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Rotational Kinetics

Rotational Kinematics

Problems

Rotational Kinematics, page 2

page 1 of 2

In this section we will use our new definitions for rotational variables to generate kinematic equations for rotational motion. In addition, we will examine the vector nature of rotational variables and, finally, relate linear and angular variables.

Kinematic Equations

Because our equations defining rotational and translational variables are mathematically equivalent, we can simply substitute our rotational variables into the kinematic equations we have already derived for translational variables. We could go through the formal derivation of these equations, but they would be the same as those derived in One-Dimensional Kinematics. Thus we can simply state the equations, alongside their translational analogues:


v f = v o + at   σ f = σ o + αt  
x f = x o + v o t + at 2   μ f = μ o + σ o t + αt 2  
v f 2 = v o 2 + 2ax   σ f 2 = σ o 2 +2αμ  
x = (v o + v f)t   μ = (σ o + σ f)t  

These equations for rotational motion are used identically as the corollary equations for translational motion. In addition, like translational motion, these equations are only valid when the acceleration, α , is constant. These equations are frequently used and form the basis for the study of rotational motion.

Relationships Between Rotational and Translational Variables

Now that we have established both equations for our variables and kinematic equations relating them, we can also relate our rotational variables to translational variables. This can sometimes be confusing. It is easy to think that because a particle is engaged in rotational motion, it is not also defined by translational variables. Simply remind yourself that no matter what path a given particle is traveling in, it always has a position, velocity and acceleration. The rotational variables we generated do not substitute for these traditional variables; instead, they simplify calculations involving rotational motion. Thus we can relate our rotational and translational variables.

Translational and Angular Displacement

Recall from our definition of angular displacement that:

μ = s/r

Implying that

s = μr    

Thus the displacement, s , of a particle in rotational motion is given by the angular displacement multiplied by the radius of the particle from the axis of rotation. We can differentiate both sides of the equation with respect to time:

=

Thus

v = σr    

Translational and Angular Velocity

Just as linear displacement is equal to angular displacement times the radius, linear velocity is equal to angular velocity times the radius. We can relate α and a , by the same method we used before: differentiating with respect to time.

= r

Translational and Angular Acceleration

We must be careful in relating translationa and angular acceleration because only gives us the change in velocity with respect to time in the tangential direction. We know from Dynamics that any particle traveling in a circle experiences a radial force equal to . We must therefore generate two different expressions for the linear acceleration of a particle in rotational motion:


a T = αr  
a R =  
  = σ 2 r  

These two equations may seem a bit confusing, so we shall examine them closely. Consider a particle moving around a circle with a constant speed. The rate at which the particle makes a revolution about the axis is constant, so α = 0 and a T = 0 . However, the particle is being constantly accelerated towards the center of the circle, so a R is nonzero, and varies with the square of the angular velocity of the particle.

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