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Imagine a spinning wheel or a rotating fan-these are everyday examples of rotational motion. Unlike linear motion, where objects move along a straight path, rotational motion involves objects turning about an axis. To understand this motion deeply, we need to explore how quantities like displacement, velocity, and acceleration translate from straight-line movement to rotation.
In linear motion, mass measures how much an object resists changes in its motion. In rotational motion, a similar concept exists called the moment of inertia, which measures how much an object resists changes in its rotational motion. Think of it as rotational mass. The larger the moment of inertia, the harder it is to start or stop spinning the object.
This section will build your understanding from the ground up, starting with basic angular quantities, moving through the forces that cause rotation, and culminating in how energy behaves in rotating systems. By the end, you'll be equipped to solve a variety of problems involving rotational motion, a key topic in competitive exams.
Just as linear motion uses displacement, velocity, and acceleration, rotational motion uses angular displacement, angular velocity, and angular acceleration.
Angular displacement is the angle through which a point or line has been rotated in a specified direction about a specified axis. It is measured in radians (rad). One full rotation corresponds to \(2\pi\) radians.
Angular velocity is the rate of change of angular displacement with time. It tells us how fast an object is rotating and in which direction. It is measured in radians per second (rad/s).
Angular acceleration is the rate of change of angular velocity with time, measured in radians per second squared (rad/s²). It indicates how quickly the rotational speed is changing.
These angular quantities are vectors. Their direction is along the axis of rotation, determined by the right-hand rule: curl the fingers of your right hand in the direction of rotation; your thumb points in the direction of the angular velocity vector.
Just like linear motion, rotational motion under constant angular acceleration follows similar equations:
Here, \(\omega_0\) is the initial angular velocity, \(\omega\) is the angular velocity after time \(t\), \(\alpha\) is the angular acceleration, and \(\theta\) is the angular displacement.
In linear motion, force causes acceleration. In rotational motion, the equivalent of force is torque.
Torque (\(\tau\)) is the measure of the turning effect of a force applied at a distance from the axis of rotation. It depends on three factors:
Mathematically, torque is given by:
The moment of inertia quantifies how the mass of a body is distributed relative to the axis of rotation. It plays the role of mass in rotational dynamics. The farther the mass is from the axis, the larger the moment of inertia.
For a system of particles, moment of inertia is defined as:
For continuous bodies, this becomes an integral over the mass distribution.
Just as force causes linear acceleration, torque causes angular acceleration. The rotational equivalent of Newton's second law is:
Different shapes have different moments of inertia depending on their mass distribution and axis of rotation. Here are some commonly used formulas:
| Body | Axis | Moment of Inertia \(I\) |
|---|---|---|
| Thin Rod | About center, perpendicular to length | \( \frac{1}{12} M L^2 \) |
| Thin Rod | About end, perpendicular to length | \( \frac{1}{3} M L^2 \) |
| Solid Disc | About center, perpendicular to face | \( \frac{1}{2} M R^2 \) |
| Thin Ring | About center, perpendicular to plane | \( M R^2 \) |
| Solid Sphere | About center | \( \frac{2}{5} M R^2 \) |
When the axis of rotation is not through the center of mass, the parallel axis theorem helps calculate the moment of inertia:
This theorem applies to flat planar objects lying in the xy-plane. It states:
Step 1: Identify the given quantities:
Radius, \(r = 0.5\, m\); Linear velocity, \(v = 10\, m/s\)
Step 2: Use the relation between linear and angular velocity:
\( v = \omega r \Rightarrow \omega = \frac{v}{r} \)
Step 3: Substitute values:
\( \omega = \frac{10}{0.5} = 20\, \text{rad/s} \)
Answer: The angular velocity of the wheel is \(20\, \text{rad/s}\).
Step 1: Calculate torque using \(\tau = r F \sin \theta\). Since force is perpendicular, \(\sin 90^\circ = 1\).
