Rotational Dynamics

Rotational dynamics is the study of accelerated rotational motion.

Relationship between torque and Angular Acceleration

Let's Consider an object of mass m rotating in a circular path of radius r⊥ with an angular acceleration of α under the influence of a tangential force Ft. The torque acting on the object about the center of the circle is equal to the product of the tangential force and the perpendicular distance between the force and the center of the circle: τ = Ft r⊥ . And from Newton's second law: Ft = mat. But at = r⊥ α . Therefore

τ = mr⊥ 2α

The product mr⊥ 2 is called the moment of inertial of the object about the given axis and denoted by I.

I = mr⊥ 2

The unit of measurement for moment of inertia is kg m 2. The relationship between torque and angular acceleration is that torque is proportional to angular acceleration with the constant of proportionality being the moment of inertia of the object about the given axis. Moment of inertia of an object is a constant for a given axis of rotation but different for different axes of rotation.

τ = Iα

For a system involving more than one particle, the moment of inertia of the system is equal to the sum of the moment of inertias of the individual particles.

I = ∑i mi ri⊥ 2 = m1 r1⊥ 2 + m2 r2⊥ 2 + . . .

  • Calculate the moment of inertia of the system about the y-axis.

    Solution: If the y-axis is the axis of rotation, then the perpendicular distance between the axis and a particle is equal to the absolute value of the x-coordinate of the particle.

    m1 = 6 kg ; r1⊥ = 2 m ; m2 = 4 kg ; r2⊥ = 2 m ; m3 = 5 kg ; r3⊥ = 0 m ; Ix = ?

    Iy = m1 r1⊥ 2 + m2 r2⊥ 2 + m3 r3⊥ 2 = ( 6 * 2 2 + 4 * 2 2 + 5 * 0 2 ) kg m 2 = 40 kg m 2

  • If the system is rotating about the y-axis with an angular acceleration of 7 rad ⁄ s 2, calculate the torque acting on these system of particles.

    Solution: Iy = 40 kg m 2 ; α = 7 rad ⁄ s 2 ; τ = ?

    τ = Iy α = 40 * 7 N m = 280 N m

    • Obtaining moment of inertia of solid objects requires the use of calculus. But formulas for the moment of inertia of common shapes such as cylinder and sphere may be found in physics text books. For example the moment of inertia of a spherical object of mass M and radius R about an axis through the center of the sphere is given by Isphere = 2MR 2 ⁄ 5 .

    Rotational Kinetic Energy

    Let's consider an object of mass m revolving in a circular path of radius r⊥ with a speed of v. This motion can be looked at as linear motion in the circular trajectory or rotational motion about the center. Its linear kinetic energy which should be equal to the rotational kinetic energy is given by KE = mv 2 ⁄ 2 . But the linear speed v can be expressed in terms of the angular speed ω as v = r⊥ ω; and the kinetic energy may be written as KE = mr⊥ 2ω 2 ⁄ 2. The expression mr⊥ 2 is the moment of inertia, I, of the object. Therefore the rotational kinetic energy ( KErot ) of the object is given by the following expression.

    KErot = Iω 2 ⁄ 2

    Kinetic Energy of a Rolling Object

    A rolling object has both translational and rotational kinetic energy because it rotates as it moves. The kinetic energy of a rolling object is the sum of its translational ( KEtra ) and rotational kinetic energy ( KErot ). If a rolling object is moving with a speed of v and rotating with angular speed ω, its kinetic energy ( KErol ) is given by

    KErol = KEtra + KErot = Mv 2 ⁄ 2 + Iω 2 ⁄ 2

    The translational speed, v, and rotational speed, ω, are related: v = Rω. R is the radius of rotation.

    Example: A spherical object of radius 0.2 m is rolling down a 10 m 30° inclined plane. Calculate its speed by the time it reaches the ground.

    Solution: The forces acting on the object are gravity and friction. Eventhough friction is non-conservative, it doesn't contribute to the work done because each particle of the sphere is only instantaneously in contact with the plane. Thus the principle of conservation of mechanical energy can be applied. Let the origin of the coordinate system be fixed at the ground.

    yf = 0 ; vi = ωi = 0 ; yi = 10 * sin 30° m = 5 m ; R = 0.2 ; vf = ?

    mvi 2 ⁄ 2 + Iωi 2 ⁄ 2 + m|g|yi = mvf 2 ⁄ 2 + Iωf 2 ⁄ 2 + m|g|yf

    m|g|yi = mvf 2 ⁄ 2 + ( 2mR 2 ⁄ 5 )(vf ⁄ R ) 2 ⁄ 2

    vf = √( 10|g|yi ⁄ 7 ) = 10 * 9.8 * 5 ⁄ 7 m ⁄ s = m ⁄ s

    Angular Momentum

    Angular momentum is a physical quantity used as a measure of the rotational motion an object has. It is defined to be the product of the moment of inertia of the object and its angular speed.

    L = Iω

    L is the angular momentum of an object of moment of inertia I rotating with an angular speed ω.

    Principle of Conservation of Angular Momentum

    Since τnet = Iα and α = Δω ⁄ Δt, τnet = IΔω ⁄ Δt = Δ(Iω) ⁄ Δt. Torque is equal to the rate of change of angular momentum with time.

    τnet = ΔL ⁄ Δt

    If the net torque acting on a system is zero, it follows that ΔL = 0 or Li = Lf. That is, the angular momentum of the system does not change. The Principle of conservation of angular momentum states that if the net torque acting on a system is zero, then its angular momentum is conserved.

    If

    τnet = 0

    Then

    Li = Lf

    or

    Iiωi = Ifωf