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System of Particles and Rotational Motion - Kinematics of Rotational Motion

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Angular Displacement (θ\theta): The angle through which a point or line has been rotated about a specified axis. It is measured in radians (radrad).

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Angular Velocity (ω\omega): The rate of change of angular displacement with respect to time. ω=dθdt\omega = \frac{d\theta}{dt}. Its SI unit is rad/srad/s.

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Angular Acceleration (α\alpha): The rate of change of angular velocity with respect to time. α=dωdt=d2θdt2\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}. Its SI unit is rad/s2rad/s^2.

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Relation between Linear and Angular Kinematics: For a particle moving in a circle of radius rr, linear velocity v=rωv = r\omega and tangential acceleration at=rαa_t = r\alpha.

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Equations of Rotational Motion: These are analogous to linear kinematic equations, applicable when angular acceleration α\alpha is constant.

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Uniform Circular Motion: A special case where α=0\alpha = 0, meaning ω\omega remains constant. Here, centripetal acceleration exists (ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r), but tangential acceleration is zero.

📐Formulae

ω=dθdt\omega = \frac{d\theta}{dt}

α=dωdt\alpha = \frac{d\omega}{dt}

v=rωv = r\omega

at=rαa_t = r\alpha

ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

ω=ω0+αt\omega = \omega_0 + \alpha t

θ=ω0t+12αt2\theta = \omega_0 t + \frac{1}{2} \alpha t^2

ω2=ω02+2αθ\omega^2 = \omega_0^2 + 2\alpha \theta

θn=ω0+α2(2n−1)\theta_n = \omega_0 + \frac{\alpha}{2}(2n - 1)

💡Examples

Problem 1:

A ceiling fan is rotating at 600 rpm600\text{ rpm} (rotations per minute). When it is switched off, it comes to rest in 10 s10\text{ s}. Find the angular acceleration α\alpha assuming it to be uniform.

Solution:

Given initial frequency ν0=600 rpm=60060=10 rev/s\nu_0 = 600\text{ rpm} = \frac{600}{60} = 10\text{ rev/s}. Initial angular velocity ω0=2πν0=2π×10=20π rad/s\omega_0 = 2\pi\nu_0 = 2\pi \times 10 = 20\pi\text{ rad/s}. Final angular velocity ω=0\omega = 0 (at rest). Time t=10 st = 10\text{ s}. Using the equation: ω=ω0+αt\omega = \omega_0 + \alpha t 0=20π+α(10)0 = 20\pi + \alpha(10) 10α=−20π10\alpha = -20\pi α=−2π rad/s2\alpha = -2\pi\text{ rad/s}^2

Explanation:

The negative sign indicates angular retardation. We first convert frequency from rpm to rev/s and then calculate angular velocity in rad/s before applying the kinematic equation.

Problem 2:

A wheel starts from rest and accelerates with a constant angular acceleration of 4 rad/s24\text{ rad/s}^2. Calculate the angular displacement after 5 s5\text{ s} and the number of revolutions made.

Solution:

Given ω0=0\omega_0 = 0, α=4 rad/s2\alpha = 4\text{ rad/s}^2, and t=5 st = 5\text{ s}. Using θ=ω0t+12αt2\theta = \omega_0 t + \frac{1}{2} \alpha t^2 θ=0(5)+12(4)(5)2\theta = 0(5) + \frac{1}{2}(4)(5)^2 θ=2×25=50 rad\theta = 2 \times 25 = 50\text{ rad} To find the number of revolutions (nn): n=θ2π=502×3.14n = \frac{\theta}{2\pi} = \frac{50}{2 \times 3.14} n≈7.96 revolutionsn \approx 7.96\text{ revolutions}

Explanation:

Since acceleration is constant, we use the second equation of rotational motion to find θ\theta. To convert radians to revolutions, we divide by 2π2\pi because one full revolution equals 2π2\pi radians.