Angular Momentum 10-1

Chapter 10

Angular momentum of a particle

In this chapter we introduce the rotational couterpart to the linear momentum ~p. It is called the angular momentum. For a point particle, it is represented by the vector ~`. For a massive body rotating about a fixed axis, it is represented by the ~ vector L. Consider a particle with momentum ~p at a position ~r relative to the origin O of an inertial reference frame, then the angular momentum of the particle with respect to the origin O is defined to be: ~` = ~r × ~p

(1)

z p

r

r x p y

O

x

As with all cross-products, we can write this definition in terms of a determinant: ~` =

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

ˆı ˆ kˆ x y z px py pz

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

The magnitude of the angular momentum can be written as ` = rp sin θ. ` = (r sin θ)p = r⊥ p

or

` = r(p sin θ) = rp⊥

where r⊥ is the moment arm, similar to the way we defined it in the torque equation τ = r⊥ F . 1

Exercise 3:

Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

We can relate the angular momentum ~` to the torque ~τ by taking a time-derivative of the angular momentum. d d~p d~r d~` ~ = ~τ ~ + ~v × m~v = ~r × F = (~r × ~p) = ~r × + × ~p = ~r × F {z } | dt dt dt dt =~0

In general, we can write that the sum of the torque vectors is X

~τ =

d~` dt

(2)

This can also be written as three scalar equations. X

10-2

τx =

d`x dt

X

τy =

d`y dt

X

τz =

d`z dt

(3)

System of Particles

In this section, we extend our definition of angular momentum to include more ~ to be the than one particle. In this case, we define the total angular momentum L sum of the individual angular momenta ~`i . ~ = ~`1 + ~`2 + · · · = L

N X

~`i

(4)

i=1 P

We can also relate the total torque ~τext to the time-derivative of the total angular momentum (similar to the previous section). ~ d`1 d`2 dL = + + ··· = dt dt dt

N X

d`i = i=1 dt

N X

i=1

~τi =

X

~τext

So, finally we have the following equation demonstrating Newton’s 2nd law of motion for rotational motion: X

~τext = 2

~ dL dt

(5)

N.B. When you calculate the angular momentum of a point-like object (or a small number of objects), you must calculate them about a point. 10-3

Angular Momentum and Angular Velocity

The previous section provides our segue into this section where we define the angular momentum for an extended mass, and more precisely, a rigid body rotating about a fixed axis. First, recall our definition for a point-like particle is ~` = ~r × ~p. If the particle is rotating about the z axis, and we focus our attention to the z component of the angular momentum, we find that `z = ` sin θ = rp sin θ where θ is the angle made with respect to the z axis. If we define R as the radius of the circle the particle is traveling about the z axis, then R = r sin θ. Now we can write `z as: `z = rp sin θ = r(mv) sin θ = (mv)(r sin θ) = (mRω)(R) = mR2 ω = Iz ω

(6)

where Iz is the moment of inertia of the particle about the z axis. This technique can be extended to a massive, rigid body rotating about the z axis and we find that: Lz = Iz ωz

(7)

This equation can be written in the x and y direction and combined to form one vector equation. ~ = I~ω L

(8)

However, in Eq. 8, one should take careful notice as to the definition for the moment of inertia I. It is no longer a scalar, but instead a tensor (or matrix) that can be written as:  

~ =  L 

Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz

    

ω ~

Because of this complicated interpretation for Eq. 8, its use will be delayed until a junior-level mechanics course. In the meantime, we will use Eq. 7 where Iz is a scalar quantity. 3

Exercise 10:

A wheel of radius 24.7 cm, moving initially at 43.3 m/s, rolls to a stop in 225 m. Calculate (a) its linear acceleration and (b) its angular acceleration. (c) The wheel’s rotational inertia is 0.155 kg·m2 . Calculate the torque exerted by rolling friction on the wheel.

Let’s examine the acceleration of two masses (m1 and m2 ) attached by a massless string over a pulley with mass M . The pulley is supported by a vertical force to the ceiling F and the masses m1 and m2 are relased (assume m1 > m2 ) and the system begins to accelerate in the +x direction. What is the acceleration?

I=

M

R

1 MR2 2

+x

m2

+x

a=?

m1

10-4

Conservation of Angular Momentum

In this section, we investigate another utility of angular momentum, namely, it is conserved when the sum of the external torques are equal to zero. If

X

~τext = ~0

then

~i = L ~f L

(9)

The vector form of this conservation law implies that there are three scalar equations describing the conservation law in all three directions x, y, and z. More precisely, we can state: 4

Any component of the angular momentum will be constant if the corresponding component of the torque is zero. For example, if the sum of the torques are the z axis are zero, then Lz = Iω Exercise 16:

10-5

and

Ii ωi = If ωf

(10)

Astronomical observations show that from 1870 to 1900 the length of the day increased by about 6.0 × 10−3 s. (a) What corresponding fractional change in the Earth’s angular velocity resulted? (b) Suppose that the cause of this change was a shift of molten material in the Earth’s core. What resulting fractional change in the Earth’s rotational inertia could account for the answer to part (a)?

The Spinning Top

Exercise 26:

A top is spinning at 28.6 rev/s about an axis making an angle of 34.0o with the vertical. Its mass is 492 g and its rotational inertia is 5.12 × 10−4 kg·m2 . The center of mass is 3.88 cm from the pivot point. The spin is clockwise as seen from above. Find the magnitude (in rev/s) and direction of the angular velocity of precession.

5