Convert rpm to rad/s:
Initial angular speed: \(\omega_0 = 600 \text{ rpm} = \frac{600 \times 2\pi}{60} = 20\pi \text{ rad/s}\)
Final angular speed: \(\omega = 1200 \text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi \text{ rad/s}\)
Find Angular Acceleration:
Using \(\alpha = \dfrac{\omega - \omega_0}{t}\):
$$\alpha = \frac{40\pi - 20\pi}{10} = 2\pi \text{ rad/s}^2$$
Find Total Angular Displacement:
Using \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\):
$$\theta = 20\pi(10) + \frac{1}{2}(2\pi)(10)^2$$
$$\theta = 200\pi + 100\pi = 300\pi \text{ rad}$$
Convert radians to revolutions:
$$n = \frac{\theta}{2\pi} = \frac{300\pi}{2\pi} = \mathbf{150 \text{ revolutions}}$$
Shortcut Method (Average rpm):
Since acceleration is uniform, average rpm \(= \dfrac{600+1200}{2} = 900 \text{ rpm} = 15 \text{ rev/s}\)
Revolutions in 10 s \(= 15 \times 10 = \mathbf{150}\) ✓
Rotational motion occurs when a body spins about a fixed axis. Every particle of the body traces a circular path around this axis. Unlike translational motion (where the entire body moves from one point to another), in pure rotational motion the axis itself stays stationary. Examples include a ceiling fan spinning, a wheel rotating about its axle, and a flywheel in an engine.
The key quantities in rotational motion are angular displacement (θ, in radians), angular velocity (ω, in rad/s), and angular acceleration (α, in rad/s²). These are the rotational analogues of linear displacement, velocity, and acceleration respectively.
Angular velocity (ω) is defined as the rate of change of angular displacement with time. Mathematically, \(\omega = d\theta/dt\). The SI unit is rad/s. In practical applications like motors and engines, rotation speed is often expressed in rpm (revolutions per minute). The conversion is:
\(\omega \text{ (rad/s)} = \text{rpm} \times \dfrac{2\pi}{60}\)
One revolution corresponds to an angular displacement of 2π radians. Therefore, N revolutions per minute means N × 2π radians per minute, which equals N × 2π/60 radians per second.
Angular acceleration (α) is the rate of change of angular velocity. Mathematically, \(\alpha = d\omega/dt\). When angular acceleration is constant (uniform), we can use the rotational kinematic equations. Angular acceleration is positive when angular speed increases and negative (angular deceleration) when it decreases. The SI unit is rad/s².
In this problem, the flywheel speeds up uniformly from 600 rpm to 1200 rpm in 10 seconds — a classic uniform angular acceleration problem perfectly suited for kinematic equations.
For motion with constant angular acceleration, the three standard kinematic equations (analogous to linear kinematics) are:
\(\omega = \omega_0 + \alpha t\)
\(\theta = \omega_0 t + \dfrac{1}{2}\alpha t^2\)
\(\omega^2 = \omega_0^2 + 2\alpha\theta\)
In these equations, θ is in radians, ω and ω₀ are in rad/s, α is in rad/s², and t is in seconds. Never mix units — always convert rpm to rad/s before using these equations.
Every quantity and equation in linear motion has a direct rotational analogue. Understanding this analogy deeply simplifies NEET problems:
📌 Linear displacement (s) → Angular displacement (θ, rad)
📌 Linear velocity (v, m/s) → Angular velocity (ω, rad/s)
📌 Linear acceleration (a, m/s²) → Angular acceleration (α, rad/s²)
📌 Mass (m, kg) → Moment of inertia (I, kg·m²)
📌 Force (F, N) → Torque (τ, N·m)
📌 Linear momentum (p = mv) → Angular momentum (L = Iω)
This one-to-one analogy means that if you understand linear kinematics thoroughly, rotational kinematics follows the exact same logic — just with different variable names and units.
A flywheel is a heavy, massive rotating disc or wheel specifically designed to store rotational kinetic energy. Due to its large moment of inertia, it resists sudden changes in angular speed — making it ideal for applications that require smooth and steady power output. Flywheels are used in internal combustion engines, locomotives, power plants, and even in spacecraft attitude control systems.
The kinetic energy stored in a rotating flywheel is \(KE = \frac{1}{2}I\omega^2\), where I is the moment of inertia and ω is the angular velocity. The larger the moment of inertia and the higher the angular speed, the more energy is stored. When the engine delivers power in pulses (as in a piston engine), the flywheel absorbs energy during the power stroke and releases it during other strokes, thereby maintaining a near-constant angular speed.
For a particle moving in a circle of radius r, the relationship between linear and angular quantities is: linear velocity \(v = r\omega\), linear acceleration (tangential) \(a_t = r\alpha\), and centripetal acceleration \(a_c = r\omega^2 = v^2/r\). These relationships are crucial for problems involving points on a rotating body where both rotational and translational aspects are asked.
The most common mistakes students make: (1) Forgetting to convert rpm to rad/s before applying formulas. (2) Confusing angular displacement (in radians) with number of revolutions — always divide θ by 2π to get revolutions. (3) Using linear kinematic formulas with rpm directly — this gives wrong answers. (4) Not recognising that the shortcut of using average angular velocity only works for uniform (constant) angular acceleration.