Integrating for Angular Motion

Hello! Welcome back to our study of rotational kinematics.

In the last lesson, we established the rigorous vector relationships between linear and angular quantities, proving that $v = ω \times r$ and that the total acceleration is the sum of tangential and centripetal components, $a = (α \times r) + (ω \times (ω \times r))$ . These are the fundamental differential relationships of circular motion.

Today, we address the learning outcome: Solve problems involving time-varying angular velocity and acceleration using integration. We will build upon the differential forms to develop a systematic, calculus-based approach for solving problems where angular acceleration is not constant. This will directly parallel the methods you mastered in Module 2 for 1D linear motion, where we dealt with $a (t)$ , $a (v)$ , and $a (x)$ . The mathematical structure is identical, providing a powerful demonstration of how fundamental principles in calculus apply across different domains of physics.

1. The Fundamental Kinematic Relationships in Calculus Form

Just as linear motion is governed by the relationships between position, velocity, and acceleration, rotational motion is governed by an analogous set of differential equations for angular position ( $θ$ ), angular velocity ( $ω$ ), and angular acceleration ( $α$ ).

The core relationships are:

Angular velocity as the derivative of position:
$ω = \frac{d θ}{d t}$
Angular acceleration as the derivative of velocity:
$α = \frac{d ω}{d t} = \frac{d ^{2} θ}{d t ^{2}}$

From these two fundamental definitions, we can derive a third, extremely useful relationship for situations where quantities are known as a function of angular position rather than time. Using the chain rule from calculus:
$α = \frac{d ω}{d t} = \frac{d ω}{d θ} \cdot \frac{d θ}{d t}$
Since $ω = d θ / d t$ , this gives us:
3. The "timeless" kinematic relationship:
$α = ω \frac{d ω}{d θ}$

This set of three equations is the complete foundation for solving any problem in 1D rotational kinematics, no matter how the acceleration varies.

Circular Motion

The following resource from Unacademy provides a concise summary of these calculus forms, which are standard in JEE preparation materials.

In the PDF, locate the subsection titled 'Calculus form of equation of motion' within the 'Kinematics of Circular Motion' section. It's on page 4. Note how it directly lists the three key relationships we've just discussed.

2. Solving Problems with Integration: A Systematic Approach

Just as we did for linear motion, we can now use integration to solve for the motion when $α$ is not constant. We will classify problems based on the functional dependence of $α$ .

Case 1: $α$ is a function of time, $α (t)$

This is the most direct case. We perform definite integrals with respect to time.

From $α (t) = d ω / d t$ , we separate variables and integrate:
$\int_{ω_{0}}^{ω (t)} d ω^{'} = \int_{0}^{t} α (t^{'}) d t^{'} ⟹ ω (t) = ω_{0} + \int_{0}^{t} α (t^{'}) d t^{'}$
Once we have $ω (t)$ , we can integrate again to find $θ (t)$ :
$\int_{θ_{0}}^{θ (t)} d θ^{'} = \int_{0}^{t} ω (t^{'}) d t^{'} ⟹ θ (t) = θ_{0} + \int_{0}^{t} ω (t^{'}) d t^{'}$

Example: A flywheel's angular acceleration is given by $α (t) = (12 t - 4)$ rad/s². If it starts from rest with an initial angular position $θ_{0} = 3$ rad, find its angular velocity and position at $t = 2$ s.

Solution:

Find $ω (t)$ : $ω_{0} = 0$ .
$ω (t) = 0 + \int_{0}^{t} (12 t^{'} - 4) d t^{'} = [6 t^{'2} - 4 t^{'}]_{0}^{t} = 6 t^{2} - 4 t$
At $t = 2$ s, $ω (2) = 6 (2)^{2} - 4 (2) = 24 - 8 = 16$ rad/s.
Find $θ (t)$ : $θ_{0} = 3$ .
$θ (t) = 3 + \int_{0}^{t} (6 t^{'2} - 4 t^{'}) d t^{'} = 3 + [2 t^{'3} - 2 t^{'2}]_{0}^{t} = 3 + 2 t^{3} - 2 t^{2}$
At $t = 2$ s, $θ (2) = 3 + 2 (2)^{3} - 2 (2)^{2} = 3 + 16 - 8 = 11$ rad.

Case 2: $α$ is a function of angular velocity, $α (ω)$

Here, we use $α = d ω / d t$ and separate variables to relate $ω$ and $t$ .
$d t = \frac{d ω}{α ( ω )} ⟹ \int_{0}^{t} d t^{'} = t = \int_{ω_{0}}^{ω} \frac{d ω ^{'}}{α ( ω ^{'} )}$
This integration gives us $t$ as a function of $ω$ . We may then need to invert this function to find $ω (t)$ .

The following resource contains an advanced problem of this type, which is excellent for deep practice.

Circular Motion

Now, let's tackle a more advanced problem where angular acceleration depends on angular velocity. The following resource presents and solves a problem where \alpha is proportional to the square root of \omega.

Find the problem on page 15. A body rotates with deceleration \alpha = -k\sqrt{\omega}. The goal is to find the average angular velocity over the whole time of rotation. First, try to solve it yourself: use separation of variables to find the time T it takes for the body to stop (i.e., for \omega to go from \omega_0 to 0). Then, calculate the total angle rotated \Delta heta and find the average velocity (\langle\omega angle = \Delta heta/T). Study the provided solution to verify your method.

