Game Development Reference

In-Depth Information

Finally, integrating over
z
, we obtain

22

ρπRh

h

C

=

=

.

(14.28)

z

12

M

4

Thus, the center of mass of the cone is given by

C

=

0, 0,

h

4

.

14.2.2 Angular Momentum and Torque

Recall that the linear momentum
p
of a particle having mass
m
moving at a ve-

locity
v
is given by

p
. Just as angular velocity is the rotational analog of

linear velocity, there exists a quantity called angular momentum that serves as

the rotational analog of linear momentum.

Suppose that a particle of mass
m
is rotating about some axis with an angular

velocity of

=

m

()

()

ω
and that the position of the particle is given by the function

r

t

.

()

The
angular momentum

L

t

of the particle is defined to be

()

()

()

L

t

=×

r

t

p
,

t

(14.29)

()

()

where

t

=

mt

v

is the linear momentum of the particle.

p

Differentiating both sides of Equation (14.29) gives us

()

()

()

()

.

()

L

t

=× +×

r

t

p r

t

t

p

t

(14.30)

()

()

r
and

()

()

Since

r

t

=

v

t

, the vectors

t

p

t

point in the same direction, so the

()

()

cross product

r

t

×

p

t

is zero. Thus,

()

()

()

()

.

()

L

t

=× =×

r

t

p r

t

t

mt

v

(14.31)

v
is equal to the net force

()

()

The vector

mt

F

t

acting on the particle, so we can

write

()

()

()

L

t

=×

r

t

F

t

.

(14.32)

()

The quantity on the right side of Equation (14.32) is called the
torque

τ

being applied to the particle:

()

()

()

τ r

t

=×

t

F
.

t

(14.33)

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