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System of Particles and Rotational Motion - Dynamics of Rotational Motion

Grade 11CBSEPhysics

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

🔑Concepts

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Torque (Moment of Force): The rotational analogue of force, defined as the cross product of the position vector and the force vector: τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}.

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Angular Momentum: For a particle, it is defined as L⃗=r⃗×p⃗\vec{L} = \vec{r} \times \vec{p}, where p⃗\vec{p} is linear momentum. For a rigid body, L=IωL = I\omega.

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Newton's Second Law in Rotation: The rate of change of angular momentum of a system is equal to the external torque acting on it, represented as τ⃗ext=dL⃗dt\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}.

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Work and Power: The work done by a torque τ\tau in rotating a body through an angle dθd\theta is dW=τdθdW = \tau d\theta. Power is the rate of doing work, P=τωP = \tau \omega.

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Conservation of Angular Momentum: If the total external torque on a system is zero (τ⃗ext=0\vec{\tau}_{ext} = 0), the total angular momentum L⃗\vec{L} remains constant. Thus, I1ω1=I2ω2I_1\omega_1 = I_2\omega_2.

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Rigid Body Equilibrium: A rigid body is in mechanical equilibrium if both the net external force is zero (∑F⃗=0\sum \vec{F} = 0 for translational equilibrium) and the net external torque is zero (∑τ⃗=0\sum \vec{\tau} = 0 for rotational equilibrium).

📐Formulae

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

τ=rFsin⁡θ\tau = rF \sin \theta

L⃗=r⃗×p⃗=m(r⃗×v⃗)\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})

τ⃗=dL⃗dt\vec{\tau} = \frac{d\vec{L}}{dt}

τ=Iα\tau = I \alpha

L=IωL = I \omega

W=∫θ1θ2τ dθW = \int_{\theta_1}^{\theta_2} \tau \, d\theta

P=τ⃗⋅ω⃗P = \vec{\tau} \cdot \vec{\omega}

Krot=12Iω2K_{rot} = \frac{1}{2} I \omega^2

💡Examples

Problem 1:

A torque of 150 N m150 \, \text{N m} is applied to a grindstone whose moment of inertia is 5 kg m25 \, \text{kg m}^2. Calculate the angular acceleration produced.

Solution:

Given: τ=150 N m\tau = 150 \, \text{N m}, I=5 kg m2I = 5 \, \text{kg m}^2. Using the formula τ=Iα\tau = I \alpha, we get: α=τI\alpha = \frac{\tau}{I} α=1505=30 rad/s2\alpha = \frac{150}{5} = 30 \, \text{rad/s}^2

Explanation:

The relationship between torque and angular acceleration is analogous to F=maF=ma. Here, torque replaces force and moment of inertia replaces mass.

Problem 2:

A ballet dancer is spinning at 2 rev/s2 \, \text{rev/s} with her arms outstretched. When she pulls her arms in, her moment of inertia decreases from 4.0 kg m24.0 \, \text{kg m}^2 to 1.6 kg m21.6 \, \text{kg m}^2. Find her new angular velocity.

Solution:

Applying the principle of conservation of angular momentum (I1ω1=I2ω2I_1\omega_1 = I_2\omega_2): Initial state: I1=4.0 kg m2I_1 = 4.0 \, \text{kg m}^2, ω1=2 rev/s\omega_1 = 2 \, \text{rev/s}. Final state: I2=1.6 kg m2I_2 = 1.6 \, \text{kg m}^2, ω2=?\omega_2 = ? 4.0×2=1.6×ω24.0 \times 2 = 1.6 \times \omega_2 8.0=1.6ω28.0 = 1.6 \omega_2 ω2=8.01.6=5 rev/s\omega_2 = \frac{8.0}{1.6} = 5 \, \text{rev/s}

Explanation:

Since no external torque acts on the dancer, the angular momentum is conserved. Reducing the moment of inertia results in an increase in angular velocity.

Problem 3:

A flywheel has an initial angular momentum of 850 kg m2/s850 \, \text{kg m}^2/\text{s}. Due to an opposing torque, its angular momentum decreases to 275 kg m2/s275 \, \text{kg m}^2/\text{s}. Calculate the change in angular momentum using vertical subtraction.

Solution:

Change in angular momentum ΔL=Linitial−Lfinal\Delta L = L_{initial} - L_{final}: 850−275575\begin{array}{r} 850 \\ - 275 \\ \hline 575 \end{array} So, ΔL=575 kg m2/s\Delta L = 575 \, \text{kg m}^2/\text{s}.

Explanation:

The change in angular momentum is the difference between the initial and final states, which is also equal to the impulse of the torque.

Dynamics of Rotational Motion Class 11 Notes & Examples