Welcome, my dear students, to a deep dive into the fascinating world of rotational motion! For anyone aspiring for JEE, mastering rotational dynamics is not just important; it's absolutely crucial. It’s a topic that beautifully connects various concepts and often tests your understanding of vectors, kinematics, and even calculus.
Today, we're going to build a solid foundation by understanding the
angular variables that describe rotational motion and, more importantly, how they relate to the
linear variables we've already mastered in translational kinematics. Think of it like learning a new language – we're going from the 'letters and words' of linear motion to the 'sentences and paragraphs' of rotational motion.
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Understanding Rotational Motion: Why New Variables?
Imagine a merry-go-round. Every point on it moves in a circle. While each point has its own linear velocity and acceleration, the entire merry-go-round rotates as a single unit. Describing the motion of each point individually using only linear variables would be incredibly tedious. This is where angular variables come in handy – they describe the motion of the *entire body* as it rotates.
For example, when a wheel rotates, all points on it rotate through the same angle in the same amount of time. This *angular displacement* is the same for all points, even though points farther from the center travel a greater linear distance. This commonality makes angular variables powerful tools.
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1. Angular Displacement ($ heta$)
Just as linear displacement (Δx) describes the change in position along a straight line,
angular displacement ($Delta heta$) describes the change in orientation of a body as it rotates about an axis.
*
Definition: It's the angle swept out by a line connecting the axis of rotation to any point on the body.
*
Units: The standard SI unit is the
radian (rad). We often use degrees (°) or revolutions (rev) as well.
*
Conversion: 1 revolution = 360° = 2π radians.
*
Vector Nature:
* For
infinitesimal (very small) angular displacements, we can treat them as vectors. The direction of this vector is given by the
right-hand thumb rule. If you curl the fingers of your right hand in the direction of rotation, your thumb points in the direction of the angular displacement vector, which lies along the axis of rotation.
* For
large angular displacements, they are generally
not considered vectors because vector addition is commutative (
A +
B =
B +
A), but successive large rotations are generally not commutative. For example, rotating an object 90° about the x-axis and then 90° about the y-axis gives a different final orientation than doing it in the reverse order. However, for 2D rotation about a fixed axis, we often simply assign a sign (+ve for counter-clockwise, -ve for clockwise) to indicate direction.
Relation to Linear Displacement (Arc Length):
For a point P at a distance 'r' from the axis of rotation, if the body undergoes an angular displacement $Delta heta$, the point P covers a linear distance along the arc (arc length), denoted by 's'.
$$s = r Delta heta$$
Important Note: This relation holds true only when $Delta heta$ is measured in radians!
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2. Angular Velocity ($omega$)
Analogous to linear velocity (v) being the rate of change of linear displacement,
angular velocity ($omega$) is the rate of change of angular displacement.
*
Average Angular Velocity ($omega_{avg}$):
$$omega_{avg} = frac{Delta heta}{Delta t}$$
Where $Delta heta$ is the angular displacement over a time interval $Delta t$.
*
Instantaneous Angular Velocity ($omega$): This is the angular velocity at a specific instant.
$$omega = lim_{Delta t o 0} frac{Delta heta}{Delta t} = frac{d heta}{dt}$$
*
Units: The standard SI unit is
radians per second (rad/s). Other common units include revolutions per minute (rpm) or revolutions per second (rps).
*
Conversion: 1 rpm = 2π/60 rad/s = π/30 rad/s.
*
Vector Nature: Angular velocity is a true vector. Its direction is also given by the right-hand thumb rule, pointing along the axis of rotation in the direction of the angular displacement vector (for positive rotation).
*
Relation to Frequency (f) and Time Period (T):
If a body completes 'f' revolutions per second, its angular velocity is:
$$omega = 2pi f$$
And since frequency is the reciprocal of the time period (f = 1/T):
$$omega = frac{2pi}{T}$$
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3. Angular Acceleration ($alpha$)
Just as linear acceleration (a) is the rate of change of linear velocity,
angular acceleration ($alpha$) is the rate of change of angular velocity.
*
Average Angular Acceleration ($alpha_{avg}$):
$$alpha_{avg} = frac{Deltaomega}{Delta t}$$
*
Instantaneous Angular Acceleration ($alpha$):
$$alpha = lim_{Delta t o 0} frac{Deltaomega}{Delta t} = frac{domega}{dt} = frac{d^2 heta}{dt^2}$$
*
Units: The standard SI unit is
radians per second squared (rad/s²).
