Newton’s second law written in the form
expresses the close relation between force and linear momentum for a single particle. We have seen enough of the parallelism between linear and angular quantities to be pretty sure that there is also a close relation between torque and angular momentum. Guided by Eq. 11-22, we can even guess that it must be
Equation 11-23 is indeed an angular form of Newton’s second law for a single particle:
The (vector) sum of all the torques acting on a particle is equal to the time rate of change of the angular momentum of that particle.
Equation 11-23 has no meaning unless the torques 
and the angular momentum 
are defined with respect to the same origin.
Proof of Equation 11-23
We start with Eq. 11-18, the definition of the angular momentum of a particle:
where 
is the position vector of the particle and 
is the velocity of the particle. Differentiating* each side with respect to time t yields
However, 
is the acceleration 
of the particle, and 
is its velocity . Thus, we can rewrite Eq. 11-24 as
Now 
(the vector product of any vector with itself is zero because the angle between the two vectors is necessarily zero). This leads to
We now use Newton’s second law 
to replace 
with its equal, the vector sum of the forces that act on the particle, obtaining
Here the symbol Σ indicates that we must sum the vector products 
for all the forces. However, from Eq. 11-14, we know that each one of those vector products is the torque associated with one of the forces. Therefore, Eq. 11-25 tells us that
This is Eq. 11-23, the relation that we set out to prove.
* In differentiating a vector product, be sure not to change the order of the two quantities (here and ) that form that product. (See Eq. 3-28.)
CHECKPOINT 5 The figure shows the position vector 
of a particle at a certain instant, and four choices for the direction of a force that is to accelerate the particle. All four choices lie in the xy plane. (a) Rank the choices according to the magnitude of the time rate of change 
they produce in the angular momentum of the particle about point O, greatest first.(b) Which choice results in a negative rate of change about O?
In Fig. 11-14, a penguin of mass m falls from rest at point A, a horizontal distance D from the origin O of an xyz coordinate system. (The positive direction of the z axis is directly outward from the plane of the figure.)
(a) What is the angular momentum of the falling penguin about O?
Solution: One Key Idea here is that we can treat the penguin as a particle, and thus its angular momentum is given by Eq. 11-18 
, where is the penguin’s position vector (extending from O to the penguin) and 
is the penguin’s linear momentum. The second Key Idea is that the penguin has angular momentum about O even though it moves in a straight line, because vector rotates about O as the penguin falls.
To find the magnitude of , we can use any one of the scalar equations derived from Eq. 11-18—namely, Eqs. 11-19 through 11-21. However, Eq. 11-21 (ℓ = r⊥ mv) is easiest because the perpendicular distance r⊥ between O and an extension of vector is the given distance D. A third Key Idea is an old one: The speed of an object that has fallen from rest for a time t is v = gt. We can now write Eq. 11-21 in terms of given quantities as
To find the direction of , we use the right-hand rule for the vector product 
in Eq. 11-18. Mentally shift until its tail is at the origin, and then use the fingers of your right hand to rotate into through the smaller angle between the two vectors. Your outstretched thumb then points into the plane of the figure, indicating that the product 
and thus also are directed into that plane, in the negative direction of the z axis. We represent with an encircled cross 
at O. The vector changes with time in magnitude only; its direction remains unchanged.
(b) About the origin O, what is the torque on the penguin due to the gravitational force 
Solution: One Key Idea here is that the torque is given by Eq. 11-14 
, where now the force is 
. An associated Key Idea is that causes a torque on the penguin, even though the penguin moves in a straight line, because rotates about O as the penguin moves.
To find the magnitude of , we can use any one of the scalar equations derived from Eq. 11-14—namely, Eqs. 11-15 through 11-17. However, Eq. 11-17 (τ = r⊥F) is easiest because the perpendicular distance r⊥ between O and the line of action of is the given distance D. So, substituting D and using mg for the magnitude of , we can write Eq. 11-17 as
Fig. 11-14 A penguin falls vertically from point A. The torque and the angular momentum of the falling penguin with respect to the origin O are directed into the plane of the figure at O.
Using the right-hand rule for the vector product 
in Eq. 11-14, we find that the direction of is the negative direction of the z axis, the same as .
The results we obtained in parts (a) and (b) must be consistent with Newton’s second law in the angular form of Eq. 11-23 
. To check the magnitudes we got, we write Eq. 11-23 in component form for the z axis and then substitute our result ℓ = Dmgt. We find
which is the magnitude we found for . To check the directions, we note that Eq. 11-23 tells us that and 
must have the same direction. So and must also have the same direction, which is what we found.
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