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2.8. CLASSICAL MECHANICS

91

2.8.3Mechanical springs

Many instruments make use of springs to translate force into motion, or vice-versa. The basic “Ohm’s Law” equation for a mechanical spring relating applied force to spring motion (displacement) is called

Hooke’s Law 13:

F = −kx

Where,

F = Force generated by the spring in newtons (metric) or pounds (British)

k = Constant of elasticity, or “spring constant” in newtons per meter (metric) or pounds per foot (British)

x = Displacement of spring in meters (metric) or feet (British)

Hooke’s Law is a linear function, just like Ohm’s Law is a linear function: doubling the displacement (either tension or compression) doubles the spring’s force. At least this is how springs behave when they are displaced a small percentage of their total length. If you stretch or compress a spring more substantially, the spring’s material will become strained beyond its elastic limit and either yield (permanently deform) or fail (break).

The amount of potential energy stored in a tensed spring may be predicted using calculus. We know that potential energy stored in a spring is the same as the amount of work done on the spring, and work is equal to the product of force and displacement (assuming parallel lines of action for both):

Ep = F x

Thus, the amount of work done on a spring is the force applied to the spring (F = kx) multiplied by the displacement (x). The problem is, the force applied to a spring varies with displacement and therefore is not constant as we compress or stretch the spring. A mathematician would say that the spring’s force is a function of x because the force varies as x varies. Thus, in order to calculate the amount of potential energy stored in the spring (Ep = F x), we must calculate the amount of energy stored over infinitesimal amounts of displacement (F dx, or kx dx) and then add those bits of energy

R

up ( ) to arrive at a total:

Z

Ep = kx dx

13Hooke’s Law may be written as F = kx without the negative sign, in which case the force (F ) is the force applied on the spring from an external source. Here, the negative sign represents the spring’s reaction force to being displaced (the restoring force). A spring’s reaction force always opposes the direction of displacement: compress a spring, and it pushes back on you; stretch a spring, and it pulls back. A negative sign is the mathematically symbolic way of expressing the opposing direction of a vector.

92

CHAPTER 2. PHYSICS

We may evaluate this integral using the power rule (x is raised to the power of 1 in the integrand):

Ep = 12 kx2 + E0

Where,

Ep = Energy stored in the spring in joules (metric) or foot-pounds (British)

k = Constant of elasticity, or “spring constant” in newtons per meter (metric) or pounds per foot (British)

x = Displacement of spring in meters (metric) or feet (British)

E0 = The constant of integration, representing the amount of energy initially stored in the spring prior to our displacement of it

For example, if we take a very large spring with a constant k equal to 60 pounds per foot and displace it by 4 feet, we will store 480 foot-pounds of potential energy in that spring (i.e. we will do 480 foot-pounds of work on the spring).

Graphing the force-displacement function on a graph yields a straight line (as we would expect, because Hooke’s Law is a linear function). The area accumulated underneath this line from 0 feet to 4 feet represents the integration of that function over the interval of 0 to 4 feet, and thus the amount of potential energy stored in the spring:

400

 

 

 

 

 

300

 

 

 

 

 

Force

 

 

 

 

 

(pounds) 200

 

 

 

 

 

(F)

 

 

 

 

 

100

 

 

 

 

 

 

Work = 480 foot-pounds

 

 

0

2

3

4

5

6

1

 

 

Displacement

(x)

 

 

 

 

(feet)

 

 

Note how the geometric interpretation of the shaded area on the graph exactly equals the result predicted by the equation Ep = 12 kx2: the area of a triangle is one-half times the base times the height. One-half times 4 feet times 240 pounds is 480 foot-pounds.

2.8. CLASSICAL MECHANICS

93

2.8.4Rotational motion

Rotational motion may be quantified in terms directly analogous to linear motion, using di erent symbols and units.

