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Space, Time and Motion - Rigid Body Mechanics

Grade 11IBPhysics

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

🔑Concepts

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A rigid body is an extended object in which the distance between any two given points remains constant, regardless of external forces acting on it.

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Torque (τ\tau), or the moment of a force, is the rotational equivalent of force. It is defined as the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force: τ=rFsin⁡θ\tau = rF \sin \theta.

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The Moment of Inertia (II) is a measure of an object's resistance to rotational acceleration, analogous to mass in linear motion. It depends on the distribution of mass relative to the axis: I=∑miri2I = \sum m_i r_i^2.

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Newton's Second Law for Rotation states that the net external torque acting on a system is proportional to the angular acceleration: ∑τ=Iα\sum \tau = I\alpha.

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For a rigid body to be in static equilibrium, two conditions must be met: the vector sum of all external forces must be zero (∑F⃗=0\sum \vec{F} = 0) and the vector sum of all external torques about any axis must be zero (∑τ⃗=0\sum \vec{\tau} = 0).

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Angular Momentum (LL) is the rotational analog of linear momentum. For a rigid body rotating about a fixed axis, L=IωL = I\omega.

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The Principle of Conservation of Angular Momentum states that if the net external torque acting on a system is zero, the total angular momentum of the system remains constant (I1ω1=I2ω2I_1\omega_1 = I_2\omega_2).

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Rotational Kinetic Energy is the energy an object possesses due to its rotation: Ek=12Iω2E_k = \frac{1}{2}I\omega^2.

📐Formulae

τ=rFsin⁡θ\tau = rF \sin \theta

I=∑mr2I = \sum mr^2

τnet=Iα\tau_{net} = I\alpha

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

L=IωL = I\omega

Erot=12Iω2E_{rot} = \frac{1}{2}I\omega^2

W=τΔθW = \tau \Delta \theta

💡Examples

Problem 1:

A uniform thin rod of length L=2.0 mL = 2.0\text{ m} and mass M=3.0 kgM = 3.0\text{ kg} is pivoted at one end. A force of 15 N15\text{ N} is applied perpendicularly at the other end. Calculate the initial angular acceleration α\alpha of the rod. (Moment of inertia of a rod pivoted at one end is I=13ML2I = \frac{1}{3}ML^2)

Solution:

  1. Calculate II: I=13(3.0)(2.0)2=4.0 kg m2I = \frac{1}{3}(3.0)(2.0)^2 = 4.0\text{ kg m}^2.
  2. Calculate τ\tau: τ=rF=(2.0)(15)=30 N m\tau = rF = (2.0)(15) = 30\text{ N m}.
  3. Use τ=Iα\tau = I\alpha: α=τI=304.0=7.5 rad s−2\alpha = \frac{\tau}{I} = \frac{30}{4.0} = 7.5\text{ rad s}^{-2}.

Explanation:

The torque is calculated using the full length of the rod as the lever arm because the force is at the end. We then apply the rotational version of Newton's second law.

Problem 2:

An ice skater is spinning with an angular velocity of 5.0 rad s−15.0\text{ rad s}^{-1} with her arms extended. Her moment of inertia in this position is 4.5 kg m24.5\text{ kg m}^2. She then pulls her arms in, decreasing her moment of inertia to 1.5 kg m21.5\text{ kg m}^2. What is her new angular velocity ωf\omega_f?

Solution:

  1. Use Conservation of Angular Momentum: Iiωi=IfωfI_i\omega_i = I_f\omega_f.
  2. Substitute values: (4.5)(5.0)=(1.5)(ωf)(4.5)(5.0) = (1.5)(\omega_f).
  3. Solve for ωf\omega_f: ωf=22.51.5=15.0 rad s−1\omega_f = \frac{22.5}{1.5} = 15.0\text{ rad s}^{-1}.

Explanation:

Since no external torque acts on the skater, angular momentum is conserved. Decreasing the moment of inertia results in a proportional increase in angular velocity.

Rigid Body Mechanics - Revision Notes & Key Formulas | IB Grade 11 Physics