*
Vector Nature: Angular acceleration is also a true vector. Its direction is the same as the change in angular velocity ($Deltaomega$). If the body is speeding up, $alpha$ is in the same direction as $omega$. If it's slowing down, $alpha$ is in the opposite direction to $omega$.
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Kinematic Equations for Constant Angular Acceleration
If a body undergoes constant angular acceleration ($alpha$), we can derive a set of kinematic equations analogous to those for constant linear acceleration.
Let:
* $ heta_0$ = initial angular position
* $omega_0$ = initial angular velocity
* $ heta$ = final angular position
* $omega$ = final angular velocity
* $alpha$ = constant angular acceleration
* t = time interval
1.
Angular Velocity as a function of time:
$$omega = omega_0 + alpha t$$
(Parallel to v = v₀ + at)
2.
Angular Displacement as a function of time:
$$ heta = heta_0 + omega_0 t + frac{1}{2} alpha t^2$$
(Parallel to x = x₀ + v₀t + ½at²)
Often, we take $ heta_0 = 0$, so $Delta heta = omega_0 t + frac{1}{2} alpha t^2$.
3.
Angular Velocity squared as a function of displacement:
$$omega^2 = omega_0^2 + 2 alpha ( heta - heta_0)$$
(Parallel to v² = v₀² + 2a(x - x₀))
JEE/CBSE Focus: These equations are fundamental. Make sure you can apply them to various scenarios involving uniformly accelerated rotation. The key is to identify the given variables and what needs to be found, just like in linear kinematics.
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Relations Between Linear and Angular Variables: The Bridge!
This is where rotational kinematics truly connects with translational kinematics. For any point on a rotating rigid body, its linear motion is directly linked to the body's angular motion.
Consider a point P rotating in a circle of radius 'r' about a fixed axis.
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A. Linear Speed (v) and Angular Speed ($omega$)
We know that for an arc length 's', $s = r heta$.
Differentiating both sides with respect to time 't':
$$frac{ds}{dt} = frac{d(r heta)}{dt}$$
Since 'r' is constant for a given point on a rigid body:
$$frac{ds}{dt} = r frac{d heta}{dt}$$
We know $ds/dt$ is the linear speed (v) and $d heta/dt$ is the angular speed ($omega$).
Therefore:
$$mathbf{v = romega}$$
The direction of this linear velocity is always
tangential to the circular path at that point.
Vector Form (JEE Advanced):
The linear velocity vector
v of a particle located by position vector
r from the origin (on the axis of rotation) is given by the cross product:
$$mathbf{v = omega imes r}$$
This cross product correctly gives both the magnitude (rω) and the direction (tangential, perpendicular to both
ω and
r). The direction of
ω is along the axis of rotation (using the right-hand rule).
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B. Linear Acceleration (a) and Angular Acceleration ($alpha$)
This is a bit more involved because a point moving in a circle has two components of linear acceleration:
1.
Tangential Acceleration ($a_t$): This component is responsible for changing the magnitude of the linear velocity.
$$a_t = frac{dv}{dt}$$
Substitute $v = romega$:
$$a_t = frac{d(romega)}{dt} = r frac{domega}{dt}$$
Since $domega/dt = alpha$:
$$mathbf{a_t = ralpha}$$
The direction of $a_t$ is always tangential to the circular path. If the body is speeding up, $a_t$ is in the same direction as
v. If it's slowing down, $a_t$ is opposite to
v.
Vector Form (JEE Advanced):
$$mathbf{a_t = alpha imes r}$$
This correctly gives the magnitude (rα) and direction (tangential, perpendicular to both
α and
r).
2.
Centripetal (or Radial) Acceleration ($a_c$): This component is responsible for changing the direction of the linear velocity, keeping the particle moving in a circle. It always points towards the center of the circle.
$$mathbf{a_c = frac{v^2}{r}}$$
Substitute $v = romega$:
$$mathbf{a_c = frac{(romega)^2}{r} = romega^2}$$
The direction of $a_c$ is always radial, pointing towards the center of the circular path.
Vector Form (JEE Advanced):
$$mathbf{a_c = omega imes v = omega imes (omega imes r)}$$
This vector triple product can be expanded using the BAC-CAB rule, but conceptually, it confirms the direction towards the center.