Case 3: $α$ is a function of angular position, $α (θ)$

For this case, we use the chain rule identity $α = ω (d ω / d θ)$ . Separating variables gives:
$ω d ω = α (θ) d θ$
Integrating both sides yields:
$\int_{ω_{0}}^{ω} ω^{'} d ω^{'} = \int_{θ_{0}}^{θ} α (θ^{'}) d θ^{'} ⟹ \frac{1}{2} ω^{2} - \frac{1}{2} ω_{0}^{2} = \int_{θ_{0}}^{θ} α (θ^{'}) d θ^{'}$
This equation, relating angular velocity to angular position, is the rotational analogue of the work-energy theorem for a one-dimensional force.

Example: A pendulum bob is released from rest at an angle $θ_{0}$ . Its angular acceleration is approximately given by $α = - (g / L) sin θ$ . Find its angular velocity $ω$ when it passes through the vertical position ( $θ = 0$ ).

Solution:

Set up the integral: We start from rest ( $ω_{0} = 0$ ) at $θ_{0}$ .
$\frac{1}{2} ω^{2} - 0 = \int_{θ_{0}}^{0} (- \frac{g}{L} sin θ^{'}) d θ^{'}$
Solve the integral:
$\frac{1}{2} ω^{2} = - \frac{g}{L} [- cos θ^{'}]_{θ_{0}}^{0} = \frac{g}{L} [cos (0) - cos (θ_{0})] = \frac{g}{L} (1 - cos θ_{0})$
Final Result:
$ω^{2} = \frac{2 g}{L} (1 - cos θ_{0}) ⟹ ∣ ω ∣ = \frac{2 g}{L} (1 - cos θ_{0})$
This is a classic result from mechanics, derived here purely from rotational kinematics.

Test your understanding!

A particle on a circular path starts from rest. Its angular acceleration is constant, $α_{0}$ . At what time $t$ will the magnitude of its centripetal acceleration $a_{c}$ be equal to the magnitude of its tangential acceleration $a_{t}$ ?

Show answer

Tangential Acceleration: For a constant angular acceleration $α_{0}$ and radius $r$ , the tangential acceleration is also constant:
$a_{t} = α_{0} r$
Centripetal Acceleration: This component depends on the angular velocity $ω$ . We must first find $ω (t)$ by integration.
$ω (t) = ω_{0} + \int_{0}^{t} α_{0} d t^{'} = 0 + α_{0} t = α_{0} t$
Now, we find the centripetal acceleration as a function of time:
$a_{c} (t) = ω (t)^{2} r = (α_{0} t)^{2} r = α_{0}^{2} t^{2} r$
Equate the magnitudes: We set $a_{t} = a_{c} (t)$ .
$α_{0} r = α_{0}^{2} t^{2} r$
Assuming $α_{0} \neq = 0$ and $r \neq = 0$ , we can simplify:
$1 = α_{0} t^{2} ⟹ t^{2} = \frac{1}{α _{0}} ⟹ t = \frac{1}{α _{0}}$

This problem required you to first find the time-dependent angular velocity using integration before you could analyze the other physical quantities. The provided solution to a similar problem can be found in the Unacademy resource (LINK) on page 13.

3. Synthesis: Connecting Linear and Rotational Kinematics

Let's work through an example that synthesizes the concepts of this lesson with the vector formalisms from the previous one. This problem involves a pulley system where the linear motion of a block dictates the rotational motion of the pulley.

3. Curvilinear motion

The following example from the 'Dynamics' textbook by J. L. Meriam and L. G. Kraige is a perfect application of our new skills. It connects the linear acceleration of a block to the angular acceleration of a pulley, which you will then integrate.

Read through Example 3.2. Pay close attention to how the constant linear acceleration of the block is used to find the constant angular acceleration of the pulley. Then, observe how the solution integrates \alpha to find \omega and subsequently finds the velocity and acceleration vectors of a point P on the pulley's rim. This connects everything from today's and the previous lesson.

The key steps in that example are:

Constraint: The linear acceleration of the string, $a$ , is related to the tangential acceleration of the pulley's rim by $a = a_{t} = α r$ . Since $a$ is constant ( $g /4$ ), $α$ must also be constant ( $α = a / r = g / (4 r)$ ).
Integration for $ω (t)$ : With $α$ constant and starting from rest, $ω (t) = \int_{0}^{t} α d t^{'} = α t$ .
Vector Calculation: At a given time $t$ , the velocity and acceleration of a point P with position vector $r_{P}$ are found using the full vector equations:
- $v_{P} = ω \times r_{P}$
- $a_{P} = (α \times r_{P}) + (ω \times (ω \times r_{P}))$

This example beautifully illustrates how a problem that seems complex can be broken down systematically using the principles we've developed.

Conclusion

In this lesson, we established a complete framework for solving 1D rotational kinematics problems using calculus. By recognizing the direct analogy to linear motion, we can confidently tackle any problem where angular acceleration varies.

Key Takeaways:

The rotational kinematic equations ( $ω = d θ / d t$ , $α = d ω / d t$ , and $α = ω d ω / d θ$ ) form the basis for solving all problems.
Problems involving non-constant angular acceleration can be solved by classifying them based on the functional form of $α$ :
- For $α (t)$ , we integrate with respect to time.
- For $α (ω)$ , we separate variables $d ω$ and $d t$ .
- For $α (θ)$ , we use $α d θ = ω d ω$ and integrate.
These integration techniques allow us to find $ω$ and $θ$ at any point in time or space, which can then be used to find linear quantities like velocity and the components of acceleration.

Preview of the Next Lesson:
We have now fully explored rotational motion in a fixed plane. In the next lesson, we will move to the most general case: spatial circular motion. This involves applying the 3D vector forms of $ω$ and $α$ to situations where the plane of rotation itself might be moving or oriented arbitrarily in space, demanding a full command of the vector cross product.

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