The rotational equivalent of linear force (F ) is torque (τ ). Linear force and rotational torque are both vector quantities, mathematically related to one another by the radial distance separating the force vector from the centerline of rotation. To illustrate with a string pulling on the circumference of a wheel:

Rotational torque

Linear force

 

 

(τ) vector

(F) vector

 

 

 

 

Right-hand rule

 

for vector cross-products

right angle

 

C

A

right angle

Radius vector

B

 

 

 

(r)

(r)

 

 

radius

 

 

 

 

 

C = A × B

 

This relationship may be expressed mathematically as a vector cross-product, where the vector directions are shown by the right-hand rule (the first vector ~r is the direction of the index finger,

~

the second vector F is the direction of the middle finger, and the product vector ~τ is the direction of the thumb, with all three vectors perpendicular to each other):

~

~τ = ~r × F

Labeling force, radius, and torque as vectors is the formally correct way of noting the variables in a mechanical system such as this, and is the way college students studying physics typically learn

~

the calculation of torque. In less academic settings, the force vector (F ) is typically labeled as a force along the line of action, and the radius vector (~r) is called the moment arm, with the line of action and moment arm always being perpendicular to each other.

The proper unit of measurement for torque is the product of the force unit and distance unit. In the metric system, this is customarily the Newton-meter (N-m). In the British system, this is customarily the foot-pound (ft-lb) or alternatively the pound-foot (lb-ft). Note that while these are the exact same units as those used to express work, they are not the same types of quantities. Torque

~

is a vector cross-product, while work is a dot-product (W = F ·~x). The cross-product of two vectors

94

CHAPTER 2. PHYSICS

is always another vector14, while the dot-product of two vectors is always a scalar (direction-less) quantity. Thus, torque always has a direction, whereas work or energy does not.

An example calculation applied to a hand wrench turning a bolt appears here:

Direction of torque

Axis of rotation

 

 

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 0

 

 

 

.

 

feet

 

 

 

Moment

 

 

 

 

 

 

61

 

 

 

 

 

 

 

arm

 

τ = r × F

 

 

 

 

 

 

 

 

τ = (0.61 ft) × (58 lb)

 

 

Line action

 

τ = 35.38 lb-ft

 

 

 

 

 

 

of

F = 58 lb

90o

With the radius and force vectors at right angles to each other, torque is simply the product of both. In many non-academic settings, torque is calculated this way as a scalar quantity, with the direction of rotation determined by observation rather than by strict adherence to the right-hand rule of vector cross products. In this example, we see the magnitude of torque as the simple product of 58 pounds force and 0.61 feet of moment arm (35.38 lb-ft of torque), with the torque direction obviously counter-clockwise as viewed from the head of the bolt.

14Technically, it is a pseudovector, because it does not exhibit all the same properties of a true vector, but this is a mathematical abstraction far beyond the scope of this book!

2.8. CLASSICAL MECHANICS

95

If we apply the same force to the wrench handle at a di erent angle (not perpendicular to the handle), the resulting torque will be less. The radius vector (moment arm), however, will still remain perpendicular to the force vector (line of action) – it just decreases in length. To determine the placement of the radius vector, all one must do is draw a line straight from the axis of rotation perpendicular to the line of action, then use trigonometry to calculate its magnitude:

Direction of torque

Axis of rotation

 

 

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

15o

=

0.59

feet

 

 

 

 

 

Moment

 

 

 

 

 

arm

 

 

 

 

Wrench

handle

 

 

 

 

 

 

 

=

0.

 

 

 

τ = r × F

 

 

 

61

feet

 

 

 

 

 

 

τ = (0.59 ft) × (58 lb)

 

 

 

 

of

action

 

τ = 34.22 lb-ft

 

 

 

Line

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

F = 58 lb

90o

A very practical example of torque is in the action of meshing gears, transferring mechanical power from one gear to another. Each gear e ectively acts as a wheel, the point of contact between gear teeth acting to transfer force perpendicular to the radius of each gear (wheel). Thus, torque applied to one gear becomes a linear force at the meshing teeth, which translates into another torque at the second gear:

Gear #1

Gear #2

 

The ratio of torques between two meshing gears is equal to the ratio of gear teeth:

τ1 = n1 τ2 n2

Where,

τ1 = Torque of first gear τ2 = Torque of second gear

n1 = Number of teeth on first gear n2 = Number of teeth on second gear

96

CHAPTER 2. PHYSICS

For example, if a small gear having 28 teeth meshes with a larger gear having 75 teeth, the torque multiplication factor from the small gear to the large gear will be 75:28, or 2.679 to 1. A torque of 40 lb-ft applied to the small gear will result in a torque of 107.1 lb-ft or torque generated at the large gear. This ratio of gear teeth is called the gear ratio.