Total Linear Acceleration (a):
The total linear acceleration of a point on a rotating body is the vector sum of its tangential and centripetal components.
$$mathbf{a = a_t + a_c}$$
Since $a_t$ and $a_c$ are always perpendicular to each other, the magnitude of the total acceleration is:
$$mathbf{|a| = sqrt{a_t^2 + a_c^2} = sqrt{(ralpha)^2 + (romega^2)^2}}$$
$$mathbf{|a| = r sqrt{alpha^2 + omega^4}}$$
This table summarizes the crucial relationships:
Linear Variable |
Angular Variable |
Relation (Scalar Form) |
Relation (Vector Form - JEE Advanced) |
|---|
Displacement (s) |
Displacement (θ) |
s = rθ |
N/A (for large angles) |
Velocity (v) |
Velocity (ω) |
v = rω |
v = ω × r |
Tangential Acc. ($a_t$) |
Acceleration (α) |
$a_t$ = rα |
$a_t$ = α × r |
Centripetal Acc. ($a_c$) |
Velocity (ω) |
$a_c$ = rω² |
$a_c$ = ω × v |
JEE Focus: Pay close attention to the vector relationships. Problems in JEE Advanced often require you to use the cross product definitions, especially when the axis of rotation is not aligned with a principal axis or when dealing with 3D rotation.
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Examples to Solidify Understanding
Let's walk through some examples to see these concepts in action.
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Example 1: Basic Conversions and Tangential Speed
Problem: A ceiling fan rotates at a constant speed of 150 revolutions per minute (rpm).
(a) What is its angular velocity in rad/s?
(b) If a point on the tip of a blade is 0.75 m from the center, what is its linear speed?
Solution:
(a) Angular Velocity in rad/s:
Given: $omega_{rpm}$ = 150 rpm
To convert rpm to rad/s, we use the conversion factor: 1 rpm = $frac{2pi}{60}$ rad/s.
$$omega = 150 ext{ rev/min} imes frac{2pi ext{ rad}}{1 ext{ rev}} imes frac{1 ext{ min}}{60 ext{ s}}$$
$$omega = 150 imes frac{2pi}{60} ext{ rad/s}$$
$$omega = frac{300pi}{60} ext{ rad/s}$$
$$omega = 5pi ext{ rad/s} approx 15.71 ext{ rad/s}$$
The angular velocity of the fan is
5π rad/s.
(b) Linear Speed of the blade tip:
Given: Radius r = 0.75 m
We use the relation v = rω.
$$v = 0.75 ext{ m} imes 5pi ext{ rad/s}$$
$$v = 3.75pi ext{ m/s} approx 11.78 ext{ m/s}$$
The linear speed of the blade tip is
3.75π m/s.
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Example 2: Uniformly Accelerated Rotation
Problem: A grinding wheel, initially at rest, is accelerated uniformly at 2.0 rad/s² for 10 seconds.
(a) What is its final angular velocity?
(b) How many revolutions does the wheel make during this time?
(c) What is the tangential acceleration of a point on its rim 0.2 m from the center?
(d) What is the total acceleration of that point at t = 10 s?
Solution:
Given:
Initial angular velocity ($omega_0$) = 0 rad/s (starts from rest)
Angular acceleration ($alpha$) = 2.0 rad/s²
Time (t) = 10 s
Radius (r) = 0.2 m
(a) Final angular velocity ($omega$):
Using the equation: $omega = omega_0 + alpha t$
$$omega = 0 + (2.0 ext{ rad/s}^2)(10 ext{ s})$$
$$omega = 20 ext{ rad/s}$$
The final angular velocity is
20 rad/s.
(b) Revolutions made ($Delta heta$):
Using the equation: $Delta heta = omega_0 t + frac{1}{2}alpha t^2$
$$Delta heta = (0)(10) + frac{1}{2}(2.0 ext{ rad/s}^2)(10 ext{ s})^2$$
$$Delta heta = frac{1}{2}(2.0)(100) ext{ rad}$$
$$Delta heta = 100 ext{ rad}$$
Now, convert radians to revolutions: 1 revolution = 2π radians.
$$ ext{Revolutions} = frac{100 ext{ rad}}{2pi ext{ rad/rev}} = frac{50}{pi} ext{ rev} approx 15.92 ext{ revolutions}$$
The wheel makes approximately
15.92 revolutions.
