You are viewing an old version of this page. View the current version.

Compare with Current View Page History

« Previous Version 36 Next »

Unknown macro: {table}
Unknown macro: {tr}
Unknown macro: {td}
Error formatting macro: live-template: java.lang.NullPointerException
Unknown macro: {td}

Torque (One-Dimensional)

An interaction which has the potential to produce a change in the rotational velocity of a system about a specified axis.


MotivationforConcept"> Motivation for Concept

ConditionsforOne-DimensionalTorque"> Conditions for One-Dimensional Torque

DefinitionofTorque"> Definition of Torque

CrossProduct"> Cross Product

Defining the torque resulting from a force requires two pieces of information:

  1. The force applied (magnitude and direction).
  2. The position (magnitude and direction) of the force's application with respect to the axis of rotation about which the torque is to be calculated.

The torque is most succinctly defined by using the vector cross product:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[\tau_

Unknown macro: {z}

\equiv \vec

Unknown macro: {r}

\times \vec

Unknown macro: {F}

]\end

where τ is the torque, r is the position of the point of application of the force with respect to the axis of rotation, and F is the force. Note that since the torque is assumed to lie in the +z or -z direction, we have specified its vector nature with the z subscript rather than a vector arrow.


Magnitude"> Magnitude

In two dimensions, this formula is equivalent to:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ |\tau_

Unknown macro: {z}

| = rF\sin\theta]\end

where the angle θ is the angle between the position vector and the force vector.

That angle should technically be found by extending the position vector and then taking the smaller angle to the force as shown here.

The given vectors.

Step 1: Extend position vector.

Step 2: Choose the smaller angle to the force.

It is often helpful to make use of the fact that the sine of supplementary angles is the same, and therefore the closer angle between the position vector and force can also be used without extension as shown here.

The given vectors.

Simply take the smaller angle between them.


Direction"> Direction

Note that our second form of the equation for torque gives the magnitude only. To define the direction, the vectors r and F must be examined to determine the sense of the rotation. Since we are restricted to one-dimensional torques, we can describe the sense of the rotation as clockwise or counterclockwise. Examples of forces producing clockwise and counterclockwise torques are shown here.

Counterclockwise

Clockwise

When constructing free-body diagrams for systems in which torques are of interest, it is important to draw them from the perspective of someone looking along the axis of rotation and to assign a mathematical sign (+ or -) to each sense of rotation. This is usually done as shown below.

Counterclockwise rotation defined to be positive.

Clockwise rotation defined to be positive.

If you are using explicit vector cross products to calculate the torque, then technically you must choose the positive sense of rotations to correspond with the positive z-axis and you must take care to construct a right-handed coordinate system. If you fail to do this, you will encounter sign errors when computing your cross products (the mathematical technique used to compute cross products assumes that a right-handed coordinate system is employed).



ParsingtheMagnitude"> Parsing the Magnitude

It is sometimes useful to associate the sinθ portion of the formula for the magnitude of the one-dimensional torque with either the force or the position. These two possible associations lead to two terms that are often used in describing the rotational effects of a force: tangential force and moment arm.

TangentialForce"> Tangential Force

Suppose we group the formula for the magnitude of the torque in the following way:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ |\tau_

Unknown macro: {z}

| = r (F\sin\theta) \equiv r F_

Unknown macro: {perp}

]\end

where we have defined the tangential component of F as F sinθ. This name is chosen because, as shown in the pictures below, F sinθ is the size of the part of F that is directed perpendicular to r.


MomentArm"> Moment Arm

If we instead group the formula as:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ |\tau_

Unknown macro: {z}

|= F (r \sin\theta) \equiv F r_

Unknown macro: {perp}

] \end

where we have defined a perpendicular component of r. In this case, we give the component the special name of moment arm. The moment arm can be thought of as the closest distance of approach of the line of action of the force to the axis of rotation, as is shown in the pictures below.

Error formatting macro: live-template: java.lang.NullPointerException
  • No labels