As gears multiply torque (τ ), they divide rotational speed (ω). Thus, the 75:28 tooth gear set creates a multiplication of torque from the small gear to the large gear, and an identical reduction ratio of speed from the small gear to the large gear. Given this ratio, the small gear will have to be turned 2.679 revolutions in order to make the large gear turn just one revolution.

We may express gear speeds as another ratio of gear teeth, reciprocated in relation to torque:

ω1 = n2 ω2 n1

Where,

ω1 = Rotational speed of first gear ω2 = Rotational speed of second gear n1 = Number of teeth on first gear n2 = Number of teeth on second gear

In a set of meshed gears, the smaller gear will have the least torque and the greatest speed; the larger gear will have the greatest torque and the least speed.

This is precisely how gear sets are used in industry: to transform torque and speed in mechanical power systems. The complementary e ects of a gear set on torque and speed is analogous to the complementary e ects that a transformer has on AC voltage and current: a step-up transformer (having more turns of wire in the secondary coil than in the primary coil) will multiply voltage but reduce (divide) current, both by the same turns ratio.

2.8. CLASSICAL MECHANICS

97

Every quantity of force and motion which may be expressed in linear form has a rotational equivalent. As we have seen, torque (τ ) is the rotational equivalent of force (F ). The following table contrasts equivalent quantities for linear and rotational motion (all units are metric, shown in italic font):

Linear quantity, symbol, and unit

Rotational quantity, symbol, and unit

Force (F ) N

Torque (τ ) N-m

 

 

Linear displacement (x) m

Angular displacement (θ) radian

Linear velocity (v) m/s

Angular velocity (ω) rad/s

 

 

Linear acceleration (a) m/s2

Angular acceleration (α) rad/s2

Mass (m) kg

Moment of Inertia (I) kg-m2

Familiar equations for linear motion have rotational equivalents as well. For example, Newton’s Second Law of motion states, “The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to the object’s mass.” We may modify this law for rotational motion by saying, “The angular acceleration of an object is directly proportional to the net torque acting upon it and inversely proportional to the object’s moment of inertia.” The mathematical expressions of both forms of Newton’s Second Law are as follows:

F = ma

τ = Iα

The calculus-based relationships between displacement (x), velocity (v), and acceleration (a) find parallels in the world of angular motion as well. Consider the following formula pairs, linear motion on the left and angular motion on the right:

 

v =

dx

 

(Velocity as the time-derivative of displacement)

ω =

 

 

 

 

dt

 

dt

 

 

 

 

 

 

 

 

 

a =

 

dv

(Acceleration as the time-derivative of velocity)

α =

 

 

 

 

dt

dt

 

 

 

 

 

 

 

 

a =

d2x

 

 

(Acceleration as the second time-derivative of displacement)

 

α =

d2θ

dt2

 

 

dt2

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CHAPTER 2. PHYSICS

An object’s “moment of inertia” represents its angular inertia (opposition to changes in rotational velocity), and is proportional to the object’s mass and to the square of its radius. Two objects having the same mass will have di erent moments of inertia if there is a di erence in the distribution of their mass relative to radius. Thus, a hollow tube will have a greater moment of inertia than a solid rod of equal mass, assuming an axis of rotation in the center of the tube/rod length:

axis of rotation

axis of rotation

 

m = 300 kg

m = 300 kg

 

 

I = 1.5 kg-m2

I = 2.8 kg-m2

 

 

 

 

 

This is why flywheels15 are designed to be as wide as possible, to maximize their moment of inertia with a minimum of total mass.

The formula describing the amount of work done by a torque acting over an angular displacement is remarkably similar to the formula describing the amount of work done by a force acting over a linear displacement:

W = F x

W = τ θ

The formula describing the amount of kinetic energy possessed by a spinning object is also similar to the formula describing the amount of energy possessed by a linearly-traveling object:

Ek =

1

mv2

Ek =

1

2

2

2

 

 

 

 

15A “flywheel” is a disk on a shaft, designed to maintain rotary motion in the absence of a motivating torque for the function of machines such as piston engines. The rotational kinetic energy stored by an engine’s flywheel is necessary to give the pistons energy to compress the gas prior to the power stroke, during the times the other pistons are not producing power.