(c) Tangential acceleration ($a_t$) of a point on the rim:
Using the relation: $a_t = ralpha$
$$a_t = (0.2 ext{ m})(2.0 ext{ rad/s}^2)$$
$$a_t = 0.4 ext{ m/s}^2$$
The tangential acceleration is
0.4 m/s².
(d) Total acceleration ($a$) of that point at t = 10 s:
At t = 10 s, we know:
$omega$ = 20 rad/s
$alpha$ = 2.0 rad/s²
r = 0.2 m
First, calculate the centripetal acceleration ($a_c$):
$$a_c = romega^2$$
$$a_c = (0.2 ext{ m})(20 ext{ rad/s})^2$$
$$a_c = (0.2)(400) ext{ m/s}^2$$
$$a_c = 80 ext{ m/s}^2$$
The tangential acceleration is $a_t = 0.4 ext{ m/s}^2$ (from part c).
Now, find the magnitude of the total acceleration:
$$a = sqrt{a_t^2 + a_c^2}$$
$$a = sqrt{(0.4)^2 + (80)^2}$$
$$a = sqrt{0.16 + 6400}$$
$$a = sqrt{6400.16} approx 80.001 ext{ m/s}^2$$
The total acceleration of the point is approximately
80.001 m/s². Notice how the centripetal acceleration dominates when the angular speed is high.
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Example 3: Vector Approach (JEE Advanced)
Problem: A particle is moving in a circle of radius 3 m in the xy-plane with its center at the origin. Its angular velocity is
ω = 2
k rad/s. Find the velocity of the particle when its position vector is
r = (3
i + 0
j) m. If it also has an angular acceleration
α = -1
k rad/s², find its total linear acceleration at that instant.
Solution:
Given:
ω = 2
k rad/s
r = 3
i m (position vector at the instant)
α = -1
k rad/s²
(a) Linear Velocity (v
):
Using the vector cross product:
v =
ω ×
r
$$mathbf{v} = (2mathbf{k}) imes (3mathbf{i})$$
$$mathbf{v} = 6 (mathbf{k} imes mathbf{i})$$
Recall the cyclic order for cross products:
i ×
j =
k,
j ×
k =
i,
k ×
i =
j.
$$mathbf{v} = 6mathbf{j} ext{ m/s}$$
The velocity of the particle is
6j m/s. This means it's moving in the positive y-direction, which is tangential to the circle at (3,0).
(b) Total Linear Acceleration (a
):
Total acceleration has two components: tangential (
a_t) and centripetal (
a_c).
Tangential Acceleration (a_t
):
Using the vector cross product:
a_t =
α ×
r
$$mathbf{a_t} = (-1mathbf{k}) imes (3mathbf{i})$$
$$mathbf{a_t} = -3 (mathbf{k} imes mathbf{i})$$
$$mathbf{a_t} = -3mathbf{j} ext{ m/s}^2$$
The tangential acceleration is in the negative y-direction, indicating the particle is slowing down (since
v is in +y direction).
Centripetal Acceleration (a_c
):
Using the vector cross product:
a_c =
ω ×
v
$$mathbf{a_c} = (2mathbf{k}) imes (6mathbf{j})$$
$$mathbf{a_c} = 12 (mathbf{k} imes mathbf{j})$$
Recall:
k ×
j = -
i.
$$mathbf{a_c} = 12 (-mathbf{i})$$
$$mathbf{a_c} = -12mathbf{i} ext{ m/s}^2$$
The centripetal acceleration is in the negative x-direction, which points towards the origin (center of the circle) from the position (3,0). This is correct.
Total Acceleration (a
):
$$mathbf{a} = mathbf{a_t} + mathbf{a_c}$$
$$mathbf{a} = (-3mathbf{j}) + (-12mathbf{i})$$
$$mathbf{a} = -12mathbf{i} - 3mathbf{j} ext{ m/s}^2$$
The total acceleration of the particle at that instant is
(-12i - 3j) m/s².
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Conclusion
Mastering angular variables and their connection to linear motion is a cornerstone of rotational dynamics. It allows us to analyze complex motions systematically. Remember to always keep the units consistent (especially radians for calculations involving 'r'), understand the vector nature of angular quantities, and correctly identify the tangential and centripetal components of linear acceleration. With these tools, you are well-equipped to tackle a wide range of problems in rotational motion, from basic conceptual questions to advanced JEE-level challenges! Keep practicing, and you'll find rotational motion to be one of the most rewarding topics in physics.