Math Review Night Vector Product, Angular Momentum, and Torque

Summary: Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors

Direction: determined by the Right-Hand-Rule

( ) ( )sin sin sin (0 )θ θ θ θ π× = = = ≤ ≤A B A B A B A B

Properties of Cross Products

( )

( )

c c c

× = − ×

× = × = ×

+ × = × + ×

A B B A

A B A B A B

A B C A C B C

Cross Product of Unit Vectors

  Unit vectors in Cartesian coordinates ( )ˆ ˆ ˆ ˆ| || | sin 2 1

ˆ ˆ ˆ ˆ| || | sin(0) 0

π× = =

× = =

i j i j

i i i j

ˆ ˆ ˆ ˆˆ

ˆ ˆ ˆ ˆˆ

ˆ ˆˆ ˆ ˆ

× = × =

× = × =

× = × =

i j k i i 0

j k i j j 0

k i j k k 0

Components of Cross Product

x y z x y zˆ ˆ ˆ ˆˆ ˆA A A , B B B= + + = + +A i j k B i j k

ˆ ˆ ˆ( ) ( ) ( )

ˆ ˆ ˆy z z y z x x z x y y x

x y z

x y z

A B A B A B A B A B A B

A A AB B B

× = − + − + −

=

A B i j k

i j k

Checkpoint Problem: Vector Product

Find a unit vector perpendicular to

and .

A = i + j− k

B = −2i − j+ 3k

Cross Product of Unit Vectors   Unit vectors in cylindrical coordinates

r × θ = kθ × k = rk × r = θ

A ×B = −

B ×A

r × r = θ × θ = k × k =0

Angular Momentum of a Point Particle

  Point particle of mass m moving with a velocity

  Momentum

  Fix a point S

  Vector from the point S to the location of the object

  Angular momentum about the point S

  SI Unit

v

m=p v

Sr

S S= ×L r p

[kg ⋅m2 ⋅ s-1]

Cross Product: Angular Momentum of a Point Particle

Magnitude:

a)  moment arm

b)  Perpendicular momentum

S S= ×L r p

sinS S θ=L r p

, sinS Sr θ⊥ = r

,S Sr ⊥=L p

, sinSp θ⊥ = p S S p⊥=L r

Angular Momentum of a Point Particle: Direction

Direction: Right Hand Rule

Worked Example: Angular Momentum and Cross Product

A particle of mass m = 2 kg moves with a uniform velocity

At time t, the position vector of the particle with respect ot the point S is

Find the direction and the magnitude of the angular momentum about the origin, (the point S) at time t.

v = 3.0 m ⋅ s-1 i + 3.0 m ⋅ s-1 j

rS = 2.0 m i + 3.0 m j

Solution: Angular Momentum and Cross Product

The angular momentum vector of the particle about the point S is given by:

The direction is in the negative direction, and the magnitude is

LS = rS × p = rS × m v

= (2.0m i + 3.0m j) × (2kg)(3.0m ⋅ s−1i + 3.0m ⋅ s−1j)

= 12kg ⋅m2 ⋅ s−1 k +18kg ⋅m2 ⋅ s−1(−k)

= −6.0kg ⋅m2 ⋅ s−1 k.

,

,

× =

× = −

× = × =

i j k

j i k

i i j j 0

2 16.0 kg m s .S−= ⋅ ⋅L

k

Angular Momentum and Circular Motion of a Point Particle:

Fixed axis of rotation: z-axis

Angular velocity

Velocity

Angular momentum about the point S

ˆˆ ˆR Rω ω= × = × =v r k r θω

2ˆ ˆ ˆS S S m Rmv RmR mRω ω= × = × = = =L r p r v k k k

ˆω≡ kω

Checkpoint Problem: angular momentum of dumbbell

A dumbbell is rotating at a constant angular speed about its center (point A). How does the angular momentum about the point B compared to the angular momentum about point A, (as shown in the figure)?

