When studying how objects rotate, it quickly becomes necessary to figure out how a given force results in a change in the rotational motion. The tendency of a force to cause or change rotational motion is called torque, and it's one of the most important concepts to understand in resolving rotational motion situations.

### What Is Torque?

Torque (also called moment — mostly by engineers) is calculated by multiplying force and distance.

The SI units of torque are newton-meters, or N*m (even though these units are the same as Joules, torque isn't work or energy, so should just be newton-meters).

In calculations, torque is represented by the Greek letter tau: *τ*.

Torque is a vector quantity, meaning it has both a direction and a magnitude. This is honestly one of the trickiest parts of working with torque because it is calculated using a vector product, which means you have to apply the right-hand rule. In this case, take your right hand and curl the fingers of your hand in the direction of rotation caused by the force. The thumb of your right hand now points in the direction of the torque vector. (This can occasionally feel slightly silly, as you're holding your hand up and pantomiming in order to figure out the result of a mathematical equation, but it's the best way to visualize the direction of the vector.)

The vector formula that yields the torque vector * τ* is:

=τr×F

The vector ** r** is the position vector with respect to an origin on the axis of rotation (This axis is the

*τ*on the graphic). This is a vector with a magnitude of the distance from where the force is applied to the axis of rotation. It points from the axis of rotation toward the point where the force is applied.

The magnitude of the vector is calculated based upon *θ*, which is the angle difference between ** r** and

*, using the formula:*

**F**τ=rFsin(θ)

### Special Cases of Torque

A couple of key points about the above equation, with some benchmark values of *θ*:

*θ*= 0° (or 0 radians) - The force vector is pointing out in the same direction as. As you might guess, this is a situation where the force will not cause any rotation around the axis ... and the mathematics bears this out. Since sin(0) = 0, this situation results in**r***τ*= 0.*θ*= 180° (or*π*radians) - This is a situation where the force vector points directly into. Again, shoving toward the axis of rotation isn't going to cause any rotation either and, once again, the mathematics supports this intuition. Since sin(180°) = 0, the value of the torque is once again**r***τ*= 0.*θ*= 90° (or*π*/2 radians) - Here, the force vector is perpendicular to the position vector. This seems like the most effective way that you could push on the object to get an increase in rotation, but does the mathematics support this? Well, sin(90°) = 1, which is the maximum value that the sine function can reach, yielding a result of*τ*=*rF*. In other words, a force applied at any other angle would provide less torque than when it is applied at 90 degrees.

- The same argument as above applies to cases of
*θ*= -90° (or -*π*/2 radians), but with a value of sin(-90°) = -1 resulting in the maximum torque in the opposite direction.

### Torque Example

Let's consider an example where you're applying a vertical force downward, such as when trying to loosen the lug nuts on a flat tire by stepping on the lug wrench. In this situation, the ideal situation is to have the lug wrench perfectly horizontal, so that you can step on the end of it and get the maximum torque. Unfortunately, that doesn't work. Instead, the lug wrench fits onto the lug nuts so that it is at a 15% incline to the horizontal. The lug wrench is 0.60 m long until the end, where you apply your full weight of 900 N.

What is the magnitude of the torque?

What about direction?:Applying the "lefty-loosey, righty-tighty" rule, you will want to have the lug nut rotating to the left - counter-clockwise - in order to loosen it. Using your right hand and curling your fingers in the counter-clockwise direction, the thumb sticks out. So the direction of the torque is away from the tires ... which is also direction you want the lug nuts to ultimately go.

To begin calculating the value of the torque, you have to realize that there's a slightly misleading point in the above set-up. (This is a common problem in these situations.) Note that the 15% mentioned above is the incline from the horizontal, but that's not the angle *θ*. The angle between * r* and

*has to be calculated. There's a 15° incline from the horizontal plus a 90° distance from the horizontal to the downward force vector, resulting in a total of 105° as the value of*

**F***θ*.

That's the only variable that requires set-up, so with that in place we just assign the other variable values:

*θ*= 105°*r*= 0.60 m*F*= 900 N

τ=rFsin(θ) =

(0.60 m)(900 N)sin(105°) = 540 × 0.097 Nm = 520 Nm

Note that the above answer involved maintaining only two significant figures, so it is rounded.

### Torque and Angular Acceleration

The above equations are particularly helpful when there is a single known force acting on an object, but there are many situations where a rotation can be caused by a force that cannot easily be measured (or perhaps many such forces). Here, the torque often isn't calculated directly, but can instead be calculated in reference to the total angular acceleration, *α*, that the object undergoes. This relationship is given by the following equation:

Στ=Iα

where the variables are:

- Σ
τ- The net sum of all torque acting on the objectI- the moment of inertia, which represents the object's resistance to a change in angular velocityα- angular acceleration