\( \tau = 0.4 \times 15 \times 1 = 6\, \text{N·m} \)
Step 2: Calculate moment of inertia of rod about pivot (end):
\( I = \frac{1}{3} M L^2 = \frac{1}{3} \times 2 \times (0.4)^2 = \frac{1}{3} \times 2 \times 0.16 = 0.107\, \text{kg·m}^2 \)
Step 3: Use Newton's second law for rotation \(\tau = I \alpha\) to find angular acceleration:
\( \alpha = \frac{\tau}{I} = \frac{6}{0.107} \approx 56.07\, \text{rad/s}^2 \)
Answer: Torque is \(6\, \text{N·m}\) and angular acceleration is approximately \(56.07\, \text{rad/s}^2\).
Step 1: Moment of inertia of one disc about its center:
\( I_{disc} = \frac{1}{2} M R^2 = \frac{1}{2} \times 3 \times (0.2)^2 = 0.06\, \text{kg·m}^2 \)
Step 2: For the second disc, use parallel axis theorem to find moment of inertia about the first disc's center:
Distance between centers, \(d = 0.5\, m\)
\( I = I_{cm} + M d^2 = 0.06 + 3 \times (0.5)^2 = 0.06 + 0.75 = 0.81\, \text{kg·m}^2 \)
Step 3: Total moment of inertia is sum of both:
\( I_{total} = 0.06 + 0.81 = 0.87\, \text{kg·m}^2 \)
Answer: The total moment of inertia about the axis through the first disc's center is \(0.87\, \text{kg·m}^2\).
Step 1: Initial potential energy at top:
\( U = M g h \), where \(h\) is height of incline.
Height \(h = 10 \sin \theta\). Since \(\theta\) is not given, assume vertical height \(h = 10 \times \sin \theta\). For simplicity, consider \(h = 10\, m\) (if incline angle given, use accordingly).
Step 2: At bottom, total kinetic energy is sum of translational and rotational:
\( K = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \)
For rolling without slipping, \(v = \omega R\), so:
\( K = \frac{1}{2} M v^2 + \frac{1}{2} I \frac{v^2}{R^2} = \frac{1}{2} M v^2 + \frac{1}{2} \times \frac{2}{5} M R^2 \times \frac{v^2}{R^2} = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2 \)
Step 3: Using energy conservation \(M g h = \frac{7}{10} M v^2\), mass cancels out:
\( v = \sqrt{\frac{10}{7} g h} = \sqrt{\frac{10}{7} \times 9.8 \times 10} = \sqrt{140} \approx 11.83\, \text{m/s} \)
Answer: The speed of the sphere at the bottom is approximately \(11.83\, \text{m/s}\).
Step 1: Moment of inertia of solid cylinder about center:
\( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 4 \times (0.25)^2 = 0.125\, \text{kg·m}^2 \)
Step 2: Using Newton's second law for rolling objects, acceleration \(a\) is given by:
\( a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}} \)
Calculate \(\frac{I}{M R^2} = \frac{0.125}{4 \times (0.25)^2} = \frac{0.125}{4 \times 0.0625} = \frac{0.125}{0.25} = 0.5\)
Assuming incline angle \(\theta\) such that length \(L = 5\, m\) and height \(h = L \sin \theta\). For simplicity, assume \(\sin \theta = \frac{h}{L} = 1\) (vertical drop) or use given angle if available. Here, let's assume \(\sin \theta = 0.5\) (approx 30° incline).
Then, \(a = \frac{9.8 \times 0.5}{1 + 0.5} = \frac{4.9}{1.5} = 3.27\, \text{m/s}^2\)
Step 3: Velocity at bottom using \(v^2 = 2 a L\):
\( v = \sqrt{2 \times 3.27 \times 5} = \sqrt{32.7} \approx 5.72\, \text{m/s} \)
Answer: The acceleration is approximately \(3.27\, \text{m/s}^2\) and velocity at bottom is \(5.72\, \text{m/s}\).
When to use: When converting linear motion problems to rotational motion.
When to use: When the axis of rotation is not through the center of mass.
When to use: When solving problems involving rolling objects.
When to use: During calculation steps to avoid unit mismatch errors.
When to use: When determining the direction of angular quantities.
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