \theta $, but we don’t yet know $ \theta $, so:

Instead, use components of forces:

Vertical: $ T \cos \theta = mg $
Horizontal: $ T \sin \theta = \frac{mv^2}{r} $

Divide equations: $ \tan \theta = \frac{v^2}{rg} $

But $ r = L \sin \theta $, so:

$ \tan \theta = \frac{v^2}{g L \sin \theta} \Rightarrow \tan \theta \cdot \sin \theta = \frac{v^2}{g L} $

Solve numerically using substitution or trial:

$ \frac{v^2}{g L} = \frac{(2.0)^2}{9.8 \times 1.5} = \frac{4}{14.7} ≈ 0.272 $

Try $ \theta = 15^\circ $: $ \tan \theta \cdot \sin \theta = 0.267 $ — close.

$\boxed{\theta ≈ 15^\circ}$

Example :

A 0.60 kg toy car is going around a vertical loop of radius 0.80 m. What is the minimum speed the car must have at the top of the loop to stay in contact with the track?

▶️ Answer/Explanation

At the top of the loop, for the car to just stay in contact, the normal force $ N = 0 $.

So only gravity provides the centripetal force:

$ mg = \frac{mv^2}{r} \Rightarrow v^2 = rg \Rightarrow v = \sqrt{rg} $

Substitute values:

$ v = \sqrt{0.80 \times 9.8} = \sqrt{7.84} ≈ 2.80 \, \text{m/s} $

$\boxed{v_{\text{min}} = 2.80 \, \text{m/s}}$

Example:

A satellite orbits Earth in a circular path at an altitude where the radius of its orbit is $ 7.0 \times 10^6 \, \text{m} $. Given that the gravitational field strength near Earth is approximately $ g = 9.8 \, \text{m/s}^2 $, calculate:

(a) The orbital speed $ v $ of the satellite
(b) The period $ T $ of the satellite’s orbit

▶️ Answer/Explanation

(a) Orbital speed using centripetal force:

The centripetal force is provided by gravity: $ \frac{mv^2}{r} = mg $
Cancelling $ m $, we get: $ \frac{v^2}{r} = g \Rightarrow v^2 = rg $
$ v = \sqrt{rg} = \sqrt{(7.0 \times 10^6)(9.8)} ≈ \sqrt{6.86 \times 10^7} ≈ 8.28 \times 10^3 \, \text{m/s} $
$\boxed{v ≈ 8.28 \, \text{km/s}}$

(b) Orbital period:

$ T = \frac{2\pi r}{v} = \frac{2\pi \times 7.0 \times 10^6}{8.28 \times 10^3} $ $ T ≈ \frac{4.40 \times 10^7}{8.28 \times 10^3} ≈ 5315 \, \text{s} ≈ \boxed{1.48 \, \text{hours}} $

The satellite orbits Earth at a speed of $ \boxed{8.28 \, \text{km/s}} $ and completes one orbit in about $ \boxed{1.48 \, \text{hours}} $.

Describing Circular Motion Using Angular Velocity

In circular motion, a body moves along a curved path around a central point, and its motion can be described in two ways:

Using linear (tangential) quantities like linear speed $ v $, arc length $ s $
Using angular quantities like angular displacement $ \theta $, angular velocity $ \omega $

Angular velocity $ \omega $ is defined as the rate of change of angular displacement:

$ \omega = \frac{\theta}{t} \quad \text{(in radians per second)} $

The total angle covered in one full revolution is $ \theta = 2\pi \, \text{radians} $
Thus, if the object completes one revolution in time $ T $ (the period), then:

$ \omega = \frac{2\pi}{T} $

The linear speed $ v $ is related to the angular speed by:

As the arc length is $ s = r\theta $, and $ v = \frac{s}{t} $, we get:

$ v = r\omega \quad \text{(or)} \quad v = \frac{2\pi r}{T} $

$ v $ is the magnitude of the linear velocity (tangential), always perpendicular to the radius at any point.
$ \omega $ is the angular speed, same for all points on the rotating object, regardless of radius.
$ v $ increases with $ r $: points farther from the axis move faster linearly even though $ \omega $ is the same.

Example:

A rotating circular platform of radius $ 1.5 \, \text{m} $ completes one full revolution every $ 4.0 \, \text{s} $. A coin is placed at the edge of the platform.

(a) Calculate the angular velocity $ \omega $ of the platform.
(b) Find the linear speed $ v $ of the coin.
(c) How far does the coin travel in 6.0 seconds?

▶️ Answer/Explanation

(a) Angular velocity:

$ \omega = \frac{2\pi}{T} = \frac{2\pi}{4.0} = \frac{\pi}{2} ≈ 1.57 \, \text{rad/s} $

(b) Linear speed:

$ v = \omega r = 1.57 \times 1.5 = 2.36 \, \text{m/s} $

(c) Distance traveled in 6.0 seconds:

$ \text{Distance} = v \times t = 2.36 \times 6.0 = \boxed{14.2 \, \text{m}} $

Thus, the coin travels a total of $\boxed{14.2 \, \text{m}}$ in 6 seconds along the circular path.

IB DP Physics Circular motion Study Notes

IB DP Physics Circular motion Study Notes - 2025 Syllabus

IB DP Physics Circular motion Study Notes

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