5.3 Torque

Torque is defined as the moment of force, that is, the product of the distance from a reference point and the force. In terms of vectors, this is given by the cross product

𝝉=𝒓×𝑭. (5.32)

The magnitude of 𝝉 is given by

τ =|𝒓||𝑭|sinθ (5.33)
=rFtan. (5.34)

This formula tells us that for a fixed radius, the torque has maximum magnitude when θ=±π2 i.e. the force acts perpendicular to the radius. It also tells us that if the force acts along the same line as the radius (θ=0 or θ=π), then the torque is equal to 0. Since torque is defined as a moment like angular momentum, its value is relative to an arbitrary reference point. Changing the reference point will change the value of the torque. For a fixed reference point, if the same force is applied further away from the centre, the torque will be greater.

5.3.1 Changing the Origin

Let’s look at how changing the reference point changes the angular momentum and the torque. Consider the following diagram.

yx𝒓𝒑𝑭𝑨𝒅

The angular momentum and torque relative to the origin are 𝑳=𝒓×𝒑 and 𝝉=𝒓×𝑭 like before, but what are the angular momentum and torque relative to the point 𝑨? The radius vector or lever arm is now the vector 𝒅 between the points 𝑨 and 𝒓, i.e.

𝑳𝑨 =𝒅×𝒑 (5.35)
=(𝒓𝑨)×𝒑 (5.36)
𝝉𝑨 =𝒅×𝑭 (5.37)
=(𝒓𝑨)×𝑭. (5.38)
Example 5.3.

Consider a 2 kg particle at y=3 m moving to the right with velocity 𝒗=5 m s1ı^. What is its angular momentum relative to the origin and relative to the point 𝑨=3 mȷ^.

yx𝑨𝒗𝒅𝒓

The position vector is 𝒓=xı^+3 mȷ^ and the radius vector relative to 𝑨 is 𝒅=xı^. The momentum vector is 𝒑=mvxı^=10 kg m s1, so the angular momentum relative to the origin is

𝑳 =𝒓×𝒑=|ı^ȷ^𝒌^x301000| (5.39)
=30 kg m2 s1𝒌^, (5.40)

at the angular momentum relative to 𝑨 is

𝑳𝑨 =𝒅×𝒑=|ı^ȷ^𝒌^x001000| (5.41)
=0. (5.42)

This makes sense because relative to 𝑨 the motion is radial so there is no lever arm.

Example 5.4.

Let’s now look at a familiar example, uniform circular motion. First, what is the angular momentum relative to the origin for a particle travelling with angular speed ω?

Working in polar coordinates, the radius vector is 𝒓=r𝒓^ and the velocity vector is 𝒗=rω𝜽^. Then the angular momentum is 𝑳=r𝒓^×mrω𝜽^=mr2ω𝒌^. This is a constant, which we know to be the case since the centripetal force is a central force.

Now, suppose the particle is moving around a circle at y=h. What is the angular momentum relative to the origin?

The position vector is now given by 𝒓=r𝒓^+h𝒌^, so the angular momentum becomes

𝑳 =m|𝒓^𝜽^𝒌^r0h0rω0| (5.43)
=mhrω𝒓^+mr2ω𝒌^. (5.44)

The unit vector 𝒓^ changes over time, so 𝑳 is not constant (its magnitude is constant but its direction changes). This is because the centripetal force is no longer central relative to the origin.

5.3.2 Angular Version of Newton’s Second Law

Torque is measured in N m. This is dimensionally equivalent to Joules, but this does not mean that torque is a kind of energy. Energy is a scalar quantity, whereas torque is a vector, so they are really different things.

In linear dynamics, we have Newton’s second law which states 𝑭=𝒑˙. We will now show that there is a rotational equivalent of Newton II that relates torque to angular momentum. Using the product rule to differentiate 𝑳, we find

d𝑳dt =ddt(𝒓×𝒑) (5.45)
=(𝒓˙×𝒑)+(𝒓×𝒑˙) (5.46)
=(1m𝒑×𝒑)+(𝒓×𝑭) (5.47)
=𝒓×𝑭=𝝉. (5.48)

This is the same calculation we did earlier for central forces. We now have a shortcut to prove that angular momentum is conserved under the influence of a central force. Since central forces are always parallel to the radius vector, the torque is zero.

𝝉=𝒓×F(r)𝒓^=0. (5.49)

Then using the angular equivalent of Newton II, 𝑳˙=𝝉=0. So central forces exert no torque and therefore angular momentum is conserved.