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System of Particles and Rotational Motion - Angular Velocity and its Relation with Linear Velocity

Grade 11CBSEPhysics

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

🔑Concepts

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In a rigid body rotating about a fixed axis, every particle of the body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis.

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Angular Displacement (θ\theta): The angle swept by the radius vector of a particle moving in a circular path. 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. For a rigid body, all particles have the same angular velocity at any instant.

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The direction of angular velocity ω⃗\vec{\omega} is along the axis of rotation and is determined by the Right-Hand Thumb Rule.

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Relation between Linear and Angular Velocity: For a particle at a perpendicular distance rr from the axis of rotation, its linear speed vv is given by the product of the distance and the angular speed.

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Vector Form: The linear velocity v⃗\vec{v} of a particle at position r⃗\vec{r} relative to an origin on the axis of rotation is given by the cross product of ω⃗\vec{\omega} and r⃗\vec{r}.

📐Formulae

ω=dθdt\omega = \frac{d\theta}{dt}

v=rωv = r \omega

v⃗=ω⃗×r⃗\vec{v} = \vec{\omega} \times \vec{r}

ω=2πT=2πν\omega = \frac{2\pi}{T} = 2\pi \nu

α=dωdt\alpha = \frac{d\omega}{dt}

at=rαa_t = r \alpha

💡Examples

Problem 1:

A ceiling fan has blades of length 0.5 m0.5 \text{ m} and is rotating at 120 rpm120 \text{ rpm}. Calculate the angular velocity of the fan and the linear velocity of the tip of a blade.

Solution:

Given: Frequency ν=120 rpm=12060 rps=2 Hz\nu = 120 \text{ rpm} = \frac{120}{60} \text{ rps} = 2 \text{ Hz}. Radius r=0.5 mr = 0.5 \text{ m}.

  1. Angular velocity: ω=2πν=2×π×2=4π rad/s≈12.57 rad/s\omega = 2\pi \nu = 2 \times \pi \times 2 = 4\pi \text{ rad/s} \approx 12.57 \text{ rad/s}
  2. Linear velocity of the tip: v=rω=0.5×4π=2π m/s≈6.28 m/sv = r\omega = 0.5 \times 4\pi = 2\pi \text{ m/s} \approx 6.28 \text{ m/s}

Explanation:

The angular velocity is calculated using the frequency of rotation. The linear velocity of any point on the blade depends on its distance from the axis, so the tip (at maximum rr) has the maximum linear velocity.

Problem 2:

Find the linear velocity v⃗\vec{v} of a particle whose angular velocity is ω⃗=3i^−4j^+k^\vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k} and position vector is r⃗=5i^−6j^+6k^\vec{r} = 5\hat{i} - 6\hat{j} + 6\hat{k}.

Solution:

Using the vector relation v⃗=ω⃗×r⃗\vec{v} = \vec{\omega} \times \vec{r}: v⃗=∣i^j^k^3−415−66∣\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 1 \\ 5 & -6 & 6 \end{vmatrix} v⃗=i^[(−4)(6)−(1)(−6)]−j^[(3)(6)−(1)(5)]+k^[(3)(−6)−(−4)(5)]\vec{v} = \hat{i}[(-4)(6) - (1)(-6)] - \hat{j}[(3)(6) - (1)(5)] + \hat{k}[(3)(-6) - (-4)(5)] v⃗=i^[−24+6]−j^[18−5]+k^[−18+20]\vec{v} = \hat{i}[-24 + 6] - \hat{j}[18 - 5] + \hat{k}[-18 + 20] v⃗=−18i^−13j^+2k^\vec{v} = -18\hat{i} - 13\hat{j} + 2\hat{k}

Explanation:

The linear velocity is the cross product of the angular velocity vector and the position vector. The resulting vector is perpendicular to both ω⃗\vec{\omega} and r⃗\vec{r}.