Checkpoint Problem: angular momentum of a single particle

A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane.

a) Find the magnitude and the direction of

the angular momentum relative to the origin.

b) Is this angular momentum relative to the origin constant? If yes, why? If no, why is it not constant?

0L

Torque as a Vector Force exerted at a point P on a rigid body.

Vector from a point S to the point P.

S S ,P Pτ = ×r F

PF

S ,Pr

P

θ PF

,rS PS

Torque about point S due to the force exerted at point P:

Torque: Magnitude and Direction Magnitude of torque about a point S:

τ S = rF⊥ = rF sinθ

Direction of torque: Perpendicular to the plane formed by and . Determined by the Right-Hand-rule.

FP

rS ,P

where is the magnitude of the force .

FP

FP

θ FP

r

S ,PS

Pθ

FP

S rS ,PτS

Checkpoint Problem: Torque

Consider two vectors with x > 0 and with Fx > 0 and Fz > 0 . What is the direction of the cross

product ?

x=r i

×r F

F = Fxi + Fzk

Conditions for Static Equilibrium

(1) Translational equilibrium: the sum of the forces acting on the rigid body is zero.

(2) Rotational Equilibrium: the vector sum of the torques about any point S in a rigid body is zero.

total 1 2 ...= + + =F F F 0

,1 ,2 ...S Sτ τ τ= + + =totalS 0

Worked Example: Lever Law Pivoted Lever at Center of Mass in Equilibrium

Show that (Lever Law):

d1m1g = d2m2g

Worked Example: Pivoted Lever Apply Newton’s 2nd law to each body:

Third Law:

Pivot force:

Torque about pivot:

Lever Law:

N1,B − m1g = 0

N2,B − m2g = 0

N1,B = N B,1 N2,B = N B,2

Fpivot = (mB + m1 + m2 )g

Fpivot − mBg − NB,1 − NB,2 = 0

d1N1k − d2 N2k =0

d1m1g = d2m2g

Generalized Lever Law

F1 =F1 , +

F1,⊥

F2 =F2, +

F2,⊥

F1,⊥ = F1 sinθ1

F2,⊥ = F2 sinθ2 F1 ,

F1,⊥

F2 ,

F2,⊥

τ cm

total =τ cm,1 +

τ cm,2 =

0 → (d1F1,⊥ − d2 F2,⊥ )k = 0

d1F1,⊥ = d2 F2,⊥

Problem Solving Strategy Force:

1.  Identify System and draw all forces and where they act on Free Body Force Diagram

2.  Write down equations for static equilibrium of the forces: sum of forces is zero

Torque:

1.  Choose point to analyze the torque about.

2.  Choose sign convention for torque

3.  Calculate torque about that point for each force. (Note sign of torque.)

4.  Write down equation corresponding to condition for static equilibrium: sum of torques is zero

Checkpoint Problem: Lever Law Suppose a beam of length s = 1.0 m and mass m

= 2.0 kg is balanced on a pivot point that is placed directly beneath the center of the beam. Suppose a mass m1 = 0.3 kg is placed a distance d1 = 0.4 m to the right of the pivot point. A second mass m2 = 0.6 kg is placed a distance d2 to the left of the pivot point to keep the beam static.

(1) What is the force that the pivot exerts on the beam?

(2) What is the distance d2 that maintains static equilibrium?

Problem: Ankle A person of mass m is crouching with their weight evenly distributed on both tiptoes. The force on the skeletal part of the foot are shown in the diagram. The normal force N acts at the contact point between the foot and the ground. In this position, the tibia acts on the foot at the point S with a force of magnitude F and makes a angle β with the vertical. This force acts on the ankle a horizontal distance s from the point where the foot contacts the floor. The Achilles tendon also acts on the foot and is under considerable tension with magnitude T and acts at an angle α with the horizontal as shown in the figure. The tendon acts on the ankle a horizontal distance b from the point where the tibia acts on the foot. You may ignore the weight of the foot. Let g be the gravitational constant. Compute the torque about the point S due to

a)  the normal force of the floor on the foot; b)  the tendon force on the foot; c)  the force of the tibia on the foot.