krit.club logo

Space, Time and Motion - Rigid Body Mechanics

Grade 12IBPhysics

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

🔑Concepts

•

Torque τ\tau is the rotational analogue of force, defined as the product of the force and the perpendicular distance from the axis of rotation: τ=rFsin⁡θ\tau = rF\sin\theta.

•

Moment of Inertia II represents a body's resistance to angular acceleration. For a point mass, I=mr2I = mr^2, and for rigid bodies, it depends on the distribution of mass relative to the axis.

•

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

•

A rigid body is in translational equilibrium if ∑F⃗=0\sum \vec{F} = 0 and in rotational equilibrium if ∑τ⃗=0\sum \vec{\tau} = 0.

•

Angular Momentum LL of a rigid body is given by L=IωL = I\omega. In the absence of an external torque, the total angular momentum of a system remains constant (ΔL=0\Delta L = 0).

•

The total kinetic energy of a rolling object (without slipping) is the sum of its translational kinetic energy and rotational kinetic energy: Ek=12mv2+12Iω2E_k = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2.

📐Formulae

τ=rFsin⁡θ\tau = rF\sin\theta

I=∑miri2I = \sum m_i r_i^2

τnet=Iα\tau_{net} = I\alpha

L=IωL = I\omega

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

ωf=ωi+αt\omega_f = \omega_i + \alpha t

θ=ωit+12αt2\theta = \omega_i t + \frac{1}{2}\alpha t^2

v=ωrv = \omega r

💡Examples

Problem 1:

A uniform thin rod of mass M=1.5 kgM = 1.5\text{ kg} and length L=2.0 mL = 2.0\text{ m} is free to rotate about a frictionless pivot at one end. If the rod is released from a horizontal position, calculate its initial angular acceleration α\alpha. (Note: Irod=13ML2I_{rod} = \frac{1}{3}ML^2 for rotation about the end).

Solution:

The weight of the rod acts at its center of mass, which is at distance r=L2r = \frac{L}{2} from the pivot. The torque is τ=Mg(L2)\tau = Mg(\frac{L}{2}). Using τ=Iα\tau = I\alpha, we get Mg(L2)=(13ML2)αMg(\frac{L}{2}) = (\frac{1}{3}ML^2)\alpha. Solving for α\alpha: α=3g2L=3×9.812×2.0=7.36 rad s−2\alpha = \frac{3g}{2L} = \frac{3 \times 9.81}{2 \times 2.0} = 7.36\text{ rad s}^{-2}.

Explanation:

The torque is calculated using the weight of the rod acting at its geometric center. We then equate this to the product of the moment of inertia and angular acceleration.

Problem 2:

An ice skater spins with an initial angular velocity ωi=5.0 rad s−1\omega_i = 5.0\text{ rad s}^{-1} and a moment of inertia Ii=4.0 kg m2I_i = 4.0\text{ kg m}^2. She pulls her arms in, reducing her moment of inertia to If=1.6 kg m2I_f = 1.6\text{ kg m}^2. Calculate her new angular velocity ωf\omega_f.

Solution:

By conservation of angular momentum, Iiωi=IfωfI_i\omega_i = I_f\omega_f. Substituting the values: (4.0)(5.0)=(1.6)ωf(4.0)(5.0) = (1.6)\omega_f. Therefore, ωf=201.6=12.5 rad s−1\omega_f = \frac{20}{1.6} = 12.5\text{ rad s}^{-1}.

Explanation:

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

Rigid Body Mechanics - Revision Notes & Key Formulas | IB Grade 12 Physics